Sem$IIramadosss.weebly.com/uploads/5/8/7/2/58728489/chapter_7.pdfWhatto$expectin$2 nd$Sem$ 1....
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Sem II
Transcript of Sem$IIramadosss.weebly.com/uploads/5/8/7/2/58728489/chapter_7.pdfWhatto$expectin$2 nd$Sem$ 1....
Sem II
What to expect in 2nd Sem 1. 4 more Chapters 2. 4-‐5 more labs 3. Seat Assignments 4. Only Web Assign HW ( you could print them and do on
paper) 6. Cedar Point Trip – 1st week of May Pre Reqs-‐ • No more than 3 absences • No behavorial issues in class during the course of the Sem2 -‐
Ch7 -MoMENTUM! Momentum
Impulse
Conservation of Momentum in 1 Dimension
Conservation of Momentum in 2 Dimensions
Angular Momentum
Torque
Moment of Inertia
Momentum Defined
p = m v
p = momentum vector
m = mass
v = velocity vector
Momentum Facts
• p = m v
• Momentum is a vector quanVty!
• Velocity and momentum vectors point in the same direcVon.
• SI unit for momentum: kg ·∙ m /s (no special name).
• Momentum is a conserved quanVty (this will be proven later).
• A net force is required to change a body’s momentum.
• Momentum is directly proporVonal to both mass and speed.
• Something big and slow could have the same momentum as something small and fast.
Momentum Examples 10 kg
3 m /s 10 kg
30 kg ·∙ m /s
Note: The momentum vector does not have to be drawn 10 Vmes longer than the velocity vector, since only vectors of the same quanVty can be compared in this way.
5 g
9 km /s
p = 45 kg ·∙ m /s at 26º N of E
26º
Equivalent Momenta
Bus: m = 9000 kg; v = 16 m /s
p = 1.44 ·∙105 kg ·∙ m /s
Train: m = 3.6 ·∙104 kg; v = 4 m /s p = 1.44 ·∙105 kg ·∙ m /s
Car: m = 1800 kg; v = 80 m /s
p = 1.44 ·∙105 kg ·∙ m /s
conVnued on next slide
Equivalent Momenta (cont.) The train, bus, and car all have different masses and speeds, but their momenta are the same in magnitude. The massive train has a slow speed; the low-‐mass car has a great speed; and the bus has moderate mass and speed. Note: We can only say that the magnitudes of their momenta are equal since they’re aren’t moving in the same direcVon.
The difficulty in bringing each vehicle to rest-‐-‐in terms of a combinaVon of the force and Vme required-‐-‐would be the same, since they each have the same momentum.
Impulse Defined Impulse is defined as the product force acVng on an object and the Vme during which the force acts. The symbol for impulse is J. So, by definiVon:
J = F t Example: A 50 N force is applied to a 100 kg boulder for 3 s. The impulse of this force is J = (50 N) (3 s) = 150 N ·∙ s.
Note that we didn’t need to know the mass of the object in the above example.
Impulse Units J = F t shows why the SI unit for impulse is the Newton ·∙ second. There is no special name for this unit, but it is equivalent to a kg ·∙ m /s.
proof: 1 N ·∙ s = 1 (kg ·∙ m /s2) (s) = 1 kg ·∙ m /s
{ Fnet = m a shows this is equivalent to
a newton.
Therefore, impulse and momentum have the same units, which leads to a useful theorem.
Impulse -‐ Momentum Theorem
The impulse due to all forces acVng on an object (the net force) is equal to the change in momentum of the object:
Fnet t = Δ p
We know the units on both sides of the equaVon are the same (last slide), but let’s prove the theorem formally:
Fnet t = m a t = m (Δ v / t) t = m Δ v = Δ p
Imagine a car hiing a wall and coming to rest. The force on the car due to the wall is large (big F ), but that force only acts for a small amount of Vme (lijle t ). Now imagine the same car moving at the same speed but this Vme hiing a giant haystack and coming to rest. The force on the car is much smaller now (lijle F ), but it acts for a much longer Vme (big t ). In each case the impulse involved is the same since the change in momentum of the car is the same. Any net force, no majer how small, can bring an object to rest if it has enough Vme. A pole vaulter can fall from a great height without geing hurt because the mat applies a smaller force over a longer period of Vme than the ground alone would.
Stopping Time
F t = F t
Impulse -‐ Momentum Example A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who kicks it in the exact opposite direcVon at 22 m/s with an average force of 1200 N. How long are his foot and the ball in contact?
answer: We’ll use Fnet t = Δ p.
Since the ball changes direction, Δ p = m Δ v = m (vf - v0) = 1.3 [22 - (-13)] = (1.3 kg) (35 m/s) = 45.5 kg · m /s. Thus, t = 45.5 / 1200 = 0.0379 s, which is just under 40 ms.
During this contact Vme the ball compresses substanVally and then decompresses. This happens too quickly for us to see, though. This compression occurs in many cases, such as hiing a baseball or golf ball.
1. What is the momentum of a 0.25 g bug flying with a speed of 12 m/s? Ans-‐ 3.0 x 10-‐3 kg m/s 2. a) What is the momentum of a 112 kg quarterback running with a speed of 4.8 m/s?Ans-‐ 5.4 x 102 kg•m/s b) What impulse must a tackler impart to the quarterback to stop him? Ans-‐5.4 x 102 N•s c) If the tackle is completed in 1.2 s, what average force is exerted by the tackler on the quarterback? Ans–4.5 x 102 N 3. Imagine you are doing a space walk outside a space shujle, with no cable between you and the shujle. (Actually, there is one, but you cannot see it because you don't have Cablevision!) Your small maneuvering rocket pack suddenly quits on you and you are stranded in space with nothing but your $10,000 camera in your hands. What will you do to get back to the shujle?
Impulse as Area of Fnet vs. t graph Fnet (N)
t (s) 6
A variable strength net force acts on an object in the posiVve direcVon for 6 s, thereaqer in the opposite direcVon. Since impulse is Fnet t, the area under the curve is equal to the impulse, which is the change in momentum. The net change in momentum is the area above the curve minus the area below the curve. This is just like a v vs. t graph, in which net displacement is given area under the curve.
Net area = Δ p
ConservaVon of Momentum in 1-‐D Whenever two objects collide (or when they exert forces on each other without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and aqer the collision.
before: p = m1 v1 - m2 v2
after: p = - m1 va + m2 vb
m1 m2 v1 v2
(Choosing right as the + direc2on, m2 has
-‐ momentum.)
m1 m2 va vb
m1 v1 - m2 v2 = - m1 va + m2 vb
DirecVons aqer a collision On the last slide the boxes were drawn going in the opposite direcVon aqer colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direcVon, but the bat doesn’t. It doesn’t really majer, though, which way we draw the velocity vectors in “aqer” picture. If we solved the conservaVon of momentum equaVon (red box) for vb and got a negaVve answer, it would mean that m2 was sVll moving to the leq aqer the collision. As long as we interpret our answers correctly, it majers not how the velocity vectors are drawn.
m1 v1 - m2 v2 = - m1 va + m2 vb
m1 m2 v1 v2
m1 m2 va vb
Total Momentum before Collision = Total Momentum a4er Collision
Sample Problem 1
7 kg
v = 0
700 m/s
A rifle fires a bullet into a giant slab of bujer on a fricVonless surface. The bullet penetrates the bujer, but while passing through it, the bullet pushes the bujer to the leq, and the bujer pushes the bullet just as hard to the right, slowing the bullet down. If the bujer skids off at 4 cm/s aqer the bullet passes through it, what is the final speed of the bullet? (The mass of the rifle majers not.)
35 g
7 kg v = ?
35 g
4 cm/s
con2nued on next slide
7 kg
v = 0
700 m/s
35 g
7 kg v = ?
35 g
4 cm/s
p before = 7 (0) + (0.035) (700)
= 24.5 kg ·∙ m /s
Let’s choose le9 to be the + direc2on & use conserva2on of momentum, conver2ng all units to meters and kilograms.
p aqer = 7 (0.04) + 0.035 v
= 0.28 + 0.035 v
p before = p aqer 24.5 = 0.28 + 0.035 v v = 692 m/s
v came out posiVve. This means we chose the correct direcVon of the bullet in the “aqer” picture.
Sample Problem
7 kg v = 0
700 m/s
35 g
Same as the last problem except this Vme it’s a block of wood rather than bujer, and the bullet does not pass all the way through it. How fast do they move together aqer impact?
v
7. 035 kg
(0.035) (700) = 7.035 v v = 3.48 m/s Note: Once again we’re assuming a fricVonless surface, otherwise there would be a fricVonal force on the wood in addiVon to that of the bullet, and the “system” would have to include the table as well.
Proof of ConservaVon of Momentum The proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an acVon-‐reacVon pair, equal in strength, opposite in direcVon. This means the net force on the system (the two objects together) is zero, since these forces cancel out.
M m F F
force on m due to M force on M due to m
For each object, F = (mass) (a) = (mass) (Δv / t ) = (mass Δv) / t = Δp / t. Since the force applied and the contact Vme is the same for each mass, they each undergo the same change in momentum, but in opposite direcVons. The result is that even though the momenta of the individual objects changes, Δp for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system aqer.
ConservaVon of Momentum applies only in the absence of external
forces! In the first two sample problems, we dealt with a fricVonless surface. We couldn’t simply conserve momentum if fricVon had been present because, as the proof on the last slide shows, there would be another force (fricVon) in addiVon to the contact forces. FricVon wouldn’t cancel out, and it would be a net force on the system.
The only way to conserve momentum with an external force like fricVon is to make it internal by including the tabletop, floor, or the enVre Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-‐wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-‐Earth system is conserved!
Sample Problem 3
Earth M
apple
An apple is originally at rest and then dropped. Aqer falling a short Vme, it’s moving prejy fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be?
m
F
F
answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitaVonal force-‐-‐Earth. The net force on the apple-‐Earth system is zero, and momentum is conserved for it. During the fall the Earth ajains a very small speed v. So, by conservaVon of momentum:
V
v
m V = M v
Sample Problem 4
before
a9er
3 kg 15 kg 10 m/s 6 m/s
3 kg 15 kg 4.5 m/s v
A crate of raspberry donut filling collides with a tub of lime Kool Aid on a fricVonless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the aqer picture.
3 (10) -‐ 6 (15) = -‐3 (4.5) + 15 v v = -‐3.1 m/s Since v came out negaVve, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. Aqer the collision the lime Kool Aid is moving 3.1 m/s to the leq.
ConservaVon of Momentum in 2-‐D
m1 m2
v1 v2
m1 m2
va vb
θ1 θ2
θa θ b
To handle a collision in 2-‐D, we conserve momentum in each dimension separately. Choosing down & right as posiVve:
before:
px = m1 v1 cosθ1 - m2 v2 cosθ2
py = m1 v1 sinθ1 + m2 v2 sinθ2
after:
px = -m1 va cosθa + m2 vb cosθ b
py = m1 va sinθa + m2 vb sinθ b
Conservation of momentum equations:
m1 v1 cosθ1 - m2 v2 cosθ2 = -m1 va cosθa + m2 vb cosθ b
m1 v1 sinθ1 + m2 v2 sin θ2 = m1 va sinθa + m2 vb sinθ b
Conserving Momentum w/ Vectors
m1 m2
p1
p 2
m1 m2
p a p b
θ1 θ2
θa θ b
BEFORE
AFTER
This diagram shows momentum vectors, which are parallel to their respecVve velocity vectors. Note p1 + p 2
= p a + p b and p before = p aqer as conservaVon of momentum demands.
p1
p 2
p before
p a
p b
p after
Exploding Bomb
A c m e
before
a9er
m
A bomb, which was originally at rest, explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the aqer picture.
con2nued on next slide
Exploding Bomb (cont.) Since the momentum of the bomb was zero before the explosion, it must be zero aqer it as well. Each piece does have momentum, but the total momentum of the exploded bomb must be zero aqerwards. This means that it must be possible to place the momentum vectors Vp to tail and form a closed polygon, which means the vector sum is zero.
If the original momentum of the bomb were not zero, these vectors would add up to the original momentum vector.
2-‐D Sample Problem
0.3 kg
34 m/s 40°
152 g A mean, old dart strikes an innocent mango that was just passing by minding its own business. Which way and how fast do they move off together?
before
a9er
5 m/s
452 g
θ
v
152 (34) sin 40° = 452 v sinθ
152 (34) cos 40° -‐ 300 (5) = 452 v cosθ
Working in grams and taking le9 & down as + :
Dividing equa2ons : 1.35097 = tanθ
θ = 53.4908°
Subs2tu2ng into either of the first two equa2ons :
v = 9.14 m/s
ElasVc and InelasVc collisions
Total KE before = Total KE a4er Total KE before ≠ Total KE a4er
Types of Collisions Energy is always conserved but may change types (mv2/2, mgh, kx2/2 etc). There is only one type of
momentum (mv). We idenVfy collisions based upon their conservaVon of kine2c energy.
InelasVc • kineVc energy is NOT constant
ElasVc • kineVc energy IS constant
InelasVc Collisions These collisions are considered PERFECT when the objects collide and combine to move as one object.
InelasVc • Objects bounce but may be deformed so kineVc energy is transformed.
Perfectly InelasVc • Objects sVck together
Perfectly InelasVc Collisions:
pAi + pBi = pAB, fmAvAi +mBvBi = mA +mB( )vf
For elas9c collisions KE is transferred near perfectly. Perfect elastic is not practically possible in macroscopic world
A toy car with a mass of 120 g moves to the right with a speed of 0.59 m/s. A small child drops a 20.0-‐g piece of clay onto the car. The clay sVcks to the car and the car conVnues to the right. What is the change in speed of the car? Consider the fricVonal force between the car and the ground to be negligible.
-‐0.0843 m/s
A firecracker is tossed straight up into the air. It explodes into three pieces of equal mass just as it reaches the highest point. Two pieces move off at 140 m/s at right angles to each other. How fast is the third piece moving?
198 m/s
Body A of mass M has an original velocity of 5.5 m/s in the +x-‐direcVon toward a staVonary body (body B) of the same mass. Aqer the collision, body A has velocity components of 1.1 m/s in the +x-‐direcVon and 1.8 m/s in the +y-‐direcVon. What is the magnitude of body B's velocity aqer the collision?
4.75 m/s
Two idenVcal pucks are on an air table. Puck A has an iniVal velocity of 2.2 m/s in the posiVve x-‐direcVon. Puck B is at rest. Puck A collides elasVcally with puck B and A moves off at 1.2 m/s at an angle of +60° above the x-‐axis. What is the speed and direcVon of puck B aqer the collision? (Let angles above the +x-‐axis to be posiVve and below to be negaVve.)
1.91 m/s -‐33.0 °
In a circus trapeze act, two acrobats actually fly through the air and grab on to each other, then together grab a swinging bar. One acrobat, with a mass of 71 kg, is moving at 3 m/s at an angle of 10° above the horizontal and the other, with a mass of 80 kg, is approaching her with a speed of 1.7 m/s at an angle of 20° above the horizontal. What is the direcVon and speed of the acrobats right aqer they grab on to each other? Let the posiVve x-‐axis be in the horizontal direcVon and assume the first acrobat has posiVve velocity components in the posiVve x-‐ and y-‐direcVons.
0.775 m/s 45.5 °
