Post on 01-Apr-2015
This Week MomentumMondaybull What is momentum bull p = mv
Tuesdaybull How does momentum
change (impulse)bull I = mΔv = fΔt
Wednesdaybull Quizbull Conservation of
momentum bull pinitial=pfinal
Thursdaybull Elastic and inelastic collisions
Fridaybull Quizbull Momentum conservation in 2D
Monday bull Problem Solvingbull Integrating our knowledge
Today What is Momentum
Momentum in the Vernacular
bull In everyday experience momentum is the amount ldquounfrdquo an object has
So what factors affect the momentum of an object
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
A car will certainly hurt more whyBecause it is more massive (more mass)
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Today What is Momentum
Momentum in the Vernacular
bull In everyday experience momentum is the amount ldquounfrdquo an object has
So what factors affect the momentum of an object
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
A car will certainly hurt more whyBecause it is more massive (more mass)
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum in the Vernacular
bull In everyday experience momentum is the amount ldquounfrdquo an object has
So what factors affect the momentum of an object
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
A car will certainly hurt more whyBecause it is more massive (more mass)
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
A car will certainly hurt more whyBecause it is more massive (more mass)
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
What affects Momentum
Which has more ldquounfrdquondash A biker going at 20 mphndash A car going at 20 mph
A car will certainly hurt more whyBecause it is more massive (more mass)
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
What affects Momentum
Which has more ldquounfrdquondash A car going at 10 mphndash A car going at 2 mph
The faster car will have more ldquounfrdquo whyBecause faster things are harder to stop
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum Defined
Momentum is the product of mass and velocity
This is normally written p = m x vBolded letters denote vectors
What are the units of momentump = m x vm kg v ms
p kg bull ms kilogram meters per second
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
p=mv
Which has more momentum my car or me
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
p=mv
Which has more momentum my car or me
vcar=0vme= something like 1mspcar = mcar vcar = mcar 0 = 0pme = mme vme = mme something like 1ms = more than 0
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Homework
bull Review for quiz tomorrowbull This powerpoint is on the websitebull Visit the physics classroom for more practice
on momentum
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Todayrsquos Schedule (Jan 34)
bull How does momentum change (impulse)bull I = mΔv = fΔt
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
How Does Momentum of an Object Change
p=mvConsider Δp=ΔmvWhat does this mean
Why is this not a change in momentum of the object
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
How Does Momentum of an Object Change
p=mvConsider Δp=mΔvWhat does this mean
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
How Does Momentum of an Object Change
p=mvConsider Δp=mΔv
This means that velocity is changing Unlike Δm Δv does not imply that the object is
falling apart or clumping together
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Introducing Impulse
bull Δp is know as impulse (I)
bull Think of impulse as a change from the default path (Momentum would keep carrying me this direction
but I changed it)bull I had a sudden impulse to ldquordquo
(I was suddenly did something that was not planned)
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Δp = I = mΔv
A bowling ball (5kg) moving at 5ms bowls through a set of bowling pins Right before the ball falls down the shoot it is going 3ms
What is the impulse the bowling pins provide to a bowling ball
I = mΔvI = m(vfinal ndash vinitial)I = 5(3-5)=-10kgms
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Other Ways to Find Impulse
I = mΔv
Remember a = ΔvΔt therefore Δv = aΔt
Substituting in we getI = maΔt
Remember F=maSubstituting in we get
I=FΔt
I = Δp = mΔv = FΔt
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
I = Δp = mΔv = FΔt
I want to open a cracked door by throwing a ball at it If I have two balls of equal mass one bouncy and the other clay which of the two should I throw to achieve my goal
A)The bouncy oneB)The clay oneC)It doesnrsquot matter they are the sameZ) I have no clue
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Bouncy ball Clay ball
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Homework
bull Period 3 ndash Prepare for the quiz tomorrowndash Review the physics classroom ldquoThe Impulse-
Momentum Change Theoremrdquo (link on website)
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Todayrsquos Schedule (Jan 4)
bull Quiz on What is momentumbull Conservation of momentum
pinitial=pfinal
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Two-Minute Problem
A loaded truck collides with a car causing a huge damage to the car Which of the following is true about the collision
A The force on the truck is greater than the force on the carB The force on the car is greater than the force on the truckC The force on the truck is the same in magnitude as the force on the carD The car and truck accelerate in the same directionE During the collision the truck has greater acceleration than the carZ I have no idea
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Conservation of Momentum
Recall Newtons Third lawEvery action has an equal and opposite reactionF1=-F2
When I push on the desk it pushes back on me with equal force in the opposite direction
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Steel ball demoI = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Steel ball demo
Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1
F12 = -F21
Compare the time of contact for Ball 1 with Ball 2to the time of contact for Ball 2 with Ball 1
Δt12 = Δt21
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt
F1 = -F2
Δt1 = Δt2= Δt
I1= F1 ΔtI2 = F2 Δt = - F1 Δt
I1=-I2
In any interaction momentum gain of one object is equal to the loss of momentum from another
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Quick checkAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherA)120 kgmsB)240 kgmsC)0 kgmsD)-120 kgmsZ) I have no idea
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Total Momentum of a System is conserved
If there are no outside forces acting on an system the momentum of that system remains constant it is conserved
This is the property momentum conservation
ptotal = Σp = p1 + p2 + p3 + hellip
ptotalinitial = ptotalfinal
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum Conservation
A 10 kg object moves at a constant velocity 2 ms to the right and collides with a 4 kg object moving at a velocity 5 ms to the left Which of the following statements is correct
A The total momentum before and after the collision is 20 kgmsB The total momentum before and after the collision is 40 kgmsC The total momentum before and after the collision is 10 kgmsD The total momentum before and after the collision is 30 kgmsE The total momentum before and after the collision is zero
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
A freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg After the collision the car A is moving at 3 ms in the same direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum Conservationptotal = Σp = p1 + p2 + p3 + hellip
ptotailnitial = ptotalfinal
The same situation as beforeA freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 24000 kg However after the collision the car A is moving at 3 ms in the opposite direction What is the velocity of car B after the collision
A 1 ms B 3 ms C 5 ms D 7 ms D 11 ms
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
ptotal = Σp = p1 + p2 + p3 + hellipptotailnitial = ptotalfinal
Time to get tricky you may need a calculator
A loaded freight car A with a mass of 24000 kg moves at a constant velocity of 8 ms on a horizontal railroad track and collides with an empty stationary car B with a mass of 8000 kg After the collision the cars stick to each other and moves like one object What is the velocity of two cars after the collision
A 2 ms B 4 ms C 6 ms D 8 ms D 12 ms
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Homework
bull Periods 2 and 3 ndashWe do not have a quiz tomorrow however I
expect you to be able to do the practice quiz on the website If you donrsquot feel comfortable with it take extra care reviewing the physics classroom
ndash Review the Physics Classroom ldquoMomentum Conservation Principlerdquo (link on website)
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Todayrsquos Schedule (Jan 5)
bull Summary of Equationsbull Schedule for the next week and a half
- Test on Jan 13th (Friday next week)bull Todayrsquos material Review momentum
equation uses
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Momentum in a Nutshellbull p = mvbull Δp = I = mΔv = FΔtbull ptotal = Σp = p1 + p2 + p3 + hellipbull ptotailnitial = ptotalfinal
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
p=mv
What is the momentum of a bee that weighs 10 grams and flies at 2 ms
10g=01kg p=mv=01x2= 02kgms
How does that compare to a tortoise that weighs 1kg and moves at 05ms
p=mv=1x05= 05kgms
The tortoise has more momentum
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
I = Δp = mΔv = FΔtIn football a field goal is kicked with the football initially at rest The
football (300g) is kicked at 25ms What was the impulse
300g=3kg I = mΔv = m(Vfinal-Vinitial) = 3(25-0) = 75kgms
The players foot was in contact with the ball for 1 seconds What is the average force during the time of contact
I = FΔt
75 = F 1
F= 75 N
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
ptotal = Σp = p1 + p2 + p3 + hellipAn astronaut (80kg) in space kicks off of his space shuttle at 15ms What is the impulse that he provides to the space shuttle (Assume that away from the space shuttle is the positive direction)
What is the impulse that the space shuttle provides to him
What is the total change in momentum of the space shuttle and the astronaut togetherbull 120 kgmsbull 240 kgmsbull 0 kgmsbull -120 kgmsZ) I have no idea
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
ptotailnitial = ptotalfinal
ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotal final = m₁f x v ₁f + m ₂f x v₂f
ptotailnitial = ptotalfinal
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
No Official Homework
We will be working on more difficult problems from here on out
If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems
Review kinematics if the projectile motion was difficult
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Egg Throw
(bring a jacket)
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Other Similar Impulse Applications
bull Climbing ropesbull Airbagsbull Circus Netsbull Diving
GolfBaseball following through on your swings
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Sample Test Question
Ball A of mass 010 kg is sliding at 14 ms on the horizontal tabletop of negligible friction shown above It makes a head-on collision with a target ball B of mass 050 kg at rest at the edge of the table As a result of the collision the Ball A rebounds sliding backwards at 070 ms immediately after the collision
(a) Calculate the speed of Ball B immediately after the collision
(b) Calculate the horizontal displacement d
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
(a) Calculate the speed of the 050 kg target ball immediately after the collision We use conservation of momentum to solve this one
The tabletop is 120 m above a level horizontal floor The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
(b) Calculate the horizontal displacement dThis is a projectile motion problem We know the ballrsquos horizontal velocity and the height of the table so we can easily find the horizontal distance it travels as it fallsTime to fall
The ball has a horizontal velocity of 042 ms (which we just figured out) so the distance d is simply
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
2 Dimensional MomentumWhich of the following are possible final momenta for the case below
A
B
C
D
E
F
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Full 2D Problem on Board
Ball 1 (1kg) travelling at 5ms collides with Ball 2 (4kg) travelling at 3msAfter the collision Ball 1 is travelling 1ms at 60 degrees above the x-axis
What is the velocity (direction and speed) of Ball 2
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Tricky Extension
In another experiment on the same table the target ball B is replaced by target ball C of mass 010 kg The incident ball A again slides at 14 ms as shown above left but this time makes a glancing collision with the target ball C that is at rest at the edge of the table The target ball C strikes the floor at point P which is at a horizontal displacement of 015 m from the point of the collision and at a horizontal angle of 30 from the + x-axis as shown above right (c) Calculate the speed v of the target ball C immediately after the collision
(d) Calculate the y-component of incident ball As momentum immediately after the collision
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Tricky Extension
(c) Calculate the speed v of the target ball C immediately after the collisionThis sounds very hard but itrsquos actually quite simpleWe know the distance C traveled (15 m) and we know how long it is in the air before it hits ndash same as the previous problem time Using the horizontal distance and the time to fall we can find the horizontal velocity which is the velocity it began with which is the velocity right after the collision Find v
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Tricky Extension
(d) Calculate the y-component of incident ball As momentum immediately after the collision
We know that momentum is conserved in the x and y directions So we can sum momentum in the y direction We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Homeworkbull Read the lab prep sheet if you have not alreadybull There will be a lab tomorrow Decide on the details of
your experiment bull How much mass will you have on Cart B bull What will the initial velocities be bull Do you want both carts to be moving or just have one hit the other
Tomorrow you will come in and perform your experiment be ready Review the calculations and how you will take the data
bull Do any one of the three worksheet problems posted on the website
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Practice Test Questions
RocketTurbine
Try out the problems in the worksheet section
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
RocketsA rocket engine works by shooting particles out at incredibly high speeds How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 ms Exhaust is ejected at a speed of 1333 ms (Assume the mass of the rocket does not change significantly the speed of the exhaust is constant and that gravity is negligable)
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
RocketsCan we use momentum conservation
Yes there are no outside forces on our system since we are ignoring gravity
Lets setup our momentum conservationInitial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
RocketsFinal momentum rocket = m rocket x v rocketfinal = 5 50 = 250 kgms
Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved the exhaust must have a momentum equal and opposite to the rocket p exhaust = -250 = mexhaust x vexhaustnote that the v in this equation is negative because the exhaust is ejected downward This cancels the negative momentum resulting in a positive mass Vexhaust=1333msX = 2501333 = 0187546887 kg = 188 grams
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Turbine ProblemA stream of water strikes a stationary turbine blade as the drawing illustrates The incident water stream has a velocity of +180 ms while the exiting water stream has a velocity of -180 ms The mass of water per second that strikes the blade is 250 kgs Find the magnitude of the average force exerted on the water by the blade
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
Which sort of problem is this
A Total momentum
B Impulse
C Conservation of momentum
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own
Turbine ProblemVelocityi = +180 ms Velocityf = -180 ms Water enters at 250 kgs Find average force on the water
B Impulse
We are dealing with change in momentum and forces over a period of time
Try solving now on your own