AP Physics 1 Quiz: Change In Momentum And Impulse
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Change In Momentum And ImpulseQuestion 1 of 20

A constant force of 3 N3\,\text{N} acts downward on a glider for 2.0 s2.0\,\text{s}. What is the magnitude of the glider's momentum change?

1.5 kg\cdotm/s1.5\,\text{kg\cdot m/s}
6 kg\cdotm/s6\,\text{kg\cdot m/s}
3 kg\cdotm/s3\,\text{kg\cdot m/s}
5 kg\cdotm/s5\,\text{kg\cdot m/s}
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AP Physics 1 Quiz

AP Physics 1 Quiz: Change In Momentum And Impulse

Practice Change In Momentum And Impulse 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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This quiz focuses on Change In Momentum And Impulse, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.

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Question 1

A constant force of 3 N3\,\text{N} acts downward on a glider for 2.0 s2.0\,\text{s}. What is the magnitude of the glider's momentum change?

  1. 1.5 kg\cdotm/s1.5\,\text{kg\cdot m/s}
  2. 6 kg\cdotm/s6\,\text{kg\cdot m/s} (correct answer)
  3. 3 kg\cdotm/s3\,\text{kg\cdot m/s}
  4. 5 kg\cdotm/s5\,\text{kg\cdot m/s}
Explanation: This question assesses magnitude of momentum change from impulse in AP Physics 1. Impulse is constant force times time, representing the total momentum alteration. It equals Δp, with magnitude independent of direction for this query. The 3 N downward for 2.0 s gives a magnitude of 6 kg·m/s. Choice C is a distractor, possibly from dividing instead of multiplying. Apply J = F Δt directly for momentum change magnitudes in constant force situations.

Question 2

A 2.0 kg2.0\,\text{kg} cart is initially at rest. A constant rightward force of 6.0 N6.0\,\text{N} acts for 0.50 s0.50\,\text{s}. What is the cart's final momentum?

  1. 12 kg ⁣⋅ ⁣m/s12\,\text{kg}\!\cdot\!\text{m/s} to the right
  2. 3.0 kg ⁣⋅ ⁣m/s3.0\,\text{kg}\!\cdot\!\text{m/s} to the right (correct answer)
  3. 6.0 N6.0\,\text{N} to the right
  4. 3.0 N ⁣⋅ ⁣s3.0\,\text{N}\!\cdot\!\text{s} to the left
Explanation: This problem requires applying the impulse-momentum theorem to find final momentum. The impulse delivered is J = F·Δt = (6.0 N)(0.50 s) = 3.0 N·s to the right. Since the cart starts from rest (initial momentum = 0), the change in momentum equals the final momentum: Δp = pf - pi = pf - 0 = 3.0 kg·m/s. Therefore, the final momentum is 3.0 kg·m/s to the right. Choice A incorrectly multiplies force by mass instead of time, while choice C gives force instead of momentum. Remember that impulse (N·s) and momentum (kg·m/s) have equivalent units, and for objects starting from rest, final momentum equals the impulse delivered.

Question 3

A tennis racket exerts an average force of 200 N200\,\text{N} on a ball for 0.010 s0.010\,\text{s}. What is the ball's momentum change magnitude?

  1. 200 kg\cdotm/s200\,\text{kg\cdot m/s}
  2. 2.0 kg\cdotm/s2.0\,\text{kg\cdot m/s} (correct answer)
  3. 20 kg\cdotm/s20\,\text{kg\cdot m/s}
  4. 0.010 kg\cdotm/s0.010\,\text{kg\cdot m/s}
Explanation: This question tests impulse and momentum change magnitude in AP Physics 1. Impulse is the average force times contact time, measuring momentum transfer. The theorem states this equals Δp, so magnitude is F Δt. The 200 N for 0.010 s gives a magnitude of 2.0 kg·m/s. Choice A is a distractor, likely from omitting time multiplication. For average force problems, multiply force by time to find momentum change reliably.

Question 4

A constant force of 7 N7\,\text{N} downward acts on a toy for 0.30 s0.30\,\text{s}. What is the toy's momentum change?

  1. 2.1 N ⁣\cdot ⁣s2.1\,\text{N\!\cdot\!s} upward
  2. 7 N7\,\text{N} downward
  3. 0.30 N ⁣\cdot ⁣s0.30\,\text{N\!\cdot\!s} downward
  4. 2.1 N ⁣\cdot ⁣s2.1\,\text{N\!\cdot\!s} downward (correct answer)
Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse measures the effect of force over time, J = F Δt. This is identical to the change in momentum, Δp = m Δv = J. The 7 N downward for 0.30 s leads to a momentum change of 2.1 N·s downward. Choice B provides the force without time, a typical oversight. Always multiply force by time interval to determine Δp effectively.

Question 5

A 1.5 kg1.5\,\text{kg} cart moving right receives a constant impulse of 2.4 N ⁣⋅ ⁣s2.4\,\text{N}\!\cdot\!\text{s} to the left. What is the cart's Δp⃗\Delta \vec p?

  1. +2.4 kg ⁣⋅ ⁣m/s+2.4\,\text{kg}\!\cdot\!\text{m/s} (rightward)
  2. −1.6 kg ⁣⋅ ⁣m/s-1.6\,\text{kg}\!\cdot\!\text{m/s} (leftward)
  3. −2.4 kg ⁣⋅ ⁣m/s-2.4\,\text{kg}\!\cdot\!\text{m/s} (leftward) (correct answer)
  4. −2.4 N-2.4\,\text{N} (leftward)
Explanation: This problem directly states that an impulse is delivered and asks for change in momentum. By the impulse-momentum theorem, change in momentum equals impulse: Δp = J. Since the impulse is 2.4 N·s to the left (opposite the cart's rightward motion), Δp = -2.4 kg·m/s (leftward). The negative sign indicates leftward direction in our coordinate system. Choice A incorrectly uses positive sign for rightward, while choice D gives units of force instead of momentum. Remember that impulse and change in momentum are always equal—they're the same physical quantity expressed in equivalent units.

Question 6

A 1.0 kg1.0\,\text{kg} cart receives a rightward impulse of +0.60 N⋅s+0.60\,\text{N}\cdot\text{s}. If it was initially moving left, what is the direction of Δp⃗\Delta \vec p?

  1. Left, because the cart was moving left initially
  2. Right, because impulse and Δp⃗\Delta \vec p have the same direction (correct answer)
  3. Zero, because impulses cancel initial momentum
  4. Cannot be determined without the cart's speed
Explanation: This problem examines the direction of momentum change in AP Physics 1. Impulse is force over time, dictating the direction of momentum shift. Impulse equals Δp⃗, so directions match regardless of initial velocity. The rightward +0.60 N·s impulse means Δp⃗ is rightward. Choice A is a distractor, confusing initial motion with Δp direction. Focus on impulse direction alone to determine Δp⃗ in impulse-related questions.

Question 7

A ball is initially at rest. A constant force of 6 N6\,\text{N} acts upward on it for 0.20 s0.20\,\text{s}. What is the magnitude of the impulse delivered?

  1. 1.2 N\cdots1.2\,\text{N\cdot s} (correct answer)
  2. 6 N6\,\text{N}
  3. 0.20 s0.20\,\text{s}
  4. 30 N\cdots30\,\text{N\cdot s}
Explanation: This question assesses understanding of impulse and its relation to change in momentum in AP Physics 1. Impulse is the integral of force over time, but for a constant force, it simplifies to force multiplied by time. This impulse is equivalent to the change in momentum of the object, as per the impulse-momentum theorem. Here, the upward force of 6 N6 \, \text{N} for 0.20 s0.20 \, \text{s} results in an impulse magnitude of 1.2 N⋅s1.2 \, \text{N}\cdot\text{s}, independent of the ball's mass since initial velocity is zero. Choice D is a distractor as it might result from multiplying force by time incorrectly or confusing units. Remember to compute impulse directly as F ΔtF \, \Delta t for constant forces to determine momentum changes accurately.

Question 8

A hockey puck experiences a constant force of 5 N5\,\text{N} to the right for 0.40 s0.40\,\text{s}. What is Δp⃗\Delta \vec p?

  1. 2.0 N ⁣\cdot ⁣s2.0\,\text{N\!\cdot\!s} to the right (correct answer)
  2. 5.0 N5.0\,\text{N} to the right
  3. 0.08 N ⁣\cdot ⁣s0.08\,\text{N\!\cdot\!s} to the right
  4. 2.0 N ⁣\cdot ⁣s2.0\,\text{N\!\cdot\!s} to the left
Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse occurs when a force acts over time, quantified as J=FΔtJ = F \Delta t. It directly corresponds to the momentum change, Δp=J\Delta p = J, including the vector direction. The 5 N5 \, \text{N} right for 0.40 s0.40 \, \text{s} results in Δp\Delta p of 2.0 N⋅s2.0 \, \text{N}\cdot\text{s} right for the puck. Choice C might come from dividing force by time instead of multiplying, a calculation error. Consistently use J=FΔt=ΔpJ = F \Delta t = \Delta p to verify answers in impulse problems.

Question 9

A tennis ball experiences a constant force of magnitude 40 N40\ \text{N} to the left for 0.020 s0.020\ \text{s}. What is the magnitude of the impulse on the ball?

  1. 40 N40\ \text{N}
  2. 2.0 N⋅s2.0\ \text{N}\cdot\text{s}
  3. 0.80 N⋅s0.80\ \text{N}\cdot\text{s} (correct answer)
  4. 800 N⋅s800\ \text{N}\cdot\text{s}
Explanation: This problem tests understanding of change in momentum and impulse. Impulse is defined as the product of force and the time interval over which it acts: J = F·Δt. Given F = 40 N and Δt = 0.020 s, the impulse magnitude is J = (40 N)(0.020 s) = 0.80 N·s. The impulse-momentum theorem tells us that impulse equals the change in momentum of the object. Choice A incorrectly gives just the force value without considering time. To find impulse, always multiply the force magnitude by the time duration, ensuring your units are N·s.

Question 10

A 1.5 kg1.5\ \text{kg} block experiences a constant 8 N8\ \text{N} force to the right for 0.25 s0.25\ \text{s}. What is the magnitude of Δp⃗\Delta \vec p?

  1. 32 kg⋅m/s32\ \text{kg}\cdot\text{m/s}
  2. 2.0 kg⋅m/s2.0\ \text{kg}\cdot\text{m/s} (correct answer)
  3. 0.50 kg⋅m/s0.50\ \text{kg}\cdot\text{m/s}
  4. 8 kg⋅m/s8\ \text{kg}\cdot\text{m/s}
Explanation: This problem tests understanding of change in momentum and impulse. The impulse is J = F·Δt = (8 N)(0.25 s) = 2.0 N·s. By the impulse-momentum theorem, the magnitude of the change in momentum equals the magnitude of the impulse: |Δp| = 2.0 kg·m/s. The mass of the block is not needed when calculating impulse from force and time. Choice A incorrectly multiplies all three values (mass, force, and time) together. Remember that change in momentum equals force times time, not mass times force times time.

Question 11

A 0.25 kg0.25\ \text{kg} toy car experiences a constant force of 12 N12\ \text{N} forward for 0.10 s0.10\ \text{s}. What is the car's change in momentum?

  1. 1.2 kg⋅m/s1.2\ \text{kg}\cdot\text{m/s} forward (correct answer)
  2. 12 kg⋅m/s12\ \text{kg}\cdot\text{m/s} forward
  3. 0.30 kg⋅m/s0.30\ \text{kg}\cdot\text{m/s} forward
  4. 1.2 kg⋅m/s1.2\ \text{kg}\cdot\text{m/s} backward
Explanation: This problem tests understanding of change in momentum and impulse. Impulse equals force multiplied by time: J = F·Δt = (12 N)(0.10 s) = 1.2 N·s forward. The impulse-momentum theorem states that this impulse equals the change in momentum, so Δp = 1.2 kg·m/s forward. The mass of the car is not used in this calculation when force and time are given. Choice B incorrectly uses just the force value without considering time. When calculating change in momentum from a constant force, always multiply force by time duration to get impulse.

Question 12

A 2.0 kg2.0\ \text{kg} glider receives an impulse of −4.0 N⋅s-4.0\ \text{N}\cdot\text{s} (left). What is the glider's change in momentum Δp⃗\Delta \vec p?

  1. +4.0 kg⋅m/s+4.0\ \text{kg}\cdot\text{m/s} (right)
  2. −2.0 kg⋅m/s-2.0\ \text{kg}\cdot\text{m/s} (left)
  3. −4.0 kg⋅m/s-4.0\ \text{kg}\cdot\text{m/s} (left) (correct answer)
  4. −8.0 kg⋅m/s-8.0\ \text{kg}\cdot\text{m/s} (left)
Explanation: This problem tests understanding of change in momentum and impulse. The impulse-momentum theorem states that impulse equals change in momentum: J = Δp. Since the glider receives an impulse of -4.0 N·s (left), its change in momentum is exactly Δp = -4.0 kg·m/s (left). The units N·s and kg·m/s are equivalent for momentum. The negative sign indicates leftward direction. Choice B incorrectly divides the impulse by the mass, but impulse already equals the change in momentum directly. When given impulse, it directly equals the change in momentum without any additional calculations.

Question 13

A 0.50 kg0.50\ \text{kg} cart initially moving right at 2.0 m/s2.0\ \text{m/s} experiences a constant leftward force of 3.0 N3.0\ \text{N} for 0.40 s0.40\ \text{s}. What is the cart's change in momentum Δp⃗\Delta \vec p?

  1. +1.2 N⋅s+1.2\ \text{N}\cdot\text{s} (to the right)
  2. −1.2 N⋅s-1.2\ \text{N}\cdot\text{s} (to the left) (correct answer)
  3. −3.0 N⋅s-3.0\ \text{N}\cdot\text{s} (to the left)
  4. −0.75 N⋅s-0.75\ \text{N}\cdot\text{s} (to the left)
Explanation: This problem tests understanding of change in momentum and impulse. Impulse is the product of force and time: J = F·Δt = (-3.0 N)(0.40 s) = -1.2 N·s, where negative indicates leftward direction. The impulse-momentum theorem states that impulse equals change in momentum: J = Δp. Therefore, Δp = -1.2 N·s (to the left). Choice C incorrectly multiplies force by initial velocity instead of time. When calculating impulse, always multiply force by the time duration, not by velocity or mass.

Question 14

A ball experiences a constant upward force of 6 N6\,\text{N} for 0.20 s0.20\,\text{s}. What is the impulse on the ball?

  1. 1.2 N ⁣\cdot ⁣s1.2\,\text{N\!\cdot\!s} upward (correct answer)
  2. 6 N6\,\text{N} upward
  3. 30 N ⁣\cdot ⁣s30\,\text{N\!\cdot\!s} upward
  4. 1.2 N ⁣\cdot ⁣s1.2\,\text{N\!\cdot\!s} downward
Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse represents the force applied over a specific time duration, calculated as J = F Δt. It is equivalent to the change in an object's momentum, Δp = m Δv, but when mass and velocity aren't provided, impulse directly gives Δp. Here, the 6 N upward force for 0.20 s yields an impulse of 1.2 N·s upward. Choice B incorrectly lists just the force without multiplying by time, a frequent mistake. To approach these, compute J = F Δt first and recognize it as Δp for transferable problem-solving.

Question 15

A 0.40 kg0.40\ \text{kg} ball moving right receives a constant leftward force of 2.5 N2.5\ \text{N} for 0.60 s0.60\ \text{s}. What is the impulse on the ball?

  1. 1.5 N⋅s1.5\ \text{N}\cdot\text{s} to the left (correct answer)
  2. 2.5 N2.5\ \text{N} to the left
  3. 1.5 N⋅s1.5\ \text{N}\cdot\text{s} to the right
  4. 0.24 N⋅s0.24\ \text{N}\cdot\text{s} to the left
Explanation: This problem tests understanding of change in momentum and impulse. Impulse is calculated as the product of force and time: J = F·Δt = (2.5 N)(0.60 s) = 1.5 N·s. Since the force is leftward, the impulse is 1.5 N·s to the left. The impulse-momentum theorem tells us this impulse equals the change in momentum of the ball. Choice C incorrectly assigns the wrong direction; since the force is leftward, the impulse must also be leftward. Always ensure the direction of impulse matches the direction of the applied force.

Question 16

A 0.50 kg0.50\,\text{kg} cart initially moves right at 2.0 m/s2.0\,\text{m/s}. A constant leftward force of 3.0 N3.0\,\text{N} acts for 0.40 s0.40\,\text{s}. What is the cart's change in momentum Δp⃗\Delta \vec p?

  1. +1.2 N ⁣⋅ ⁣s+1.2\,\text{N}\!\cdot\!\text{s} (to the right)
  2. −7.5 kg ⁣⋅ ⁣m/s-7.5\,\text{kg}\!\cdot\!\text{m/s} (to the left)
  3. −1.2 N ⁣⋅ ⁣s-1.2\,\text{N}\!\cdot\!\text{s} (to the left) (correct answer)
  4. −3.0 N-3.0\,\text{N} (to the left)
Explanation: This problem tests understanding of change in momentum and impulse. The impulse-momentum theorem states that impulse (J = F·Δt) equals the change in momentum (Δp). Here, a leftward force of 3.0 N acts for 0.40 s, giving impulse J = (-3.0 N)(0.40 s) = -1.2 N·s (negative because force is leftward). Since impulse equals change in momentum, Δp = -1.2 N·s = -1.2 kg·m/s to the left. Choice A incorrectly uses positive direction, while choice D gives force instead of impulse. When calculating impulse, always multiply force by time duration, then recognize that impulse and change in momentum have identical values and units.

Question 17

A 0.20 kg0.20\,\text{kg} cart initially moving right at 3.0 m/s3.0\,\text{m/s} experiences a constant leftward force of 2.0 N2.0\,\text{N} for 0.50 s0.50\,\text{s}. What is the cart's change in momentum Δp⃗\Delta \vec p?

  1. +1.0 N⋅s+1.0\,\text{N}\cdot\text{s} (to the right)
  2. −4.0 kg⋅m/s-4.0\,\text{kg}\cdot\text{m/s} (to the left)
  3. −1.0 kg⋅m/s-1.0\,\text{kg}\cdot\text{m/s} (to the left) (correct answer)
  4. −2.0 kg⋅m/s-2.0\,\text{kg}\cdot\text{m/s} (to the left)
Explanation: This problem tests the skill of calculating change in momentum and impulse in AP Physics 1. Impulse is defined as the product of a constant force and the time interval over which it acts, providing a measure of the force applied over time. The impulse-momentum theorem states that the impulse delivered to an object equals its change in momentum, so Δp⃗ = J⃗ = F⃗ Δt. In this case, the leftward force of -2.0 N acting for 0.50 s gives an impulse of -1.0 N·s, which is the change in momentum to the left. Choice A is a distractor because it incorrectly uses a positive sign, ignoring the direction of the force. Always calculate impulse as a vector, considering the direction of the force, to find the correct change in momentum.

Question 18

A hockey puck experiences a constant leftward force of 4.0 N4.0\,\text{N} for 0.25 s0.25\,\text{s}. Which best describes the puck's change in momentum?

  1. −4.0 kg ⁣⋅ ⁣m/s-4.0\,\text{kg}\!\cdot\!\text{m/s} (leftward)
  2. +1.0 N ⁣⋅ ⁣s+1.0\,\text{N}\!\cdot\!\text{s} (rightward)
  3. −1.0 N ⁣⋅ ⁣s-1.0\,\text{N}\!\cdot\!\text{s} (leftward) (correct answer)
  4. −4.0 N-4.0\,\text{N} (leftward)
Explanation: This question tests understanding that change in momentum equals impulse. The impulse is J = F·Δt = (-4.0 N)(0.25 s) = -1.0 N·s (negative for leftward). By the impulse-momentum theorem, the change in momentum Δp equals this impulse: Δp = -1.0 N·s = -1.0 kg·m/s leftward. Note that N·s and kg·m/s are equivalent units since 1 N = 1 kg·m/s². Choice A incorrectly gives -4.0 kg·m/s, likely from using force value without time. When asked for change in momentum given force and time, calculate impulse (F·Δt) and recognize this equals Δp.

Question 19

A 3.0 kg3.0\,\text{kg} cart experiences a constant 9 N9\,\text{N} force left for 0.20 s0.20\,\text{s}. What is Δp⃗\Delta \vec p?

  1. 1.8 kg ⁣\cdot ⁣m/s1.8\,\text{kg\!\cdot\!m/s} left (correct answer)
  2. 45 N ⁣\cdot ⁣s45\,\text{N\!\cdot\!s} left
  3. 1.8 kg ⁣\cdot ⁣m/s1.8\,\text{kg\!\cdot\!m/s} right
  4. 9 N9\,\text{N} left
Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is force applied across a time period, expressed as J = F Δt. It corresponds exactly to Δp, the vector change in momentum. For the cart, 9 N left for 0.20 s results in Δp of 1.8 kg·m/s left. Choice D omits time, stating just the force. Calculate impulse first and set it equal to Δp for consistent results.

Question 20

A 1.5 kg1.5\,\text{kg} cart moving left receives a constant 3.0 N3.0\,\text{N} force rightward for 0.60 s0.60\,\text{s}. What is Δp⃗\Delta \vec p?

  1. 1.8 kg ⁣\cdot ⁣m/s1.8\,\text{kg\!\cdot\!m/s} left
  2. 5.0 kg ⁣\cdot ⁣m/s5.0\,\text{kg\!\cdot\!m/s} right
  3. 1.8 kg ⁣\cdot ⁣m/s1.8\,\text{kg\!\cdot\!m/s} right (correct answer)
  4. 3.0 N3.0\,\text{N} right
Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is calculated as the constant force times the duration, J=FΔtJ = F \Delta t. This equals the change in momentum, Δp\Delta p, which is a vector pointing in the force's direction. The 3.0 N right for 0.60 s gives Δp\Delta p of 1.8 kg·m/s right, independent of initial motion. Choice D incorrectly uses force without time, missing the impulse. Apply J=FΔt=ΔpJ = F \Delta t = \Delta p routinely for any force-time scenario.