AP Physics 1 Quiz
Practice Conservation Of Linear Momentum in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Two carts on a frictionless track are connected by a compressed spring. Cart 1 has mass 0.40kg and Cart 2 has mass 0.60kg. The system is both carts, and external forces are negligible. When released, Cart 1 moves left at 3.0m/s. What is Cart 2’s velocity?
This quiz focuses on Conservation Of Linear Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.
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Two carts on a frictionless track are connected by a compressed spring. Cart 1 has mass 0.40kg and Cart 2 has mass 0.60kg. The system is both carts, and external forces are negligible. When released, Cart 1 moves left at 3.0m/s. What is Cart 2’s velocity?
Explanation: This question assesses the conservation of linear momentum in an explosion-like release of a compressed spring. In a system where external forces are negligible, the total linear momentum remains constant and is zero if the system starts at rest. When Cart 1 moves left at 3.0 m/s, its momentum is -1.2 kg·m/s (right as positive). To conserve zero total momentum, Cart 2 must have +1.2 kg·m/s, resulting in 2.0 m/s right. Option B, 3.0 m/s right, could stem from assuming equal speeds without considering mass ratios. A transferable strategy is to assume initial momentum is zero for stationary systems and solve for opposite momenta in explosions or separations.
Two carts collide on a frictionless track. Cart X (2.0kg) moves east at 1.5m/s and Cart Y (1.0kg) moves west at 0.50m/s. The system is both carts; external forces are negligible. What is the system’s total momentum after the collision?
Explanation: This question assesses the conservation of linear momentum in a collision between two carts. In a system where external forces are negligible, such as on a frictionless track, the total linear momentum remains constant before and after the collision. The initial momentum is 2.0 kg × 1.5 m/s east plus 1.0 kg × (-0.50 m/s) west, totaling +2.5 kg·m/s. This value is conserved, so the total momentum after is also +2.5 kg·m/s. Option D, 0 kg·m/s, might be selected if one ignores the directions and assumes cancellation. A transferable strategy is to calculate the net initial momentum considering directions and recognize it persists unchanged in isolated systems.
Two skaters push off each other on level ice. Skater A has mass 50kg and is initially at rest; Skater B has mass 70kg and is initially at rest. Define the system as both skaters; external forces are negligible. After pushing off, Skater A moves west at 2.8m/s. What is Skater B’s velocity?
Explanation: This question assesses the conservation of linear momentum during an interaction where two objects push off each other. In a system where external forces are negligible, such as on level ice, the total linear momentum of the system remains constant and is zero since both skaters start at rest. When Skater A moves west at 2.8 m/s, their momentum is -140 kg·m/s (taking east as positive). To conserve momentum, Skater B must have +140 kg·m/s, resulting in a velocity of 2.0 m/s east. Option A, 2.8 m/s east, might be selected if one assumes equal speeds regardless of mass differences. A transferable strategy is to define a consistent direction for positive momentum and ensure the vector sum remains constant for the isolated system.
On a frictionless horizontal track, a 0.60kg cart moving right at 2.0m/s collides and sticks to a 0.40kg cart initially at rest. Treat the two carts as the system; external forces on the system are negligible during the collision. After the collision, the carts move together. What is the speed of the combined carts immediately after the collision?
Explanation: This question assesses the skill of applying conservation of linear momentum to an inelastic collision between two carts. The total system momentum remains constant because external forces are negligible during the collision on the frictionless track. The initial momentum is (0.60 kg)(2.0 m/s) + (0.40 kg)(0 m/s) = 1.2 kg·m/s to the right. After colliding and sticking, the final momentum is (1.0 kg) v, so v = 1.2 m/s to the right. A common distractor like 2.0 m/s might come from ignoring the mass of the second cart and using only the initial velocity. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
A 0.25kg ball moving right at 8.0m/s collides and sticks to a 0.75kg cart initially at rest on a nearly frictionless track. Consider the ball+cart as the system; external forces are negligible during impact. What is the velocity of the ball-cart system immediately after the collision?
Explanation: This question assesses the skill of applying conservation of linear momentum to an inelastic collision between a ball and a cart. The total system momentum remains constant because external forces are negligible during the impact on a nearly frictionless track. The initial momentum is (0.25 kg)(8.0 m/s) + (0.75 kg)(0 m/s) = 2.0 kg·m/s to the right. After sticking, the final momentum is (1.0 kg) v, so v = 2.0 m/s to the right. A common distractor like 8.0 m/s right might come from using only the ball's initial velocity. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
Two ice skaters initially at rest push off each other on level ice. Skater A has mass 50kg and skater B has mass 75kg. Define the system as both skaters together; external forces on the system are negligible during the push. Afterward, skater A moves left at 3.0m/s. What is skater B’s velocity?
Explanation: This question assesses the skill of applying conservation of linear momentum to two skaters pushing off each other. The total system momentum remains constant because external forces are negligible during the push on level ice. The initial momentum is zero since both skaters are at rest. After pushing, the momentum of skater A is (50 kg)(-3.0 m/s) = -150 kg·m/s, so skater B's momentum must be +150 kg·m/s, giving v = 2.0 m/s to the right. A common distractor like 3.0 m/s right might come from assuming equal velocities without considering the mass difference. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
A 0.20kg cart moving right at 3.0m/s collides and sticks to a 0.30kg cart initially at rest on a level track. Treat the two carts as the system, and assume external forces on the system are negligible during the collision. What is the speed of the stuck-together carts immediately after the collision?
Explanation: This problem requires applying conservation of linear momentum to a perfectly inelastic collision. When external forces are negligible, the total momentum of a system remains constant before and after any interaction. Initially, the 0.20 kg cart has momentum (0.20 kg)(3.0 m/s) = 0.60 kg·m/s to the right, while the 0.30 kg cart at rest has zero momentum, giving a total initial momentum of 0.60 kg·m/s. After the collision, the combined mass of 0.50 kg moves with velocity v, so 0.50v = 0.60, yielding v = 1.2 m/s. Choice A (0.60 m/s) incorrectly divides the initial momentum by the combined mass without proper unit analysis. To solve momentum conservation problems, always write the equation p_initial = p_final and solve for the unknown velocity.
A 3.0kg cart moving right at 2.0m/s collides and sticks to a 1.0kg cart initially at rest on a frictionless track. The system is both carts, and external forces are negligible during the collision. What is the final velocity of the combined carts?
Explanation: This problem demonstrates conservation of linear momentum in a perfectly inelastic collision with one object initially at rest. When external forces are negligible, the total momentum of the system remains constant before and after the collision. Initially, only the 3.0 kg cart moves: p_initial = (3.0 kg)(2.0 m/s) + (1.0 kg)(0 m/s) = 6.0 kg·m/s rightward. After sticking together, the combined mass is 4.0 kg, so p_final = (4.0 kg)v_final. Choice D (6.0 m/s) incorrectly uses the momentum value as velocity. Setting p_initial = p_final: 6.0 = 4.0v_final, giving v_final = 1.5 m/s rightward. To find the final velocity in sticky collisions, divide the initial total momentum by the combined final mass.
Two identical carts on a frictionless track interact via a compressed spring. Initially, both carts are at rest and the spring is compressed between them; the system is the two carts (ignore the spring’s mass). External forces on the system are negligible. After release, cart 1 moves left at 3.0m/s. What is the velocity of cart 2?
Explanation: This problem demonstrates conservation of linear momentum in an explosion-type interaction. When external forces on a system are negligible, the total momentum remains constant throughout any internal interaction. Initially, both carts are at rest, so p_initial = 0. After the spring releases, the system momentum must still be zero: m₁v₁ + m₂v₂ = 0. Since the carts are identical and cart 1 moves left at 3.0 m/s, we have m(-3.0 m/s) + m(v₂) = 0. Choice A (3.0 m/s left) would give the same direction for both carts, violating momentum conservation. Solving: v₂ = +3.0 m/s (rightward). For explosion problems starting from rest, remember that objects must move in opposite directions with momenta that cancel out.
A 1.0kg cart moving east at 5.0m/s collides head-on and sticks with a 4.0kg cart moving west at 1.0m/s. Define the system as both carts; external forces are negligible. What is the final velocity of the combined carts?
Explanation: This question assesses the conservation of linear momentum in an inelastic head-on collision. In a system where external forces are negligible, the total linear momentum remains constant, with initial equaling final. The initial momentum is 1.0 kg × 5.0 m/s east plus 4.0 kg × (-1.0 m/s) west, totaling +1.0 kg·m/s east. After sticking, the 5.0 kg combined mass has velocity 1.0 / 5.0 = 0.20 m/s east. Option B, 0.20 m/s west, might result from switching the direction signs in the calculation. A transferable strategy is to assign consistent signs for directions and verify the net momentum direction matches the initial calculation.
A 0.20kg puck moving east at 4.0m/s collides and sticks to a 0.30kg puck initially at rest on frictionless ice. Treat the two pucks as the system, and assume external forces are negligible during the collision. Afterward, the stuck-together pucks move in the same direction. What is their final speed?
Explanation: This problem tests conservation of linear momentum in a perfectly inelastic collision. When external forces are negligible, the total momentum of a system remains constant before and after any interaction. Initially, the system has momentum from only the moving puck: p_initial = (0.20 kg)(4.0 m/s) + (0.30 kg)(0 m/s) = 0.80 kg·m/s eastward. After the collision, both pucks stick together with combined mass 0.50 kg, so p_final = (0.50 kg)v_final must equal 0.80 kg·m/s. Choice A (0.80 m/s) incorrectly uses the momentum value as the velocity. Solving for v_final: v_final = 0.80 kg·m/s ÷ 0.50 kg = 1.6 m/s. To solve momentum conservation problems, always write p_initial = p_final and carefully track the masses before and after the collision.
On frictionless ice, a 2.0kg cart moving east at 3.0m/s collides and sticks to a 1.0kg cart initially at rest. Treat the two carts as the system; external forces are negligible during the collision. What is the final velocity of the stuck-together carts?
Explanation: This question assesses the conservation of linear momentum in an inelastic collision. In a system where external forces are negligible, such as on frictionless ice, the total linear momentum of the system remains constant before and after the collision. The initial momentum is calculated as the 2.0 kg cart's momentum of 6.0 kg·m/s east, since the 1.0 kg cart is at rest. After the carts stick together, their combined mass is 3.0 kg, so the final velocity is found by dividing the conserved momentum by the total mass, yielding 2.0 m/s east. Option A, 1.0 m/s east, might result from incorrectly averaging the velocities instead of conserving momentum. A transferable strategy is to always calculate the total initial momentum of the system and set it equal to the total final momentum to solve for unknown velocities.
A 0.20kg puck slides east at 4.0m/s and makes a head-on elastic collision with a 0.30kg puck sliding west at 2.0m/s. Consider the system of both pucks; external forces are negligible. What is the system’s total momentum after the collision?
Explanation: This question assesses the conservation of linear momentum in an elastic collision. In a system where external forces are negligible, the total linear momentum of the system remains constant, equal before and after the collision. The initial momentum is 0.20 kg × 4.0 m/s east plus 0.30 kg × (-2.0 m/s) west, totaling +0.20 kg·m/s. Since momentum is conserved regardless of the collision type, the total momentum after is also +0.20 kg·m/s. Option B, 0 kg·m/s, might arise from mistakenly assuming the momenta cancel out completely without calculation. A transferable strategy is to compute the initial total momentum vectorially and recognize it must equal the final total for isolated systems.
A person of mass 60kg stands on a 20kg skateboard at rest on smooth level ground. The person jumps forward so that just after leaving the board, the person’s speed is 3.0m/s relative to the ground. Take the person+skateboard as the system; external horizontal forces are negligible during the jump. What is the skateboard’s speed just after the jump?
Explanation: This question assesses the skill of applying conservation of linear momentum to a person jumping off a skateboard. The total system momentum remains constant because external horizontal forces are negligible during the jump on smooth ground. The initial momentum is zero since the system is at rest. After jumping, the person's momentum is (60 kg)(3.0 m/s) = 180 kg·m/s forward, so the skateboard's momentum must be -180 kg·m/s, giving v = -9.0 m/s (9.0 m/s backward). A common distractor like 3.0 m/s backward might come from using equal velocities without masses. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
A 0.10kg cart moving right at 10m/s collides and sticks to a 0.40kg cart initially at rest. Consider both carts as the system and assume external forces are negligible during the collision. What is the speed of the combined carts immediately after?
Explanation: This problem requires conservation of linear momentum in a perfectly inelastic collision where one object is initially at rest. When external forces on the system are negligible, momentum before collision equals momentum after. Initial momentum is (0.10 kg)(10 m/s) + (0.40 kg)(0 m/s) = 1.0 kg·m/s right. After sticking together, the combined mass of 0.50 kg moves with velocity v, so 0.50v = 1.0, yielding v = 2.0 m/s. Choice C (10 m/s) incorrectly assumes the combined mass maintains the initial velocity of the moving cart. Remember that in inelastic collisions, the final velocity is always less than the initial velocity of the faster object.
On frictionless ice, a 60kg skater initially at rest pushes a 20kg sled so the sled moves east at 3.0m/s. Define the system as skater + sled, and assume external forces on the system are negligible during the push. What is the skater’s velocity after the push?
Explanation: This problem involves conservation of linear momentum during a push between two objects initially at rest. When external forces on a system are negligible, the total momentum before and after an interaction must be equal. The initial momentum is zero since both the skater and sled start at rest. After the push, the sled has momentum (20 kg)(3.0 m/s) = 60 kg·m/s east, so the skater must have momentum 60 kg·m/s west to maintain zero total momentum. With the skater's mass of 60 kg, the velocity is 60 kg·m/s ÷ 60 kg = 1.0 m/s west. Choice B (3.0 m/s west) incorrectly assumes equal speeds rather than equal magnitudes of momentum. Remember that in zero initial momentum problems, the momenta of the parts must be equal and opposite after separation.
A 0.50kg cart moving east at 2.0m/s collides and sticks to a 1.5kg cart moving east at 1.0m/s. Define the system as both carts; external forces are negligible. What is the final speed of the combined carts?
Explanation: This question assesses the conservation of linear momentum in an inelastic collision where both objects move in the same direction. In a system where external forces are negligible, the total linear momentum remains constant, so initial total equals final total. The initial momentum is 0.50 kg × 2.0 m/s plus 1.5 kg × 1.0 m/s, both east, totaling 2.5 kg·m/s east. After sticking, the combined 2.0 kg mass has a velocity of 2.5 / 2.0 = 1.25 m/s east. Option A, 0.75 m/s, could result from incorrectly subtracting the velocities instead of conserving momentum. A transferable strategy is to sum the individual momenta initially and divide by the total mass for inelastic collisions where objects stick together.
Two identical 0.50kg carts on a frictionless track move toward each other: cart A moves right at 1.2m/s and cart B moves left at 0.60m/s. They collide and stick together. Take both carts as the system; external forces are negligible during the collision. What is the velocity of the combined carts after the collision?
Explanation: This question assesses the skill of applying conservation of linear momentum to an inelastic collision between two identical carts. The total system momentum remains constant because external forces are negligible during the collision on the frictionless track. The initial momentum is (0.50 kg)(1.2 m/s) + (0.50 kg)(-0.60 m/s) = 0.30 kg·m/s to the right. After sticking, the final momentum is (1.0 kg) v, so v = 0.30 m/s to the right. A common distractor like 1.2 m/s right might come from ignoring the collision and using one cart's velocity. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.
On a frictionless horizontal track, a 0.50kg cart moving right at 2.0m/s collides and sticks to a 1.0kg cart initially at rest. Define the system as both carts; external forces on the system are negligible during the collision. After they stick, what is their common velocity?
Explanation: This problem tests conservation of linear momentum in a perfectly inelastic collision. When external forces are negligible (as stated), the total momentum of the system before collision equals the total momentum after collision. Initially, the 0.50 kg cart has momentum (0.50 kg)(2.0 m/s) = 1.0 kg⋅m/s to the right, while the 1.0 kg cart has zero momentum. After sticking together, the combined mass of 1.5 kg moves with velocity v, so (1.5 kg)v = 1.0 kg⋅m/s, giving v = 0.67 m/s to the right. Choice C (2.0 m/s) incorrectly assumes the stuck carts maintain the original cart's speed. To solve momentum conservation problems: identify the system, confirm external forces are negligible, then set initial total momentum equal to final total momentum.
A 0.10kg glider moving left at 5.0m/s collides with a 0.40kg glider initially at rest on an air track. After the collision, the 0.10kg glider moves right at 1.0m/s. Define the system as both gliders; external forces are negligible. What is the 0.40kg glider’s velocity after the collision?
Explanation: This question assesses the skill of applying conservation of linear momentum to a collision between two gliders. The total system momentum remains constant because external forces are negligible on the air track. The initial momentum is (0.10 kg)(-5.0 m/s) + (0.40 kg)(0 m/s) = -0.50 kg·m/s. After collision, with the 0.10 kg glider at +1.0 m/s, its momentum is (0.10 kg)(1.0 m/s) = 0.10 kg·m/s, so the 0.40 kg glider's momentum must be -0.60 kg·m/s, giving v = -1.5 m/s. A common distractor like +1.0 m/s might come from assuming symmetric velocities. A transferable strategy is to define a positive direction, calculate the total initial momentum, set it equal to the total final momentum, and solve for unknowns.