Per 9-10 Momentum Presentation.notebook 9... · 1/20/2015 · Per 910...

Per 910 Momentum_Presentation.notebook

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January 20, 2015

www.njctl.org

Momentum


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January 20, 2015

MomentumClick on the topic to go to that section

• Momentum

• Conservation of Momentum

• Inelastic Collisions and Explosions

• Elastic Collisions

• Impulse

• Momentum of a System of Objects


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January 20, 2015

Return toTable ofContents

Momentum


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Newton’s First Law tells us that – objects remain in motion with a constant velocity unless acted upon by a force.

In our experience:

• When two objects of different masses travel with the same velocity, the one with more mass is harder to stop.

• When objects of the equal masses travel with different speeds, the faster one is harder to stop.

• Define a new quantity, momentum (p), that takes these observations into account:

momentum = mass × velocity

Momentum Defined

p =mvclick here for a introductory video on

momentum from Bill Nye!

http://njc.tl/ka


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January 20, 2015

Recall that:mass is a scalar quantity

and velocity is a vector quantity

Since: momentum = mass × velocity

momentum must be a vector quantity

Momentum is a Vector Quantity


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There no specially named unit for momentum. We just use the product of the units of mass and velocity...

mass x velocity

kg⋅m/s

SI Unit for Momentum


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January 20, 2015

1 Which has more momentum?

A A large truck moving at 30 m/s

B A small car moving at 30 m/s

C Both have the same momentum.

Ans

wer


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January 20, 2015

2 What is the momentum of a 20 kg object with a velocity of +5.0 m/s?

Ans

wer


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January 20, 2015

3 What is the momentum of a 20kg object with a velocity of −5.0m/s?

Ans

wer


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4 What is the velocity of a 5.0kg object whose momentum is −15.0 kg m/s?

Ans

wer


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5 What is the mass of an object whose momentum is 35 kg⋅m/s when its velocity is 7.0 m/s?

Ans

wer


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Momentum Change & Impulse


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Suppose that there is an event that changes an object's momentum.

• from p0 the initial momentum (just before the event)

• by Δp the change in momentum

• to pf the final momentum (just after the event)

The equation for momentum change is:

Change in Momentum


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Momentum change equation:

Newton's First Law tells us that the velocity (and so the momentum) of an object won't change unless the object is affected by an external force.

When an outside force F acts on the object for a time Δt, it delivers an impulse I to the object that changes its momentum:

Where the impulse is:

Momentum Change = Impulse


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January 20, 2015

There no specially named unit for impulse. We just use the product of the units of force and time...

force x time

N⋅s

Recall that N=kg⋅m/s2, so

N⋅s=kg⋅m/s2 x s

= kg⋅m/s the same as momentum!

SI Unit for Impulse


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January 20, 2015

Effect of Collision Time on Force

Impulse = Ft = Ft change in momentum

Changing the duration (t) of an impulse by a small amount can greatly reduce the force on an object

time (seconds)

force (new

tons

)


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January 20, 2015

Real World Applications

Impulse = Ft = Ft change in momentum

• Car Design / Accidents

> airbags

> collisions headon vs walls

> crush zones

• Jumping / Landing

• Boxing / Martial Arts

• Hitting Balls Golf, Baseball...

• Catching Balls


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6 An external force of 25N acts on a system for 10s. How big is the impulse delivered to the system?

Ans

wer


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7 In the previous problem, an external force of 25N acted on a system for 10s. We found that the impulse delivered was 250 Ns. What is the magnitude of the change in momentum of the system?

Ans

wer


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January 20, 2015

8 The momentum change of an object is equal to the ______.

A force acting on it

B impulse acting on it

C velocity change of the objectD object's mass times the force acting on it

Ans

wer


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January 20, 2015

9 Air bags are use in cars because they:

A increase the force with which you hit the dashboard

B increase the duration (time) of impact in a collisionC decrease the momentum of a collision

D decrease the impulse in a collision

B Ans

wer


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January 20, 2015

10 One car crashes into a concrete barrier. Another car crashes into a collapsible barrier at the same speed. What is the difference between the 2 crashes?

A change in momentum

B force on the carC impact timeD both B & C are true

[This object is a pull tab]

Ans

wer

CWhether the wall is padded or not, the car experiences the same change in momentum, and so the same impulse. The only difference is that the impact time has increased (which in turn reduces the force experienced by the car).


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January 20, 2015

11 In order to increase the final momentum of a golf ball, we could:

A not change the speed of the golf club after the collision

B increase the force acting on it

C increase the time of contact between the club and ball

D all of the above

Ans

wer

DFollowing through, hitting the golf ball harder, and/or increasing the impact time will all result in an increase in the final momentum of the golf ball.


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January 20, 2015

12 An external force acts on an object for 0.0020 s. During that time the object's momentum increases by 400 kgm/s. What was the magnitude of the force?

Ans

wer

Δt = 0.002 s

Δp = 400 kg•m/s

I = FΔt = Δp

F = Δp/Δt

= (400 kg•m/s)/(0.002 s)

= 200,000 N


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January 20, 2015

13 A 50,000 N force acts for 0.030 s on a 2.5 kg object that was initially at rest. What is its final velocity?

*

Ans

wer


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The Momentum of a System of Objects


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If a system contains more than one object, it's total momentum is the vector sum of the momenta of those objects.

psystem = ∑ p

psystem = p1 + p2 + p3 +...

psystem = m1v1 + m2v2 + m3v3 +...


It's critically important to note that

momenta add as vectors,

not as scalars.


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January 20, 2015

psystem = m1v1 + m2v2 + m3v3 +...

In order to determine the total momentum of a system, First:• Determine a direction to be considered positive • Assign positive values to momenta in that direction • Assign negative values to momenta in the opposite direction


Then: Add the momenta to get a total.

+


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January 20, 2015

Determine the momentum of a system of two objects: m1, has a mass of 6 kg and a velocity of 13 m/s towards the east and m2, has a mass of 14 kg and a velocity of 7 m/s towards the west.

psystem = p1 + p2

psystem = m1v1 + m2v2

Example

(Choose east as positive)

m1 = 6 kgv1 = 13 m/s

m2 = 14 kgv2 = −7 m/s A

nswer psystem = m1v1 + m2v2= (6 kg)(13 m/s) + (14 kg)(−7 m/s)

= (78 kg⋅m/s) + (−98 kg⋅m/s)= −20 kg⋅m/s

psystem = m1v1 + m2v2= (6 kg)(13 m/s) + (14 kg)(−7 m/s)= (78 kg⋅m/s) + (−98 kg⋅m/s)= −20 kg⋅m/s


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January 20, 2015

14 Determine the magnitude of the momentum of a system of two objects: m1, has a mass of 6.0kg and a velocity of 20m/s north and m2, has a mass of 3kg and a velocity 20m/s south.

Ans

wer

psystem = m1v1 + m2v2= (6 kg)(20 m/s) + (3 kg)(−20 m/s)= (120 kg⋅m/s) + (−60 kg⋅m/s)= 60 kg⋅m/s (magnitude)

direction is North


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January 20, 2015

15 Determine the momentum of a system of two objects: the first has a mass of 8 kg and a velocity of 8 m/s to the east while the second has a mass of 5 kg and a velocity of 15 m/s to the west.

Ans

wer

psystem = m1v1 + m2v2= (8 kg)(8 m/s) + (5 kg)(−15 m/s)= (64 kg⋅m/s) + (−75 kg⋅m/s)= −11 kg⋅m/s (West)

m1 = 8 kgv1 = +8 m/s (East)

m2 = 5 kgv2 = −15 m/s (West)


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January 20, 2015

16 Determine the momentum of a system of 3 objects: The first has a mass of 7.0 kg and a velocity of 23 m/s north; the second has a mass of 9.0 kg and a velocity of 7 m/s north; and the third has a mass of 5.0 kg and a velocity of 42 m/s south.

Ans

wer

psystem = m1v1 + m2v2 + m3v3

= (7kg)(23m/s) + (9kg)(7m/s) +(5 kg)(−42m/s)= (161 kg⋅m/s) + (63 kg⋅m/s) + (−210 kg⋅m/s)

= 14 kg⋅m/s (North)


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Conservation of Momentum


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Some of the most powerful concepts in science are called "conservation laws".

Conservation laws:

• apply to closed systems where the objects only interact with each other and nothing else.

• enable us to solve problems without worrying about the details of an event.

Conservation Laws


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January 20, 2015

In the last unit we learned that energy is conserved.

Like energy, momentum is a conserved property of nature. It is not created or destroyed;

So in a closed system we will always have the same amount of momentum.

The only way the momentum of a system can change is if momentum is added or taken away by an outside force.

Momentum is Conserved


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January 20, 2015

To apply Conservation of Momentum, Take snapshots of a system just before and after an event. By comparing these two snapshots we can learn a lot.

We'll explore this more a little later.



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Conservation of Momentum and ImpulseRecall from our discussion of change of momentum and impulse:

When a net external force acts on an object, it imparts an impulse I to the object, changing its momentum.

This is exactly the same for a system of objects.

If there is no net external force on the system, the momentum of the system is conserved.


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January 20, 2015

Conservation of Momentum & Impulse ProofBoth the Conservation of Momentum and the concept of Impulse follow directly from Newton's Second Law:

*

F = ma

F=m(Δv/Δt)

FΔt =mΔv

FΔt = Δ(mv)

I = FΔt

I = Δp

I = pf p0p0 + I = pf

p0 = pf

where F is the net external force

since a = Δv/Δt

after multiplying both sides by Δt

since m is constant

the definition of impulse I

substituting I for FΔt, and p for mv

since Δp = pf p0

when there no net external force (F=0), I=0 so... momentum is conserved


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January 20, 2015

Conservation Laws, Collisions and Explosions

Objects in an isolated system can interact with each other in a number of ways...

• They can collide

• If they are stuck together, they can explode (push apart)

In an isolated system both momentum and total energy are conserved.

But the energy can change from one form to another.

Conservation of momentum and change in kinetic energy help us predict what happened or what will happen in one of these events.


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January 20, 2015

Collisions and ExplosionsWe differentiate collisions and explosions by the way the energy changes or does not change form.

• explosions: an object or objects break apart because potential energy stored in one or more of the objects is transformed into kinetic energy

• inelastic collisions: two objects collide and stick together converting some kinetic energy into bonding energy, heat, sound...

• elastic collisions: two objects collide and bounce off each other while conserving kinetic energy


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January 20, 2015

Collisions and Explosions Summarized

Event DescriptionMomentum Conserved

?

Kinetic Energy Conserved?

Inelastic Collision

General collision:

Objects bounce off each other

Yes

No. Some kinetic energy is converted to heat, sound...

energy

Inelastic Collision

Objects stick together Yes

No. Kinetic energy is

converted to potential energy, bonding, or heat, sound...energy

Elastic Collision

Objects bounce off each other

Yes Yes

ExplosionObject or

objects break apart

Yes

No. Release of potential energy increases kintetic

energy


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January 20, 2015

17 In _______ collisions momentum is conserved.

A Elastic

B Inelastic

C All

Ans

wer


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January 20, 2015

18 In ______ collisions kinetic energy is conserved.

A ElasticB Inelastic C All

Ans

wer


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January 20, 2015


During a collision or an explosion, measurements show that the total momentum does not change:

A BmAVA mBVB

A B

A BmAVA' mAVB'

x

the prime means "after"


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January 20, 2015

Explosions

In an explosion, one object breaks apart into two or more pieces (or coupled objects break apart) moving afterwards as separate objects.

We will assume:

• the object (or a coupled pair of objects) breaks into two pieces

• explosion is along the same line as the initial velocity

A B

Before (moving together)

pA+pB=(mA+ mB)vA B

pA'=mAvA'After (moving apart)

pB'=mBvB'


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January 20, 2015

19 A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 m/s. What is the cannon balls's velocity after the explosion?

Ans

wer

m1 = 5kg

m2 = 300kg

v1 = v2 = v = 0

v2' = 5m/s


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January 20, 2015

20 Two railcars, one with a mass of 4000 kg and the other with a mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling 6 m/s. How fast is the 6000 kg car going?


Ans

wer (m1+m2) v = m1v1'+m2v2'

0 = m1v1'+m2v2'm1v1' = m2v2'v1' = m2v2'/m1

= (4000kg)(6m/s) / (6000kg)= 4m/s

m1 = 6000kgm2 = 4000kgv1 = v2 = v = 0v2' = 6m/s


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January 20, 2015

Inelastic CollisionsIn an inelastic collision, two objects collide and stick together, moving afterwards as one object.

A B

After (moving together)

pA'+pB'=(mA+ mB)v'

A B

pA=mAvA

Before (moving towards the other)

pB=mBvB


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January 20, 2015

21 A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the system. What is the speed of the two cars after colliding?

Ans

wer

m1 = 13,500kgm2 = 25,000kgv1 = 4.5m/sv2 = 0 m/sv1' = v2' = v'


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January 20, 2015

22 A cannon ball with a mass of 100 kg flies in horizontal direction with a speed of 800 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it.

Ans

wer

m1v1+m2v2 = m1v1'+m2v2'm1v1+0 = (m1+m2) v'v' = m1v1/(m1+m2)

= (100kg)(800m/s) / (100+15,000)kg= 5.3 m/s

in same direction as cannon ball's initial velocity

m1 = 100kgm2 = 15,000kgv1 = 800m/sv2 = 0 m/sv1' = v2' = v'


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January 20, 2015

23 A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is standing still, with open arms. As they collide and hold each other, what is the speed of the couple?

Ans

wer

m1 = 40kgm2 = 70kgv1 = 5.5m/sv2 = 0 m/sv1' = v2' = v'


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January 20, 2015

Elastic CollisionsIn a elastic collision, two objects collide and bounce off each other.

Both momentum and kinetic energy are conserved.

If we know the masses and any two of the velocities, these two conservation equations enable us to calculate the other two velocities.

A B

pA=mAvA

Before (moving towards) After (moving apart)

pB=mBvB

A B

pA'=mAvA' pB'=mBvB'


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January 20, 2015

**Derivation of Elastic Collision Condition

m1v1 + m2v2 = m1v1' +m2v2'

m1v1 m1v1' = m2v2' m2v2

m1(v1 v1') = m2(v2' v2)

½m1v12 + ½m2v22 = ½m1v1'2 +½m2v2'2

m1v12 + m2v22 = m1v1'2 +m2v2'2

m1v12 m1v1'2 = m2v2'2 m2v22

m1(v12 v1'2) = m2(v2'2 v22)m1(v1 + v1')(v1 v1') = m2(v2' + v2)(v2' v2)

m1(v1 + v1')(v1 v1') = m2(v2' + v2)(v2' v2)m1(v1 v1') = m2(v2' v2)

v1 + v1' = v2' + v2

Conservation of Momentum Conservation of Kinetic Energy

v1 v2 = (v1' v'2)


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January 20, 2015

Properties of Elastic Collisions

1.For all elastic collisions, regardless of the masses of the objects, the objects separate after the collision with the same relative speed that they collided with.

2.In an elastic collision where one object is much more massive than the other, the velocity of the smaller mass after the collision will be about twice that of the projectile while the more massive object's velocity will be almost unchanged.

3.In an elastic collision between two objects of identical masses, the two objects exchange velocities.

v1 v2 = (v1' v'2)

m1 m1m2 m2

v1 v2 = 0 v'1 v'2v1 2v1

v1' = v2 and v2' = v1

v1' = v1 and v2' = 2v1


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January 20, 2015

24 Two objects have an elastic collision. Before they collide they are approaching each other with a velocity of 4m/s relative to each other. With what velocity do they go apart from one another?


Ans

wer

4m/sInitially they are approaching relative to each other at 4m/s, and afterwards they will move apart relative to each other with a velocity of 4m/s.

v1+v1'=v2+v2'

The difference in the initial velocities is the same as the difference in the final velocities therefore they will move apart relative to one another at the same velocity


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January 20, 2015

25 Two objects have an elastic collision. One object, m1, has an initial velocity of +4.0 m/s and m2 has a velocity of 3.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?

Ans

wer

v1 = 4m/sv2 = 3m/sv1' = 1m/sv2' = ?


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January 20, 2015

26 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball?

Ans

wer

v1+v1' = v2+v2'

v2' = v1+v1'v2v2' = v + v 0

v2' = 2v

(ping pong ball's speed is 2x the that of the bowling ball)

v1 = +vv2 = 0v1' = +vv2' = ?


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January 20, 2015

27 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of 2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit?

Ans

wer

v1+v1' = v2+v2'

v2' = v1+v1'v2v2' = v + v (2v)

v2' = 4v

(baseball's ball's speed doubled after being hit by the bat)

v1 = +vv2 = 2vv1' = +vv2' = ?


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January 20, 2015

28 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is 3.0m/s. What is the velocity of m1 after the collision?

Ans

wer

v1 = +6m/sv2 = 3m/sv1' = ?v2' = ?


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January 20, 2015

29 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0m/s and m2 is 3.0m/s. What is the velocity of m2 after the collision?

Ans

wer

v1 = +6m/sv2 = 3m/sv1' = ?v2' = ?


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January 20, 2015

30 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m1 after the collision?

Ans

wer

v1 = +3m/sv2 = +2m/sv1' = ?v2' = ?


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January 20, 2015

31 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +3.0m/s and m2 is +2.0m/s. What is the velocity of m2 after the collision?

Ans

wer

v1 = +3m/sv2 = +2m/sv1' = ?v2' = ?

Attachments

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