Unit 8 Momentum Review Assignment Name _________________

1. Two objects, A & B, have identical velocities. Object A has 3 times the mass of object B. Compare the momentum of each object. Justify your answer.Object A’s momentum will be 3 times greater than Object B’s. Momentum = mass x velocity, so if velocities are equal, the difference in momentum will be the difference in mass.

2. Two other objects, C and D, have identical masses. Object C has twice the velocity of object D. Compare the momentum of each object. Justify your answer.Object C’s momentum will be 2 times greater than Object D’s. Momentum = mass x velocity, so if masses are equal, the difference in momentum will be the difference in velocity.

3. While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10 -2 sec. What is the magnitude of the change in momentum of the ball? (Assume it started from rest.) Show your work!

I = Δp = Ft Δp = Ft = (132 N)(4.5 x10-2s) = 5.94 kg m/s

4. If the initial speed of the baseball in question 3 is v0 = 0.0 m/s, what will its speed be when it leaves the pitcher's hand, if there is no air resistance? Show your work.

I = Δp = Ft I = Δp I = pf – pi

I = mvf – 0vf = I / m = 5.94 kg m/s / 0.140 kg = 42.4 m/s

5. When the batter hits the ball, a net force of 1320 N, opposite to the direction of the ball's initial motion, acts on the ball for 9.0 x 10-3 s during the hit. What is the change in momentum of the ball? What is the final velocity of the ball? Show your work!

I = Δp = Ft Δp = Ft = ( - 1320 N)(9.0 x10-3s) = - 11.88 kg m/s negative sign b/c it is opposite to direction of ball’s initial positive motion

I = Δp = Ft I = Δp I = pf – pi

I = mvf – mvi

vf = (I + mvi )/ m =[-11.88 kg m/s + (0.140 kg)(42.4 m/s)] = - 42.5 m/s0.140 kg negative sign b/c it is opposite

to direction of ball’s initial positive motion

6. What force does the ball exert on the bat in the question above? Explain.Newton’s 3rd law says the force of the ball on the bat is equal to the force of the bat on the ball, but in the

opposite direction.

7. A rocket, weighing 4.36 x 104 N, has an engine that provides an upward force of 8.90 x 105 N. It reaches a maximum speed of 860 m/s.

a. Draw a force diagram for the rocket.

b. How long must the engine burn in order to reach this speed? Fnet = 846,400 N upShow your work! I = Δp = Ft Δp = Ft mass = 4.36x10 4 N = 4449 kg

pf – pi = Ft 9.8 N/kgt = mvf – mvi

Ft = (4449 kg)(860 m/s) – 0

846400 Nt = 4.5 s

8. A golf ball that weighs 0.45 N is dropped from a height of 1.0 m. Assume that the golf ball has a perfectly elastic collision with the floor.

a. Determine the time required for the ball to reach the floor. Show your work!vo = 0 x = ½ at2 + vot + xo

xo = 1 m t = √[2(x-xo)/a] = √[2(0-1m)/9.8 m/s2]x = 0 m t = 0.45 sa = -9.8 m/s2

t = ?

b. What will the instantaneous momentum of the golf ball be immediately before it strikes the floor? Show your work!Need to know velocity just before it hits the floor. Mass = 0.45 N = 0.046 kgvo = 0 v2 = vo

2 + 2a(x-xo) 9.8 N/kgxo = 1 m v = √[ vo

2 + 2a(x-xo)]x = 0 m v = √[ 02 + 2(-9.8 m/s2)(0-1m)]a = -9.8 m/s2 v = ±4.43 m/sv = ? v = -4.43 m/s (because it’s moving down)

p = mv = (0.046 kg)(-4.43 m/s) = -0.20 kg m/s

c. What will be the change in momentum , (p) from the instant before the ball collides with the floor until the instant after it rebounds from the floor? Show your work!Perfectly elastic collision means no energy will be lost, so it will leave the floor at the same speed it hit the floor. vi = vf (except opposite sign since it’s going upwards); this also means p i = pf (except opposite signs)

Δp = pf – pi = 0.20 kg m/s – (-0.20 kg m/s) = 0.40 kg m/s

d. Suppose that the golf ball was in contact with the floor for 4.0 x 10-4s. What was the average force on the ball while it was in contact with the floor? Show your work!Δp = Ft F = Δp / t = 0.40 kg m/s / 4.0 x 10-4 s = 1000 N

9. Kim holds a 2.0 kg air rifle loosely and fires a bullet of mass 1.0 g. The muzzle velocity of the bullet is 150 m/s. What is the recoil speed of the rifle? Show your work!

BEFORE AFTERm v p m v p

rifle 2.0 kg 0 0 2.0 kg ? ? bullet 0.001 kg 0 0 0.001 kg 150 m/s 0.15 kg m/s total 0 0

Since the total momentum after the collision must equal 0, the momentum of the rifle must be - 0.15 kg m/s.Solve p = mv to find the velocity of the rifle. v = p / m = (-0.15 kg m/s) / (2.0 kg) = -0.075 m/s (negative sign on velocity means rifle moves the opposite direction from the bullet)

10. If the girl in the previous question holds the rifle tightly against her body, the recoil speed is less. Explain. Calculate the new recoil speed assuming the girl has a mass of 48 kg. Show your work.Since the girl holds the rifle tightly, they act like one object, so the girl’s mass will now be added to the mass of the rifle.

BEFORE AFTERm v p m v p

girl + rifle 50 kg 0 0 50 kg ? ? bullet 0.001 kg 0 0 0.001 kg 150 m/s 0.15 kg m/s total 0 0

Since the total momentum after the collision must equal 0, the momentum of the girl + rifle must be - 0.15 kg m/s.Solve p = mv to find the velocity of the girl + rifle. v = p / m = (-0.15 kg m/s) / (50 kg) = -0.003 m/s

11. In a freight yard a train is being put together from freight cars. An empty freight car, coasting at 10 m/s, strikes a loaded car that is stationary, and the cars couple together. Each of the cars has a mass of 3000 kg when empty, and the loaded car contains 12,000 kg of canned soda (a year's supply for the Physics class). With what speed does the combination of the two cars start to move? Show your work!

BEFORE AFTERm v p m v p

empty car 3000 kg 10 m/s 30000 kg m/s loaded car 15000 kg 0 0_________

total 30000 kg m/s 30000 kg m/s

Solve p = mv to find the velocity of the coupled cars. v = p / m = (30000 kg m/s) / (18000 kg) = 1.7 m/s

12. An astronaut whose mass is 80. kg carries an empty oxygen tank with a mass of 10. kg. He throws the tank away from himself with a speed of 2.0 m/s. With what velocity does he start to move off into space? Show your work.

BEFORE AFTERm v p m v p

astronaut 80. kg 0 0 80. kg ? ? tank 10. kg 0 0 10. kg 2 m/s 20 kg m/s total 0 0

Since the total momentum after the collision must equal 0, the momentum of the astronaut must be - 20 kg m/s.Solve p = mv to find the velocity of the rifle. v = p / m = (-20 kg m/s) / (80 kg) = -0.25 m/s (negative sign on velocity means astronaut moves the opposite direction from the gas tank)

13. A tennis player returns a 30. m/s serve straight back at 25. m/s, after making contact with the ball for 0.50 s. If the ball has a mass of 0.20 kg, what is the force she exerted on the ball? Show your work.

I = Δp = Ft Δp = Ft pf – pi = FtF = mvf – mvi

t negative b/c ball is moving opposite direction from original motionF = (0.20 kg)(-25 m/s) – (0.20 kg)(30 m/s)

0.50 sF = - 22 N negative b/c ball is moving opposite direction from original motion

14. A 50. kg cart is moving across a frictionless floor at 2.0 m/s. A 70. kg boy, riding in the cart, jumps off so that he hits the floor with zero velocity.

a. What impulse did the boy give to the cart? Show your work.Newton’s 3rd law says the boy will push on the cart just as hard as the cart pushes on the boy, but in opposite directions. Since the time of the force will be the same on both objects, the impulse from the cart on the boy will equal the impulse from the boy on the cart. So if we can find the boy’s impulse, we know that the impulse on the cart is equal in magnitude but opposite in direction to the boy’s impulse.

Impulse on boy = his change in momentum = pf – pi = mvf – mvi = 0 – (70. kg)(2.0 m/s) = -140 kg m/s

Impulse on cart = 140 kg m/s

b. What was the velocity of the cart after the boy jumped? Show your work.

I = Δp I = pf – pi

I = mvf – mvi

vf = (I + mvi) / m = (140 kg m/s + (50. kg)(2.0 m/s) =4.8 m/s 50. kg

15. Two girls with masses of 50.0 kg and 70.0 kg are at rest on frictionless in-line skates. The larger girl pushes the smaller girl so that the latter rolls away at a speed of 10.0 m/s. What is the effect of the action on the larger girl? What is the impulse that each girl exerts on the other? Show your work.

18000 kg ? 30000 kg m/s

The larger girl will roll in the opposite direction, but at a lower speed.

As described in the previous problem, the impulses on each girl will be equal in magnitude but opposite in direction.

Impulse on smaller girlI = Δp I = pf – pi

I = mvf – mvi = (50.0 kg)(10.0 m/s) – 0 = 500 kg m/s

Impulse on larger girl = - 500 kg m/s

16. A 2.0 kg melon is balanced on a bald man's head. His son shoots a 50.0 g arrow at it with a speed of 30.0 m/s. The arrow passes through the melon and emerges with a speed of 18.0 m/s. Find the speed of the melon as it flies off the man's head. Show your work.

BEFORE AFTERm v p m v p

melon 2.0 kg 0 0 2.0 kg ? ? arrow 0.050 kg 30.0 m/s 1.5 kg m/s 0.050 kg 18 m/s 0.9 kg m/s total 1.5 kg m/s 1.5 kg m/s

Since the total momentum after the collision must equal 1.5 kg m/s, the momentum of the melon must be 0.6 kg m/s.Solve p = mv to find the velocity of the melon. v = p / m = (0.6 kg m/s) / (2.0 kg) = 0.3 m/s

17. Mighty Miguel has a mass of 100. kg and is running EAST towards the end zone at 9.0 m/s. Joey Gonzales (mass of 75.0 kg), runs at 12.0 m/s NORTH towards Miguel. They collide (and stick together) at the 2-yard line. Calculate their velocity after the collision. Include a direction (indicated by a specific angle). Show your work.

Miguel p = mv = (100. kg)(9.0 m/s) = 900 kg m/s EAST

Joey p = mv = (75.0 kg)(12.0 m/s) = 900 kg m/s NORTH

Add 2 momenta vectors together using Pythagorean’s theorem. Use tangent function to find angle.

900 kg m/s NORTH R = 1273 kg m/s @ 45°

900 kg m/s EAST

Now you can find their velocity by rearranging p=mv. v = p / m = (1273 kg m/s) / (100 + 75 kg)= 7.3 m/s @ 45° (angle of v = angle of momentum)