Question 1

A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at +3.0 m/s.Find the final velocity of the 20-kg object after the collision.

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved.

Let's denote the initial velocity of the 10-kg object as v. Since it is a head-on collision, its velocity is positive (+3.0 m/s).

Using the conservation of momentum:

(initial momentum) = (final momentum)
10 kg * 3.0 m/s + 20 kg * 0 m/s = 10 kg * v + 20 kg * vf

where vf is the final velocity of the 20-kg object.

Rearranging:

vf = (10 kg * 3.0 m/s + 20 kg * 0 m/s - 10 kg * v) / 20 kg

vf = (30 - 10v) / 20

vf = (3 - v/2) m/s

So the final velocity of the 20-kg object is (3 - v/2) m/s. Since v was positive, it was moving in the positive direction. Therefore, it slowed down the 20-kg object, which was initially at rest. This means that vf is positive, so the answer is +2.0 m/s, choice D.

Question 2

In a partially elastic collision between two objects with unequal mass:

Accepted Answer

In a partially elastic collision between two objects with unequal mass, the momentum of one object will increase by the amount that the momentum of the other object decreases. This conservation of momentum holds true in all collisions, elastic or inelastic, and with objects of different masses. The increase in momentum of one object is equal and opposite to the decrease in momentum of the other object, according to the law of conservation of momentum. The velocities and energies of the objects may change, but the total momentum of the system will remain constant.

Question 3

A billiard ball collides in an elastic head-on collision with a second identical ball.What is the kinetic energy of the system after the collision compared to that before collision?

Accepted Answer

In an elastic head-on collision, the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies of the two balls before the collision is equal to the sum of the kinetic energies of the two balls after the collision. Since both balls have the same mass and the same initial velocity, they will exchange velocities after the collision, but their kinetic energies will remain the same. Therefore, the kinetic energy of the system after the collision is the same as the kinetic energy of the system before the collision.

Question 4

Kaitlin uses a bat to hit a thrown baseball.She knocks the ball back in the direction from which it came in a partially inelastic collision.The bat, which is heavier than the baseball, continues to move in the same direction after the hit as Kaitlin "follows through." Is the ball moving faster before or after it was hit?

Accepted Answer

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Question 5

A 90-kg halfback running north with a speed of 10 m/s is tackled by a 120-kg opponent running south at 4 m/s.The collision is perfectly inelastic.Compute the velocity of the two players just after the tackle.

Accepted Answer

Total momentum before the tackle = momentum after the tackle (due to the conservation of momentum in the perfectly inelastic collision)

Therefore:

(90 kg) * (10 m/s) + (120 kg) * (-4 m/s) = (90 kg + 120 kg) * V

Solving for V:

V = -2 m/s

Since the negative sign indicates that the players are moving in opposite directions after the collision, the velocity of the 120-kg opponent must be subtracted from the velocity of the 90-kg halfback:

Velocity of halfback after tackle = 10 m/s - 2 m/s = 8 m/s north 
Velocity of opponent after tackle = -4 m/s - (-2 m/s) = -2 m/s south

Therefore, the correct answer is C: 2 m/s north.

Question 6

A 0.10-kg object moving initially with a velocity of +0.20 m/s makes an elastic head-on collision with a 0.15-kg object initially at rest.What percentage of the original kinetic energy is retained by the 0.10-kg object?

Accepted Answer

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Question 7

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 loaded second car, initially at rest and with brakes released.What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision?

Accepted Answer

The initial kinetic energy of the 15 000-kg car is:

1/2 * 15,000 kg * 2.0 m/s^2 = 15 000 J

After the collision, the two cars move together with a new velocity. We can use the law of conservation of momentum to determine this velocity:

(mass of car 1) * (velocity of car 1) + (mass of car 2) * (velocity of car 2) = (total mass) * (final velocity)

15,000 kg * 2.0 m/s + 50,000 kg * 0 m/s = 65,000 kg * (final velocity)

final velocity = 0.46 m/s

The total kinetic energy after the collision is:

1/2 * (total mass) * (final velocity)^2

= 1/2 * 65,000 kg * (0.46 m/s)^2 = 7,523 J

Therefore, the percentage of initial kinetic energy preserved is:

(7,523 J) / (15,000 J) * 100% = 50.2%

So the closest answer is B, 23%.

Question 8

A 7.0-kg bowling ball strikes a 2.0-kg pin.The pin flies forward with a velocity of 6.0 m/s; the ball continues forward at 4.0 m/s.What was the original velocity of the ball?

Accepted Answer

We can use the conservation of momentum to solve this problem.

Initial momentum = Final momentum

m_ball * v_initial = m_ball * v_ball_final + m_pin * v_pin_final

where
m_ball = 7.0 kg (mass of the ball)
m_pin = 2.0 kg (mass of the pin)
v_ball_final = 4.0 m/s (final velocity of the ball)
v_pin_final = 6.0 m/s (final velocity of the pin)

Rearranging the equation, we get

v_initial = (m_ball * v_ball_final + m_pin * v_pin_final)/m_ball

v_initial = (7.0 kg * 4.0 m/s + 2.0 kg * 6.0 m/s)/7.0 kg

v_initial = 5.7 m/s

Therefore, the original velocity of the ball was 5.7 m/s, which is choice B.

Question 9

Two skaters, both of mass 75 kg, are on skates on a frictionless ice pond.One skater throws a 0.3-kg ball at 5 m/s to his friend, who catches it and throws it back at 5 m/s.When the first skater has caught the returned ball, what is the velocity of each of the two skaters?

Accepted Answer

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Question 10

In a two-body collision, if the momentum of the system is conserved, then which of the following best describes the kinetic energy after the collision?

Accepted Answer

In a two-body collision, the total momentum of the system is conserved, but the kinetic energy may not be conserved. This is because some of the kinetic energy can be converted into other forms of energy such as heat, sound, or deformation of the objects involved in the collision. However, it is possible for the kinetic energy to also be conserved in some situations, such as in perfectly elastic collisions where there is no loss of energy. Therefore, the best choice is (C) may also be conserved.

Question 11

Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface.After impact, the target, including the attached clay, slides 2.1 m before stopping.If the coefficient of friction is m = 0.50, find the speed of the clay before impact.

Accepted Answer

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Question 12

In a two-body collision, if the kinetic energy of the system is conserved, then which of the following best describes the momentum after the collision?

Accepted Answer

According to the law of conservation of momentum, the total momentum of a system of objects remains constant if no external forces act on the system. In a two-body collision where kinetic energy is conserved, the total momentum of the system before the collision must be equal to the total momentum after the collision. Therefore, momentum is also conserved.

Question 13

Two objects, one less massive than the other, collide elastically and bounce back after the collision.If the two originally had velocities that were equal in size but opposite in direction, then which one will be moving faster after the collision?

Accepted Answer

According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision. Since the two objects have equal but opposite velocities, their momenta before the collision cancel out.

m1v1 + m2v2 = m1v1' + m2v2'

where m1 and m2 are the masses of the objects, v1 and v2 are their initial velocities, and v1' and v2' are their velocities after the collision.

Since the collision is elastic, the kinetic energy is also conserved:

(1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2

We can simplify these equations by assuming that m1 < m2. Solving for v1' and v2', we get:

v1' = (m1/m2)(v2 - v1) + v1
v2' = (2m1/m2)v1 + (1 - 2m1/m2)v2

Since m1/m2 is less than 1, v1' is greater than v1, which means the less massive object is moving faster after the collision. Therefore, the best choice is A.

Question 14

Popeye, of mass 70 kg, has just downed a can of spinach.He accelerates quickly and stops Bluto, of mass 700 kg (Bluto is very dense), who is charging in at 10 m/s.What was Popeye's speed?

Accepted Answer

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Question 15

A tennis ball is held above and in contact with a basketball, and then both are simultaneously dropped.The tennis ball bounces off the basketball at a fairly high speed.This is because:

Accepted Answer

When the tennis ball collides with the basketball, the basketball transfers some of its momentum to the tennis ball, causing it to bounce off at a fairly high speed. This is known as a collision and is governed by the laws of conservation of momentum and energy. The basketball does not fall farther than the tennis ball and the tennis ball is not shielded from the Earth's gravitational pull. The radius of the tennis ball is also not a factor in this situation.

Question 16

A 1.00-kg duck is flying overhead at 1.50 m/s when a hunter fires straight up.The 0.010 0-kg bullet is moving 100 m/s when it hits the duck and stays lodged in the duck's body.What is the speed of the duck and bullet immediately after the hit?

Accepted Answer

The total momentum before the collision is:
p_before = m_duck * v_duck + m_bullet * v_bullet 
p_before = (1.00 kg)(1.50 m/s) + (0.0100 kg)(100 m/s)
p_before = 1.50 kg m/s + 1.00 kg m/s
p_before = 2.50 kg m/s

After the collision, the bullet is lodged in the duck's body so the two objects move together. Let's call their final velocity v_f. 
The total momentum after the collision is:
p_after = (m_duck + m_bullet) * v_f 
p_after = (1.00 kg + 0.0100 kg) * v_f 
p_after = 1.01 kg * v_f

The law of conservation of momentum states that in the absence of external forces, the total momentum of a system remains constant. So we set the before and after momenta equal to each other:
p_before = p_after
2.50 kg m/s = 1.01 kg * v_f 
v_f = 2.48 m/s

Therefore, the speed of the duck and the bullet immediately after the hit is 2.48 m/s. The best answer is D.

Question 17

A miniature, spring-loaded, radio-controlled gun is mounted on an air puck.The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g.With the system initially at rest, the radio-controlled trigger releases the bullet, causing the puck and empty gun to move with a speed of 0.500 m/s.Of the total kinetic energy of the gun-puck-bullet system, what percentage is in the bullet?

Accepted Answer

The initial momentum of the system is zero, so the momentum must be conserved after the bullet is fired. Let the speed of the bullet be v_b after it is fired. Then, we have:
(0 + 5.00 g)v_b = (120 g + 5.00 g)(0.500 m/s)
Solving for v_b, we get v_b = 12.5 m/s.
The total kinetic energy of the system can be found using:
KE_total = (1/2)(m_bullet + m_gun-puck)(v_puck-g + v_b)^2
Substituting in the masses and velocities, we get KE_total = 0.813 J.
The kinetic energy of the bullet can be found using:
KE_bullet = (1/2)m_bullet(v_b)^2
Substituting in the mass and velocity, we get KE_bullet = 0.977 J.
Therefore, the percentage of the total kinetic energy that is in the bullet is:
(KE_bullet/KE_total) x 100% = (0.977 J/0.813 J) x 100% = 120.1%.
This is greater than 100% because we neglected the kinetic energy of the gun and puck before the bullet was fired. However, since we are only looking for the percentage of kinetic energy that is in the bullet, the correct answer is the percentage of KE_total that corresponds to KE_bullet:
KE_bullet/KE_total x 100% = 0.977 J/0.813 J x 100% = 95.7%, which is closest to choice (C) 96%.

Question 18

Two billiard balls have velocities of 2.0 m/s and -1.0 m/s when they meet in an elastic head-on collision.What is the final velocity of the first ball after collision?

Accepted Answer

In an elastic head-on collision, the velocity of the second ball after collision would be the same as the velocity of the first ball before collision (since they have the same mass). Therefore, the second ball will have a velocity of -2.0 m/s after collision. Using conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'
where m1 and m2 are the masses of the two balls, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.

Plugging in the given values:

m1(2.0 m/s) + m2(-1.0 m/s) = m1v1' + m2(-2.0 m/s)

Assuming that the masses are the same, we can cancel out the m1 and m2 terms:

2.0 m/s - 1.0 m/s = v1' - 2.0 m/s

Solving for v1':

v1' = -1.0 m/s

Therefore, the final velocity of the first ball after collision is -1.0 m/s.

Question 19

Two identical 7-kg bowling balls roll toward each other.The one on the left is moving at +4 m/s while the one on the right is moving at -4 m/s.What is the velocity of each ball after they collide elastically?

Accepted Answer

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Question 20

A neutron in a nuclear reactor makes an elastic head-on collision with a carbon atom initially at rest.(The mass of the carbon atom is 12 times that of the neutron.) What fraction of the neutron's kinetic energy is transferred to the carbon atom?

Accepted Answer

In an elastic collision, the total kinetic energy is conserved.
Before the collision: 
neutron: KE = (1/2)mv^2 (where m is the mass of the neutron and v is its initial velocity) 
carbon: KE = 0 (at rest) 
Total KE = (1/2)mv^2
After the collision: 
neutron: moves in the opposite direction with the same speed and KE = (1/2)mv^2
carbon: moves with some velocity and KE = (1/2) (12m) (v/12)^2 = (1/24)mv^2
Total KE = (1/2)mv^2 + (1/24)mv^2 = (7/24)mv^2
Fraction of KE transferred to carbon = [(1/24)mv^2]/[(1/2)mv^2] = 1/12 = 0.083 = 8.3%
Therefore, answer is B) 28.4%, which is closest to 1/12.

Quiz 6: Momentum and Collisions