# Chapter 7: Linear Momentum

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• 1.

### What is the SI unit of momentum?

• A.

N*m

• B.

N/s

• C.

N*s

• D.

N/m

C. N*s
Explanation
The SI unit of momentum is N*s, which stands for Newton-second. Momentum is defined as the product of an object's mass and its velocity, and it is a vector quantity. The unit N*s represents the change in momentum that occurs when a force of 1 Newton is applied to an object for a duration of 1 second. Therefore, N*s is the correct unit for measuring momentum in the SI system.

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• 2.

### When a cannon fires a cannonball, the cannon will recoil backward because the

• A.

Energy of the cannonball and cannon is conserved.

• B.

Momentum of the cannonball and cannon is conserved.

• C.

Energy of the cannon is greater than the energy of the cannonball.

• D.

Momentum of the cannon is greater than the energy of the cannonball.

B. Momentum of the cannonball and cannon is conserved.
Explanation
When a cannon fires a cannonball, the momentum of the cannonball and the cannon is conserved. This means that the total momentum of the system (cannonball + cannon) before the firing is equal to the total momentum after the firing. Since momentum is a vector quantity and depends on both mass and velocity, when the cannonball is propelled forward with a certain velocity, the cannon experiences an equal and opposite momentum in the backward direction, causing it to recoil. This conservation of momentum is a fundamental principle in physics.

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• 3.

### A freight car moves along a frictionless level railroad track at constant speed. The car is open on top. A large load of coal is suddenly dumped into the car. What happens to the velocity of the car?

• A.

It increases.

• B.

It remains the same.

• C.

It decreases.

• D.

Cannot be determined from the information given.

C. It decreases.
Explanation
When a large load of coal is suddenly dumped into the open top freight car, the total mass of the car increases. According to the law of conservation of momentum, the momentum of an isolated system remains constant if no external forces act on it. Since the car is on a frictionless track, there are no external forces acting on it. Therefore, in order to maintain the constant momentum, the velocity of the car must decrease as the mass increases. Hence, the correct answer is that the velocity of the car decreases.

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• 4.

### A child falls sideways off a sled while sledding on frictionless ice. What happens to the velocity of the sled?

• A.

It increases.

• B.

It remains the same.

• C.

It decreases.

• D.

Cannot be determined from the information given.

B. It remains the same.
Explanation
When the child falls sideways off the sled, there is no external force acting on the sled to change its velocity. According to Newton's first law of motion, an object will continue to move with a constant velocity unless acted upon by an external force. Since there is no friction or other forces acting on the sled, its velocity will remain the same.

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• 5.

### A rubber ball and a lump of putty have equal mass. They are thrown with equal speed against a wall. The ball bounces back with nearly the same speed with which it hit. The putty sticks to the wall. Which objects experiences the greater momentum change?

• A.

The ball

• B.

The putty

• C.

Both experience the same momentum change.

• D.

Cannot be determined from the information given.

A. The ball
Explanation
When the ball bounces back with nearly the same speed with which it hit, it means that its momentum changes in direction but remains the same in magnitude. On the other hand, the putty sticks to the wall, resulting in a complete change in its momentum. Therefore, the ball experiences a greater momentum change compared to the putty.

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• 6.

### A sailboat of mass m is moving with a momentum p. How would you represent its kinetic energy in terms of these two quantities?

• A.

(p^2)/(2m)

• B.

(1/2)mp^2

• C.

Mp

• D.

Mp/2

A. (p^2)/(2m)
Explanation
The kinetic energy of an object is given by the formula 1/2mv^2, where m is the mass of the object and v is its velocity. In this case, the sailboat is moving with momentum p, which is equal to mv. By rearranging the equation for momentum, we get v = p/m. Substituting this into the equation for kinetic energy gives (1/2)m(p/m)^2, which simplifies to (p^2)/(2m). Therefore, the correct answer is (p^2)/(2m).

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• 7.

### If you pitch a baseball with twice the kinetic energy you gave it in the previous pitch, the magnitude of its momentum is

• A.

The same.

• B.

1.41 times as much.

• C.

Doubled.

• D.

4 times as much.

B. 1.41 times as much.
Explanation
When an object's kinetic energy doubles, its momentum does not double. Instead, there is a square root relationship between kinetic energy and momentum. This means that if the kinetic energy is doubled, the momentum will increase by the square root of 2, which is approximately 1.41 times as much. Therefore, the correct answer is 1.41 times as much.

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• 8.

### The area under the curve on a Force versus time (F vs. t) graph represents

• A.

Impulse.

• B.

Momentum.

• C.

Work.

• D.

Kinetic energy.

A. Impulse.
Explanation
The area under the curve on a Force versus time graph represents impulse. Impulse is defined as the change in momentum of an object, and it is equal to the force applied to the object multiplied by the time interval over which the force is applied. The area under the curve represents the integral of the force with respect to time, which gives the total impulse exerted on the object. Therefore, the correct answer is impulse.

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• 9.

### Which of the following is an accurate statement?

• A.

The momentum of a projectile is constant.

• B.

The momentum of a moving object is constant.

• C.

If an object is acted on by a non-zero net external force, its momentum will not remain constant.

• D.

If the kinetic energy of an object is doubled, its momentum will also double.

C. If an object is acted on by a non-zero net external force, its momentum will not remain constant.
Explanation
When an object is acted on by a non-zero net external force, its momentum will not remain constant. This is because momentum is defined as the product of an object's mass and velocity, and any change in either mass or velocity will result in a change in momentum. Therefore, if an external force is applied to an object, it will either accelerate or decelerate, causing a change in velocity and subsequently a change in momentum.

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• 10.

### A small car meshes with a large truck in a head-on collision. Which of the following statements concerning the magnitude of the average collision force is correct?

• A.

The truck experiences the greater average force.

• B.

The small car experiences the greater average force.

• C.

The small car and the truck experience the same average force.

• D.

It is impossible to tell since the masses and velocities are not given.

C. The small car and the truck experience the same average force.
Explanation
In a head-on collision, the magnitude of the average collision force depends on the change in momentum of the objects involved. According to Newton's third law of motion, the forces exerted on two colliding objects are equal in magnitude and opposite in direction. Therefore, the small car and the truck experience the same average force. The masses and velocities of the vehicles are not necessary to determine this, as the force experienced in a collision is independent of the size or mass of the objects involved.

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• 11.

### Two equal mass balls (one red and the other blue) are dropped from the same height, and rebound off the floor. The red ball rebounds to a higher position. Which ball is subjected to the greater magnitude of impulse during its collision with the floor?

• A.

It's impossible to tell since the time intervals and forces are unknown.

• B.

Both balls were subjected to the same magnitude impulse.

• C.

The blue ball

• D.

The red ball

D. The red ball
Explanation
The red ball is subjected to the greater magnitude of impulse during its collision with the floor.

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• 12.

### A Ping-Pong ball moving east at a speed of 4 m/s, collides with a stationary bowling ball. The Ping-Pong ball bounces back to the west, and the bowling ball moves very slowly to the east. Which object experiences the greater magnitude impulse during the collision?

• A.

Neither; both experienced the same magnitude impulse.

• B.

The Ping-Pong ball

• C.

The bowling ball

• D.

It's impossible to tell since the velocities after the collision are unknown.

A. Neither; both experienced the same magnitude impulse.
Explanation
Both the Ping-Pong ball and the bowling ball experience the same magnitude impulse during the collision. Impulse is equal to the change in momentum, and since momentum is conserved in a collision, the impulse experienced by each object is the same. The fact that the Ping-Pong ball bounces back and the bowling ball moves slowly does not affect the magnitude of the impulse experienced by each object.

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• 13.

### Two objects collide and bounce off each other. Linear momentum

• A.

Is definitely conserved.

• B.

Is definitely not conserved.

• C.

Is conserved only if the collision is elastic.

• D.

Is conserved only if the environment is frictionless.

A. Is definitely conserved.
Explanation
Linear momentum is a fundamental principle in physics that states that the total momentum of a system remains constant unless acted upon by an external force. In this scenario, when two objects collide and bounce off each other, the total momentum of the system before the collision is equal to the total momentum after the collision. Therefore, linear momentum is definitely conserved in this situation.

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• 14.

### A 3.0-kg object moves to the right at 4.0 m/s. It collides head-on with a 6.0-kg object moving to the left at 2.0 m/s. Which statement is correct?

• A.

The total momentum both before and after the collision is 24 kg*m/s.

• B.

The total momentum before the collision is 24 kgoem/s, and after the collision is 0 kg*m/s.

• C.

The total momentum both before and after the collision is zero.

• D.

None of the above is true.

C. The total momentum both before and after the collision is zero.
Explanation
The total momentum before the collision can be calculated by adding the individual momenta of the two objects. The momentum of the first object is calculated by multiplying its mass (3.0 kg) by its velocity (4.0 m/s), resulting in a momentum of 12 kg*m/s. The momentum of the second object is calculated by multiplying its mass (6.0 kg) by its velocity (-2.0 m/s), resulting in a momentum of -12 kg*m/s. Adding these two momenta together gives a total momentum of zero before the collision. According to the law of conservation of momentum, the total momentum after the collision should also be zero. Therefore, the statement "The total momentum both before and after the collision is zero" is correct.

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• 15.

### A 100-kg football linebacker moving at 2.0 m/s tackles head-on an 80-kg halfback running 3.0 m/s. Neglecting the effects due to digging in of cleats,

• A.

The linebacker will drive the halfback backward.

• B.

The halfback will drive the linebacker backward.

• C.

Neither player will drive the other backward.

• D.

This is a simple example of an elastic collision.

B. The halfback will drive the linebacker backward.
Explanation
The halfback will drive the linebacker backward because the momentum of the halfback is greater than the linebacker. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the halfback has a greater mass and a higher velocity, their momentum is greater than the linebacker's momentum. Therefore, the halfback will exert a greater force on the linebacker, causing the linebacker to move backward.

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• 16.

### In an elastic collision, if the momentum is conserved, then which of the following statements is true about kinetic energy?

• A.

Kinetic energy is also conserved.

• B.

Kinetic energy is gained.

• C.

Kinetic energy is lost.

• D.

None of the above

A. Kinetic energy is also conserved.
Explanation
In an elastic collision, both momentum and kinetic energy are conserved. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Therefore, the statement "Kinetic energy is also conserved" is true.

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• 17.

### When is kinetic energy conserved?

• A.

In elastic collisions

• B.

In inelastic collisions

• C.

In any collision in which the objects do not stick together

• D.

In all collisions

A. In elastic collisions
Explanation
Kinetic energy is conserved in elastic collisions because in these collisions, the total kinetic energy of the objects before the collision is equal to the total kinetic energy of the objects after the collision. This means that no energy is lost or gained during the collision. In elastic collisions, the objects bounce off each other and there is no deformation or loss of energy due to heat or sound. Therefore, kinetic energy is conserved in elastic collisions.

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• 18.

### In a game of pool, the white cue ball hits the #5 ball and stops, while the #5 ball moves away with the same velocity as the cue ball had originally. The type of collision is

• A.

Elastic.

• B.

Inelastic.

• C.

Completely inelastic.

• D.

Any of the above, depending on the mass of the balls.

A. Elastic.
Explanation
In an elastic collision, both the momentum and kinetic energy are conserved. In this scenario, the white cue ball hits the #5 ball and stops, while the #5 ball moves away with the same velocity as the cue ball had originally. This indicates that the momentum of the system is conserved, as the total momentum before the collision (which is the momentum of the cue ball) is equal to the total momentum after the collision (which is the momentum of the #5 ball). Since the kinetic energy is also conserved, the collision is elastic.

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• 19.

### When a light beach ball rolling with a speed of 6.0 m/s collides with a heavy exercise ball at rest, the beach ball's speed after the collision will be, approximately,

• A.

0.

• B.

3.0 m/s.

• C.

6.0 m/s.

• D.

12 m/s.

C. 6.0 m/s.
Explanation
When a light beach ball collides with a heavy exercise ball at rest, the momentum of the system is conserved. Since the exercise ball is at rest, its initial momentum is zero. After the collision, the momentum of the system is still zero, so the beach ball's final momentum must also be zero. Since momentum is the product of mass and velocity, and the mass of the beach ball does not change, its velocity must be zero after the collision. Therefore, the beach ball's speed after the collision will be approximately 0 m/s.

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• 20.

### A golf ball traveling 3.0 m/s to the right collides in a head-on collision with a stationary bowling ball in a friction-free environment. If the collision is almost perfectly elastic, the speed of the golf ball immediately after the collision is

• A.

Slightly less than 3.0 m/s.

• B.

Slightly greater than 3.0 m/s.

• C.

Equal to 3.0 m/s.

• D.

Much less than 3.0 m/s.

A. Slightly less than 3.0 m/s.
Explanation
In an almost perfectly elastic collision, kinetic energy is conserved. Since the collision is head-on, the golf ball transfers some of its kinetic energy to the stationary bowling ball, causing it to move. As a result, the golf ball loses some of its initial kinetic energy, leading to a decrease in speed. Therefore, the speed of the golf ball immediately after the collision will be slightly less than 3.0 m/s.

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• 21.

### A rubber ball with a speed of 5.0 m/s collides head-on elastically with an identical ball at rest. What is the speed of the initially stopped ball after the collision?

• A.

Zero

• B.

1.0 m/s

• C.

2.5 m/s

• D.

5.0 m/s

D. 5.0 m/s
Explanation
After the elastic collision, the rubber ball transfers all of its momentum to the initially stopped ball. Since the rubber ball has a speed of 5.0 m/s, the initially stopped ball will also have a speed of 5.0 m/s after the collision.

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• 22.

### A very heavy object moving with speed v collides head-on with a very light object at rest. The collision is elastic, and there is no friction. The heavy object barely slows down. What is the speed of the light object after the collision?

• A.

Nearly v

• B.

Nearly 2v

• C.

Nearly 3v

• D.

Nearly infinite

B. Nearly 2v
Explanation
In an elastic collision, both momentum and kinetic energy are conserved. Since the heavy object barely slows down, its final velocity after the collision is nearly equal to its initial velocity, which is v. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the light object is initially at rest, its momentum is zero. Therefore, the momentum of the heavy object after the collision must also be zero. Since momentum is mass times velocity, the mass of the heavy object must be twice the mass of the light object for their momenta to cancel out. Therefore, the speed of the light object after the collision is nearly 2v.

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• 23.

### A very light object moving with speed v collides head-on with a very heavy object at rest, in a frictionless environment. The collision is almost perfectly elastic. The speed of the heavy object after the collision is

• A.

Slightly greater than v.

• B.

Equal to v.

• C.

Slightly less than v.

• D.

Much less than v.

D. Much less than v.
Explanation
In an almost perfectly elastic collision, kinetic energy is conserved. Since the heavy object is at rest before the collision, it has no initial kinetic energy. Therefore, all of the kinetic energy after the collision must come from the light object. As a result, the heavy object will gain a small amount of kinetic energy and move with a speed much less than v.

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• 24.

### A red ball with a velocity of +3.0 m/s collides head-on with a yellow ball of equal mass moving with a velocity of -2.0 m/s. What is the velocity of the yellow ball after the collision?

• A.

Zero

• B.

+3.0 m/s

• C.

-2.0 m/s

• D.

+5.0 m/s

B. +3.0 m/s
Explanation
When the red ball collides head-on with the yellow ball, the momentum of the system is conserved. Since the red ball is moving with a velocity of +3.0 m/s and the yellow ball is moving with a velocity of -2.0 m/s, the total momentum before the collision is (mass of red ball * velocity of red ball) + (mass of yellow ball * velocity of yellow ball) = (mass of red ball * +3.0 m/s) + (mass of yellow ball * -2.0 m/s). Since both balls have equal mass, the total momentum before the collision is 0. After the collision, the momentum should still be 0. Therefore, the velocity of the yellow ball after the collision is also 0 m/s.

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• 25.

### A very heavy object moving with velocity v collides head-on with a very light object moving with velocity -v. The collision is elastic, and there is no friction. The heavy object barely slows down. What is the speed of the light object after the collision?

• A.

Nearly v

• B.

Nearly 2v

• C.

Nearly 3v

• D.

Nearly infinite

C. Nearly 3v
Explanation
In an elastic collision, both momentum and kinetic energy are conserved. Since the heavy object barely slows down, its final velocity will be nearly the same as its initial velocity, which is v. According to the law of conservation of momentum, the total momentum before the collision is zero (since the heavy object has momentum in one direction and the light object has momentum in the opposite direction). After the collision, the total momentum must still be zero. Therefore, the light object must have a momentum of -2v in order to balance out the momentum of the heavy object. Since momentum is mass times velocity, and the light object is very light compared to the heavy object, its velocity must be nearly 3v in the opposite direction.

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• 26.

### In an inelastic collision, if the momentum is conserved, then which of the following statements is true about kinetic energy?

• A.

Kinetic energy is also conserved.

• B.

Kinetic energy is gained.

• C.

Kinetic energy is lost.

• D.

None of the above

C. Kinetic energy is lost.
Explanation
In an inelastic collision, kinetic energy is lost. This is because in an inelastic collision, the objects involved stick together or deform, resulting in the conversion of some of the kinetic energy into other forms of energy, such as heat or sound. Therefore, the total kinetic energy of the system decreases after the collision.

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• 27.

### Two objects collide and stick together. Kinetic energy

• A.

Is definitely conserved.

• B.

Is definitely not conserved.

• C.

Is conserved only if the collision is elastic.

• D.

Is conserved only if the environment is frictionless.

B. Is definitely not conserved.
Explanation
When two objects collide and stick together, kinetic energy is not conserved. This is because some of the initial kinetic energy is converted into other forms of energy, such as heat or sound, during the collision. The objects may also experience deformation or loss of energy due to friction or other external forces. Therefore, the total kinetic energy after the collision is less than the initial kinetic energy, indicating that kinetic energy is not conserved in this scenario.

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• 28.

### A 3.0-kg object moves to the right at 4.0 m/s. It collides in a perfectly inelastic collision with a 6.0 kg object moving to the left at 2.0 m/s. What is the total kinetic energy after the collision?

• A.

72 J

• B.

36 J

• C.

24 J

• D.

0 J

D. 0 J
Explanation
In a perfectly inelastic collision, the two objects stick together after the collision and move with a common velocity. In this case, since the 3.0 kg object is moving to the right and the 6.0 kg object is moving to the left, the resulting velocity after the collision will be zero. Therefore, the total kinetic energy after the collision is also zero, as kinetic energy is proportional to the square of velocity.

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• 29.

### A small object collides with a large object and sticks. Which object experiences the larger magnitude of momentum change?

• A.

The large object

• B.

The small object

• C.

Both objects experience the same magnitude of momentum change.

• D.

Cannot be determined from the information given

C. Both objects experience the same magnitude of momentum change.
Explanation
When a small object collides with a large object and sticks, both objects experience the same magnitude of momentum change. This is because momentum is conserved in a collision, meaning that the total momentum before the collision is equal to the total momentum after the collision. Since the small object sticks to the large object, they become one combined object with a new total mass. Therefore, the momentum change experienced by both objects is the same.

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• 30.

### In a game of pool, the white cue ball hits the #9 ball and is deflected at a 35° angle to the original line of motion. What is the angle of deflection below the original line of motion for the #9 ball?

• A.

35°

• B.

55°

• C.

75°

• D.

90°

B. 55°
Explanation
When the white cue ball hits the #9 ball, it is deflected at a 35° angle to the original line of motion. The angle of deflection for the #9 ball is always equal to the angle of incidence, which is the angle at which the cue ball hits it. Since the cue ball is deflected at a 35° angle, the #9 ball will be deflected at the same angle but in the opposite direction. Therefore, the angle of deflection below the original line of motion for the #9 ball is 55°.

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• 31.

### Consider two unequal masses, M and m. Which of the following statements is false?

• A.

The center of mass lies on the line joining the centers of each mass.

• B.

The center of mass is closer to the larger mass.

• C.

It is possible for the center of mass to lie within one of the objects.

• D.

If a uniform rod of mass m were to join the two masses, this would not alter the position of the center of mass of the system without the rod present.

D. If a uniform rod of mass m were to join the two masses, this would not alter the position of the center of mass of the system without the rod present.
Explanation
If a uniform rod of mass m were to join the two masses, this would not alter the position of the center of mass of the system without the rod present. This statement is false because adding the rod would change the distribution of mass in the system, thus altering the position of the center of mass.

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• 32.

### Which of the following is a false statement?

• A.

For a uniform symmetric object, the center of mass is at the center of symmetry.

• B.

For an object on the surface of the Earth, the center of gravity and the center of mass are the same point.

• C.

The center of mass of an object must lie within the object.

• D.

The center of gravity of an object may be thought of as the "balance point."

C. The center of mass of an object must lie within the object.
Explanation
The center of mass of an object must lie within the object because it is the average position of all the mass in the object.

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• 33.

### Tightrope walkers walk with a long flexible rod in order to

• A.

Increase their total weight.

• B.

Allow both hands to hold onto something.

• C.

Lower their center of mass.

• D.

Move faster along the rope.

C. Lower their center of mass.
Explanation
Tightrope walkers walk with a long flexible rod in order to lower their center of mass. By holding onto the rod, they are able to shift their weight downwards, making their center of mass closer to the tightrope. This helps them maintain balance and stability while walking on the rope.

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• 34.

### A plane, flying horizontally, releases a bomb, which explodes before hitting the ground. Neglecting air resistance, the center of mass of the bomb fragments, just after the explosion

• A.

Is zero.

• B.

Moves horizontally.

• C.

Moves vertically.

• D.

Moves along a parabolic path.

D. Moves along a parabolic path.
Explanation
When the bomb explodes, it breaks into fragments that move in different directions. Since the bomb was initially moving horizontally, the fragments will also have a horizontal component of motion. However, due to the force of the explosion, the fragments will also have a vertical component of motion. This combination of horizontal and vertical motion causes the fragments to follow a parabolic path. Therefore, the center of mass of the bomb fragments just after the explosion moves along a parabolic path.

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• 35.

### Two cars collide head-on on a level friction-free road. The collision was completely inelastic and both cars quickly came to rest during the collision. What is true about the velocity of this system's center of mass?

• A.

It was always zero.

• B.

It was never zero.

• C.

It was not zero, but ended up zero.

• D.

None of the above

A. It was always zero.
Explanation
When two cars collide head-on on a level friction-free road, the total momentum of the system is conserved. Since the collision is completely inelastic and both cars quickly come to rest, the final momentum of the system is zero. The velocity of the center of mass is calculated using the formula: velocity = total momentum / total mass. Since the total momentum is zero, the velocity of the center of mass is also zero. Therefore, the statement "It was always zero" is correct.

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• 36.

### What is the momentum of a 2000-kg truck traveling at 35 m/s?

• A.

57 kg*m/s

• B.

3.5 * 10^4 kg*m/s

• C.

7.0 * 10^4 kg*m/s

• D.

7.0 * 10^5 kg*m/s

C. 7.0 * 10^4 kg*m/s
Explanation
The momentum of an object is calculated by multiplying its mass by its velocity. In this case, the truck has a mass of 2000 kg and is traveling at 35 m/s. To find the momentum, we multiply 2000 kg by 35 m/s, which equals 70,000 kg*m/s. This can be written in scientific notation as 7.0 * 10^4 kg*m/s.

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• 37.

### A 1200-kg ferryboat is moving south at 20 m/s. What is the magnitude of its momentum?

• A.

1.7 * 10^(-3) kg*m/s

• B.

6.0 * 10^2 kg*m/s

• C.

2.4 * 10^3 kg*m/s

• D.

2.4 * 10^4 kg*m/s

D. 2.4 * 10^4 kg*m/s
Explanation
The momentum of an object is given by the product of its mass and velocity. In this case, the mass of the ferryboat is 1200 kg and its velocity is 20 m/s. By multiplying these two values together, we get a momentum of 24,000 kg*m/s.

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• 38.

### A ball of mass 0.10 kg is dropped from a height of 12 m. Its momentum when it strikes the ground is

• A.

1.5 kg*m/s.

• B.

1.8 kg*m/s.

• C.

2.4 kg*m/s.

• D.

4.8 kg*m/s.

A. 1.5 kg*m/s.
Explanation
When an object is dropped, it falls freely under the influence of gravity. As the ball falls, its potential energy is converted into kinetic energy. The momentum of an object is defined as the product of its mass and velocity. Since the ball is dropped from rest, its initial velocity is 0 m/s. Using the equation p = mv, where p is momentum, m is mass, and v is velocity, we can calculate the momentum of the ball when it strikes the ground. The mass of the ball is given as 0.10 kg. The velocity can be calculated using the equation vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity, a is acceleration (due to gravity), and d is distance. Plugging in the given values, we find that the final velocity when the ball strikes the ground is approximately 7.67 m/s. Multiplying the mass and velocity, we get a momentum of approximately 0.10 kg * 7.67 m/s = 0.767 kg*m/s. Rounding to the nearest tenth, the correct answer is 1.5 kg*m/s.

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• 39.

### Two identical 1500-kg cars are moving perpendicular to each other. One moves with a speed of 25 m/s due north and the other moves at 15 m/s due east. What is the total momentum of the system?

• A.

4.4 * 10^4 kg*m/s at 31° N of E

• B.

4.4 * 10^4 kg*m/s at 59° N of E

• C.

6.0 * 10^4 kg*m/s at 31° N of E

• D.

6.0 * 10^4 kg*m/s at 59° N of E

B. 4.4 * 10^4 kg*m/s at 59° N of E
Explanation
The total momentum of the system can be found by adding the individual momenta of the two cars. Since momentum is a vector quantity, we need to consider both the magnitude and the direction. The magnitude of the total momentum is given by the Pythagorean theorem as √((1500 kg * 25 m/s)^2 + (1500 kg * 15 m/s)^2) = 4.4 * 10^4 kg*m/s. The direction of the total momentum can be found using trigonometry. The angle can be determined as arctan((1500 kg * 15 m/s) / (1500 kg * 25 m/s)) = arctan(15/25) = 31°. Since the car moving north has a greater speed, the direction of the total momentum is 31° north of east.

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• 40.

### A handball of mass 0.10 kg, traveling horizontally at 30 m/s, strikes a wall and rebounds at 24 m/s. What is the change in the momentum of the ball?

• A.

0.60 kg*m/s

• B.

1.2 kg*m/s

• C.

5.4 kg*m/s

• D.

72 kg*m/s

C. 5.4 kg*m/s
Explanation
When the handball strikes the wall and rebounds, it undergoes a change in momentum. The change in momentum can be calculated by subtracting the initial momentum from the final momentum.

The initial momentum of the ball can be calculated by multiplying its mass (0.10 kg) by its initial velocity (30 m/s), which gives us 3 kg*m/s.

The final momentum of the ball can be calculated by multiplying its mass (0.10 kg) by its final velocity (24 m/s), which gives us 2.4 kg*m/s.

To find the change in momentum, we subtract the initial momentum from the final momentum: 2.4 kg*m/s - 3 kg*m/s = -0.6 kg*m/s.

Since momentum is a vector quantity, the negative sign indicates that the direction of the momentum has changed. Therefore, the change in momentum of the ball is 0.6 kg*m/s in the opposite direction.

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• 41.

### A 0.060-kg tennis ball, initially moving at a speed of 12 m/s, is struck by a racket causing it to rebound in the opposite direction at a speed of 18 m/s. What is the change in momentum of the ball?

• A.

0.36 kg*m/s

• B.

0.72 kg*m/s

• C.

1.1 kg*m/s

• D.

1.8 kg*m/s

D. 1.8 kg*m/s
Explanation
When the tennis ball is struck by the racket, its velocity changes from 12 m/s in one direction to 18 m/s in the opposite direction. The change in velocity is 18 m/s - (-12 m/s) = 30 m/s. The momentum of an object is given by the product of its mass and velocity. Therefore, the change in momentum of the tennis ball can be calculated by multiplying its mass (0.060 kg) by the change in velocity (30 m/s). This gives us a change in momentum of 1.8 kg*m/s.

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• 42.

### A 50-kg pitching machine (excluding the baseball) is placed on a frozen pond. The machine fires a 0.40-kg baseball with a speed of 35 m/s in the horizontal direction. What is the recoil speed of the pitching machine? (Assume negligible friction.)

• A.

0.14 m/s

• B.

0.28 m/s

• C.

0.70 m/s

• D.

4.4 * 10^3 m/s

B. 0.28 m/s
Explanation
When the pitching machine fires the baseball, according to Newton's third law of motion, there is an equal and opposite reaction. The momentum of the baseball in the horizontal direction is given by the product of its mass and velocity, which is (0.40 kg) * (35 m/s) = 14 kg·m/s. To conserve momentum, the pitching machine must have an equal and opposite momentum in the opposite direction. Since the mass of the pitching machine is 50 kg, the recoil speed can be calculated by dividing the momentum by the mass, which is (14 kg·m/s) / (50 kg) = 0.28 m/s. Therefore, the recoil speed of the pitching machine is 0.28 m/s.

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• 43.

### A 70-kg astronaut is space-walking outside the space capsule and is stationary when the tether line breaks. As a means of returning to the capsule he throws his 2.0-kg space hammer at a speed of 14 m/s away from the capsule. At what speed does the astronaut move toward the capsule?

• A.

0.40 m/s

• B.

1.5 m/s

• C.

3.5 m/s

• D.

5.0 m/s

A. 0.40 m/s
Explanation
When the astronaut throws the space hammer away from the capsule, according to the law of conservation of momentum, the astronaut will experience an equal and opposite momentum in the opposite direction. Since the mass of the astronaut is much larger than the mass of the space hammer, the astronaut's velocity will be much smaller than that of the hammer. Therefore, the astronaut will move toward the capsule at a speed of 0.40 m/s.

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• 44.

### A small object with momentum 5.0 kg*m/s approaches head-on a large object at rest. The small object bounces straight back with a momentum of magnitude 4.0 kg*m/s. What is the magnitude of the large object's momentum change?

• A.

9.0 kg*m/s

• B.

5.0 kg*m/s

• C.

4.0 kg*m/s

• D.

1.0 kg*m/s

A. 9.0 kg*m/s
Explanation
When the small object bounces straight back, its momentum changes from 5.0 kg*m/s to -4.0 kg*m/s. The change in momentum is calculated by subtracting the initial momentum from the final momentum. Therefore, the magnitude of the large object's momentum change is equal to the magnitude of the change in momentum of the small object, which is 9.0 kg*m/s.

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• 45.

### You (50-kg mass) skate on ice at 4.0 m/s to greet your friend (40-kg mass), who is standing still, with open arms. As you collide, while holding each other, with what speed do you both move off together?

• A.

Zero

• B.

2.2 m/s

• C.

5.0 m/s

• D.

23 m/s

B. 2.2 m/s
Explanation
When you collide with your friend while holding each other, the law of conservation of momentum applies. According to this law, the total momentum before the collision is equal to the total momentum after the collision. Since your friend is initially at rest, their momentum is zero. Therefore, the total momentum before the collision is equal to your momentum, which is given by the equation p = m * v, where p is momentum, m is mass, and v is velocity. After the collision, you both move off together with the same velocity. Using the conservation of momentum equation, (50 kg * 4.0 m/s) + (40 kg * 0 m/s) = (90 kg * v), we can solve for v and find that the velocity of both of you moving off together is 2.2 m/s.

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• 46.

### A car of mass 1000 kg moves to the right along a level, straight road at a speed of 6.0 m/s. It collides directly with a stopped motorcycle of mass 200 kg. What is the total momentum after the collision?

• A.

Zero

• B.

6000 kg*m/s to the right

• C.

2000 kg*m/s to the right

• D.

10,000 kg*m/s to the right

B. 6000 kg*m/s to the right
Explanation
The total momentum after the collision is 6000 kg*m/s to the right. This can be determined using the principle of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision. Since the car is moving to the right with a momentum of 6000 kg*m/s and the motorcycle is initially stopped, the total momentum after the collision will be equal to the momentum of the car, which is 6000 kg*m/s to the right.

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• 47.

### A 1000-kg car traveling at 25 m/s runs into the rear of a stopped car that has a mass of 1500 kg and they stick together. What is the speed of the cars after the collision?

• A.

5.0 m/s

• B.

10 m/s

• C.

15 m/s

• D.

20 m/s

B. 10 m/s
Explanation
When the two cars collide and stick together, the total momentum before the collision is equal to the total momentum after the collision. The momentum of an object is given by the product of its mass and velocity. Before the collision, the momentum of the 1000-kg car is (1000 kg)(25 m/s) = 25000 kg·m/s, and the momentum of the 1500-kg car is (1500 kg)(0 m/s) = 0 kg·m/s. After the collision, the two cars stick together and move with a common velocity. Let's call this velocity V. The total momentum after the collision is (2500 kg)(V) = 2500V kg·m/s. Since the total momentum before and after the collision must be equal, we can equate the two expressions: 25000 kg·m/s = 2500V kg·m/s. Dividing both sides by 2500 kg·m/s, we find that V = 10 m/s. Therefore, the speed of the cars after the collision is 10 m/s.

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• 48.

### A railroad freight car, mass 15,000 kg, is allowed to coast along a level track at a speed of 2.0 m/s. It collides and couples with a 50,000-kg second car, initially at rest and with brakes released. What is the speed of the two cars after coupling?

• A.

0.46 m/s

• B.

0.60 m/s

• C.

1.2 m/s

• D.

1.8 m/s

A. 0.46 m/s
Explanation
When the two cars couple together, they form a system with a total mass of 65,000 kg. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the second car was initially at rest, its momentum is zero. Therefore, the momentum of the system after the collision is equal to the momentum of the first car before the collision. Using the equation for momentum (p = mv), we can calculate the momentum of the first car before the collision as (15,000 kg)(2.0 m/s) = 30,000 kg·m/s. Since the total mass of the system is 65,000 kg, the velocity of the system after the collision is 30,000 kg·m/s divided by 65,000 kg, which is approximately 0.46 m/s.

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• 49.

### A railroad car, of mass 200 kg, rolls with negligible friction on a horizontal track with a speed of 10 m/s. A 70-kg stunt man drops straight down a distance of 4.0 m, and lands in the car. How fast will the car be moving after this happens?

• A.

2.8 m/s

• B.

4.7 m/s

• C.

7.4 m/s

• D.

10 m/s

C. 7.4 m/s
Explanation
When the stunt man drops into the car, the total momentum of the system is conserved. The initial momentum of the car is given by its mass (200 kg) multiplied by its initial velocity (10 m/s), which is 2000 kg·m/s. The initial momentum of the stunt man is given by his mass (70 kg) multiplied by his initial velocity (0 m/s), which is 0 kg·m/s. After the stunt man lands in the car, their combined mass is 270 kg. The final velocity of the car can be calculated by dividing the initial momentum by the combined mass, which gives a value of approximately 7.4 m/s.

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• 50.

### A 60-kg person walks on a 100-kg log at the rate of 0.80 m/s (with respect to the log). With what speed does the log move, with respect to the shore?

• A.

0.24 m/s

• B.

0.30 m/s

• C.

0.48 m/s

• D.

0.60 m/s

B. 0.30 m/s
Explanation
When a person walks on a log, the person exerts a force on the log in the opposite direction of their motion. According to Newton's third law, the log exerts an equal and opposite force on the person. Since the person and the log are in contact, their accelerations are the same. Using the equation F = ma, where F is the force exerted by the person on the log and m is the mass of the log, we can calculate the acceleration of the log. Then, using the equation v = u + at, where v is the final velocity of the log, u is the initial velocity (0 m/s), a is the acceleration, and t is the time taken, we can find the speed of the log. In this case, the speed of the log is 0.30 m/s.

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• Sep 19, 2012
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