Momentum · quiz worked solutions

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Elastic Collisions · question 1

Back to the question · AP Topic 4.3

Question: For two particles with negligible net external impulse, which expression writes the final total momentum directly in terms of their final velocities?

Answer: m₁v₁,f + m₂v₂,f.

1. Choose both particles as the system. Why: The total momentum is the sum of particle momenta.

2. Use each mass times its final velocity, then add the vectors. Why: Negligible external impulse makes this equal to the initial total, but the question asks for the final-velocity expression.

Elastic Collisions · question 2

Back to the question · AP Topic 4.4

Question: Two equal-mass particles undergo a one-dimensional head-on elastic collision. Particle 2 is initially stationary. What are the final velocities?

Answer: Particle 1 stops; particle 2 takes particle 1’s initial velocity.

1. Use conservation of momentum and kinetic energy for a head-on elastic collision. Why: Both conservation rules apply, unlike in a sticking collision.

2. For equal masses and a stationary second particle, those equations exchange velocities. Why: The moving particle transfers its motion to the initially resting one.

Elastic Collisions · question 3

Back to the question · AP Topic 4.4

Question: Ball A moving in +x undergoes a one-dimensional elastic collision with initially stationary ball B. In the position–time diagram, line b continues the original velocity and line d has its negative. Which section or line describes A after collision when m_A > m_B?

Answer: Region II only.

1. Use v_A,f = [(m_A − m_B)/(m_A + m_B)]v_A,i for stationary B. Why: This follows from momentum and kinetic-energy conservation.

2. For m_A > m_B, the coefficient lies between 0 and 1. Why: A keeps moving forward with a smaller positive position–time slope, placing it in region II.

Elastic Collisions · question 4

Back to the question · AP Topic 4.4

Question: Ball A moving in +x undergoes a one-dimensional elastic collision with initially stationary ball B. In the position–time diagram, line b continues the original velocity and line d has its negative. Which section or line describes A after collision when m_A = m_B?

Answer: Line c.

1. Set m_A = m_B in v_A,f = [(m_A − m_B)/(m_A + m_B)]v_A,i. Why: Equal masses exchange velocities in this one-dimensional elastic case.

2. Then v_A,f = 0, represented by the horizontal line c. Why: Zero slope on a position–time graph means rest.

Elastic Collisions · question 5

Back to the question · AP Topic 4.4

Question: Ball A moving in +x undergoes a one-dimensional elastic collision with initially stationary ball B. In the position–time diagram, line b continues the original velocity and line d has its negative. Which section or line describes A after collision when 0 < m_A < m_B?

Answer: Region III only.

1. For 0 < m_A < m_B, the coefficient (m_A − m_B)/(m_A + m_B) lies between −1 and 0. Why: The lighter incoming ball rebounds.

2. Choose the negative slope with magnitude smaller than the original slope. Why: That is region III, not the full −v_A,i boundary line d.

Inelastic Collisions · question 1

Back to the question · AP Topic 4.4

Question: A 100 kg ice hockey goalie catches a 160 g puck moving at 45 m / s . With what speed does the goalie return back, if no energy is lost to friction?

Answer: 0.072 m/s in the puck’s original direction.

1. Choose puck plus goalie and conserve horizontal momentum. Why: The catch is brief with negligible external horizontal impulse; kinetic energy need not be conserved.

2. Find v_f = (0.160·45)/(100 + 0.160) ≈ 0.07189 m/s. Why: They move together after the catch.

Inelastic Collisions · question 2

Back to the question · AP Topic 4.4

Question: A driver attempts to slow a runaway truck by getting in front of it and braking. If the driver is going 20 m / s and the truck is going 45 m / s , how much slower does the 23 Mg truck go once it hits the 2.3 Mg car in front of it, before braking? Assume they stick together in a brief collision with negligible external horizontal impulse.

Answer: The truck slows by 2.27 m/s.

1. Treat the collision as sticking before braking and conserve momentum. Why: The two initial velocities point in the same direction.

2. Find v_f = (23·45 + 2.3·20)/(23 + 2.3) = 42.727 m/s, so 45 − v_f = 2.273 m/s. Why: Using Mg throughout is fine because the common mass unit cancels.

Inelastic Collisions · question 3

Back to the question · AP Topic 4.4

Question: If a 5 g BB shot at 50 m / s hits a 15 g can, and sticks in it, how fast does the can go flying?

Answer: 12.5 m/s.

1. Use the BB and can as a system with negligible external impulse. Why: The embedded BB makes their final velocity common.

2. Find v_f = (5·50)/(5 + 15) = 12.5 m/s. Why: The gram factors cancel because every mass uses the same unit.

Inelastic Collisions · question 4

Back to the question · AP Topic 4.4

Question: How much energy is lost when a 50 g bullet collides with and sticks in a 10 kg block of ballistics gel, if the two move afterwards with a velocity of 0.5 m / s ? The gel block is initially at rest.

Answer: 251.24 J of kinetic energy lost.

1. First conserve momentum: (0.050)v_i = (10.050)(0.5), so v_i = 100.5 m/s. Why: Momentum finds the bullet’s initial speed even though kinetic energy is lost.

2. Compute K_i = ½(0.050)(100.5²) = 252.506 J and K_f = ½(10.050)(0.5²) = 1.256 J. Why: Both the bullet and gel contribute to final kinetic energy.

3. Subtract K_i − K_f = 251.25 J, about the keyed 251.24 J. Why: The lost kinetic energy becomes internal energy and deformation, not destroyed energy.

Inelastic Collisions · question 5

Back to the question · AP Topic 4.4

Question: A small child runs and jumps onto a swing. If the 15 kg child is moving at 8 m / s and leaps onto the 4 kg swing, how fast does the child start to swing?

Answer: 6.3 m/s.

1. Take the child and swing as the collision system. Why: The brief catch preserves tangential momentum if external tangential impulse is negligible.

2. Find v_f = (15·8)/(15 + 4) ≈ 6.316 m/s. Why: The added swing mass lowers the common speed.

Introduction to Momentum · question 1

Back to the question · AP Topic 4.3

Question: Two ice skaters are holding hands, traveling together at 8 m / s . If one ice skater has a mass of 50 kg , and the other has a mass of 70 kg , and they push off from each other, resulting in the 50 kg skater going 12 m / s , what is the new velocity of the 70 kg ice skater?

Answer: 5.14 m/s in the original travel direction.

1. Choose both skaters and use initial momentum (50 + 70)(8) = 960 kg·m/s. Why: Their push is internal and external horizontal impulse is negligible.

2. Solve 50(12) + 70v = 960, so v = 360/70 = 5.143 m/s. Why: Momentum constrains the second skater even though their kinetic energy may increase through muscular work.

Introduction to Momentum · question 2

Back to the question · AP Topic 4.3

Question: A skater and a 5 kg backpack initially move at 10 m/s. The skater throws the backpack forward at 13 m/s in the road frame and afterward moves at approximately 9.81 m/s. Neglect external horizontal impulse. Which listed value is closest to the skater’s mass?

Answer: 80 kg, nearest listed value.

1. Write (m + 5)(10) = m(9.81) + 5(13). Why: All velocities must be in the same road frame.

2. Solve 0.19m = 15, giving m ≈ 78.95 kg. Why: The given final speed is approximate, so 80 kg is the closest option.

Introduction to Momentum · question 3

Back to the question · AP Topic 4.1

Question: What has a higher magnitude of momentum, a 6 kg cannonball shot at 500 m / s , or a 225 kg motorcycle traveling at 10 m / s ?

Answer: The cannonball.

1. Calculate p_cannon = (6)(500) = 3000 kg·m/s. Why: Momentum magnitude is mass times speed.

2. Calculate p_motorcycle = (225)(10) = 2250 kg·m/s. Why: The cannonball’s much larger speed more than offsets its smaller mass.

Introduction to Momentum · question 4

Back to the question · AP Topic 4.2

Question: Three forces shown have equal magnitude and act separately on the same object for 5 s. Neglect other impulses. Which produces the largest magnitude of velocity change?

Answer: All three give equal velocity-change magnitudes.

1. Each constant force gives impulse magnitude F(5 s). Why: Only the directions differ; their magnitudes and durations match.

2. Use velocity-change magnitude = impulse magnitude/m for the same mass. Why: The velocity changes point differently, but have the same magnitude.

Introduction to Momentum · question 5

Back to the question · AP Topic 4.2

Question: An object’s momentum is the only thing that determines how hard it would hit another object. True or False?

Answer: False.

1. Stopping changes momentum, but force also depends on how long that change takes. Why: Average net force is Δp/Δt.

2. Consider energy, contact time and deformation as well. Why: Equal incoming momentum can produce very different peak forces and damage.

Introduction to Momentum · question 1

Back to the question · AP Topic 4.2

Question: A tennis player hits back a tennis ball. The ball initially moved towards the player at a velocity of 20 m / s , and was hit back with a velocity of 25 m / s . If the ball weighs 50 g , and the racquet was hitting the ball for 0.25 s , what is the force the racquet exerts on the ball?

Answer: 9 N average, away from the player.

1. Choose away from the player positive: v_i = −20 and v_f = +25 m/s. Why: The reversal means adding the speed magnitudes in Δv.

2. Find F_avg = (0.050)(25 − (−20))/0.25 = 9 N. Why: This is the average impulse-producing force in the simplified contact model.

Introduction to Momentum · question 2

Back to the question · AP Topic 4.2

Question: A baseball bat hits a 40 m / s pitch and sends the ball flying away at 30 m / s . If the collision between the bat and the 140 g ball lasted 0.05 seconds, what force does the bat exert on the ball?

Answer: 196 N average, in the outgoing direction.

1. Use v_i = −40 and v_f = +30 m/s with outgoing positive. Why: A reversal requires a 70 m/s velocity change.

2. Calculate F_avg = (0.140)(70)/0.05 = 196 N. Why: The force is an average over the contact, not its peak value.

Introduction to Momentum · question 3

Back to the question · AP Topic 4.2

Question: If a racquetball ( 40 g ) is hit against a wall with an initial speed of 30 m / s , and it collides with the wall for 0.2 s , what is its return speed if the wall exerts a 10 N force?

Answer: 20 m/s away from the wall.

1. Choose away from the wall positive, so p_i = (0.040)(−30) = −1.2 kg·m/s. Why: The initial and return velocities have opposite signs.

2. Add impulse J = (10)(0.2) = 2 kg·m/s to get p_f = 0.8 kg·m/s and v_f = 20 m/s. Why: The impulse must first cancel the incoming momentum before producing rebound.

Introduction to Momentum · question 4

Back to the question · AP Topic 4.2

Question: A 70 kg skydiver lands on cardboard boxes at 10 m/s downward and stops in 2 s. What is the magnitude of the average NET vertical force during stopping?

Answer: 350 N upward net force.

1. Take upward positive: Δv = 0 − (−10) = 10 m/s. Why: Stopping downward motion requires upward acceleration.

2. Find F_net,avg = (70)(10)/2 = 350 N. Why: The cardboard contact force is larger because it must also overcome weight; the question asks specifically for net force.

Introduction to Momentum · question 5

Back to the question · AP Topic 4.2

Question: If a Tee Ball player swings at a stationary ball with a force of 50 N , and the bat strikes the ball for 0.2 s , how fast does the 50 g ball leave the tee?

Answer: 200 m/s.

1. Calculate J = (50)(0.2) = 10 N·s. Why: The stipulated average force acts over the whole contact interval.

2. From rest, v_f = J/m = 10/0.050 = 200 m/s. Why: This is the idealized model’s result, even though an actual tee-ball hit would not use these numbers.

FFR - Kinetic Energy · question 1

Back to the question · AP Topic 3.1

Question: How much energy is contained in a 40 m / s fastball, if a baseball has a mass of 145 g ?

Answer: 116 J.

1. Convert 145 g to 0.145 kg. Why: SI energy requires kilograms.

2. Calculate K = ½(0.145)(40²) = 116 J. Why: The speed is squared, so substituting 40 only once would miss most of the energy.

FFR - Kinetic Energy · question 2

Back to the question · AP Topic 4.4

Question: A completely inelastic collision occurs between two boats. The larger boat has a mass of 5.9 Mg , and is initially moving at 4 m / s before hitting the smaller, 800 kg boat, which is initially at rest. What is the combined kinetic energy of the two boats after colliding?

Answer: 41.6 kJ.

1. Conserve the boats’ momentum: v_f = (5900·4)/(5900 + 800) ≈ 3.5224 m/s. Why: They stick together, so share the final velocity.

2. Calculate K_f = ½(6700)(3.5224²) ≈ 41,564 J. Why: Using the combined mass gives 41.6 kJ; kinetic energy is not conserved in the sticking event.

FFR - Kinetic Energy · question 3

Back to the question · AP Topic 4.4

Question: A 5 kg bowling ball moving at 5 m/s collides head-on and elastically with an initially stationary 1.53 kg pin. Treat them as particles and ignore spin. What is the pin’s kinetic energy afterward?

Answer: 44.9 J.

1. Use the head-on elastic result v_pin = (2 × 5/(5 + 1.53)) × 5 ≈ 7.657 m/s. Why: The initially stationary pin gains speed under both momentum and energy conservation.

2. Calculate K_pin = ½(1.53)(7.657²) ≈ 44.85 J. Why: The question explicitly ignores rotational energy, so only translation counts.

FFR - Kinetic Energy · question 4

Back to the question · AP Topic 3.2

Question: The source omits the box’s mass and initial speed. For an explicitly repaired classroom version, assume a 20 kg box starts from rest on a frictionless horizontal surface and a 5 N horizontal force pushes it through 10 m. What is its speed when the push ends?

Answer: 2.24 m/s.

1. Find net work W = (5)(10) = 50 J on the frictionless level surface. Why: Weight and normal are perpendicular to the displacement.

2. Use ½(20)v² − 0 = 50, giving v = √5 ≈ 2.236 m/s. Why: The repaired mass and initial-rest condition are necessary for a unique answer.

FFR - Kinetic Energy · question 5

Back to the question · AP Topic 3.1

Question: An asteroid collides with a planet, releasing an exorbitant amount of energy in doing so. If the asteroid weighs 8.2 Zg ( 8.2 ⋅ 10 18 kg ) and is moving at a rate of 11 km / s , and 99.999 % of its energy is lost during approach, what energy is contained in the impact (after the energy loss)?

Answer: 4.961 ZJ.

1. Convert 11 km/s to 11,000 m/s and compute K_i = ½(8.2×10¹⁸)(11000²) = 4.961×10²⁶ J. Why: Energy uses squared SI speed.

2. Retain 0.001% = 10⁻⁵ of that value, giving 4.961×10²¹ J = 4.961 ZJ. Why: Losing 99.999% does not mean retaining 0.001 as a decimal fraction.

Model references

These original worked steps use the stated question models. Physics references: Momentum conservation, Collisions. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.