AP Physics 1 Quiz: Change In Momentum And Impulse

What this quiz covers

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.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $1.8\,\text{kg\!\cdot\!m/s}$ left (correct answer)
2. $45\,\text{N\!\cdot\!s}$ left
3. $1.8\,\text{kg\!\cdot\!m/s}$ right
4. $9\,\text{N}$ 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 2

1. $2.0\,\text{N\!\cdot\!s}$ to the right (correct answer)
2. $5.0\,\text{N}$ to the right
3. $0.08\,\text{N\!\cdot\!s}$ to the right
4. $2.0\,\text{N\!\cdot\!s}$ 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 $J = F \Delta t$. It directly corresponds to the momentum change, $\Delta p = J$, including the vector direction. The $5 \, \text{N}$ right for $0.40 \, \text{s}$ results in $\Delta p$ of $2.0 \, \text{N\!\cdot\!s}$ right for the puck. Choice C might come from dividing force by time instead of multiplying, a calculation error. Consistently use $J = F \Delta t = \Delta p$ to verify answers in impulse problems.

Question 3

1. $1.8\,\text{kg\!\cdot\!m/s}$ left
2. $5.0\,\text{kg\!\cdot\!m/s}$ right
3. $1.8\,\text{kg\!\cdot\!m/s}$ right (correct answer)
4. $3.0\,\text{N}$ 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, $J = F \Delta t$. This equals the change in momentum, $\Delta p$, which is a vector pointing in the force's direction. The 3.0 N right for 0.60 s gives $\Delta p$ of 1.8 kg·m/s right, independent of initial motion. Choice D incorrectly uses force without time, missing the impulse. Apply $J = F \Delta t = \Delta p$ routinely for any force-time scenario.

Question 4

1. $1.2\,\text{N\!\cdot\!s}$ left (correct answer)
2. $12\,\text{N}$ left
3. $0.83\,\text{N\!\cdot\!s}$ left
4. $1.2\,\text{N\!\cdot\!s}$ right

Explanation: This question assesses the concept of change in momentum and impulse in AP Physics 1. Impulse is the integral of force over time, but for constant force, it's simply $J = F \Delta t$. It equates to the momentum shift, $\Delta p = J$, with direction opposite to the ball's initial motion here. The 12 N left for 0.10 s causes $\Delta p$ of $1.2 \, \text{N·s}$ left. Choice B lists only the force, forgetting the time multiplication. Use the formula $J = F \Delta t = \Delta p$ as a key step in all impulse-related questions.

Question 5

1. $+3.0\,\text{kg\cdot m/s}$ to the right
2. $-3.0\,\text{kg\cdot m/s}$ to the left (correct answer)
3. $-12\,\text{kg\cdot m/s}$ to the left
4. $-48\,\text{kg\cdot m/s}$ to the left

Explanation: This question assesses change in momentum from impulse in AP Physics 1. Impulse is force over time, providing the net effect that alters momentum. The impulse-momentum theorem links them directly: Δp⃗ = F⃗ Δt. The leftward -12 N force for 0.25 s yields Δp⃗ of -3.0 kg·m/s to the left, focusing only on the force applied. Choice C is a distractor, maybe from using initial momentum without the sign. Always isolate impulse calculation from initial conditions for accurate Δp⃗ determination.

Question 6

1. $-6.0\,\text{N}$ (left) (correct answer)
2. $-0.54\,\text{N}$ (left)
3. $+6.0\,\text{N}$ (right)
4. $-1.8\,\text{N}$ (left)

Explanation: This problem evaluates finding force from given impulse in AP Physics 1. Impulse equals force multiplied by time for constant forces, embodying the cumulative impact. It corresponds to change in momentum, so J⃗ = F⃗ Δt = Δp⃗. With J = -1.8 N·s over 0.30 s, F = -6.0 N to the left. Choice D is a distractor, perhaps from confusing impulse with momentum units. Rearrange the impulse formula to solve for unknowns like force in such scenarios.

Question 7

1. $\Delta \vec p$ is $3.0\,\text{kg\cdot m/s}$ to the left
2. $\Delta \vec p$ is $2.0\,\text{kg\cdot m/s}$ to the right
3. $\Delta \vec p$ is $3.0\,\text{kg\cdot m/s}$ to the right (correct answer)
4. $\Delta \vec p$ is $4.5\,\text{kg\cdot m/s}$ to the right

Explanation: This question examines the relationship between impulse and change in momentum in AP Physics 1. Impulse is force applied over time, serving as a vector quantity that changes an object's momentum. The theorem equates impulse directly to Δp⃗, meaning the change matches the impulse's magnitude and direction. Thus, a rightward impulse of 3.0 N·s causes Δp⃗ of 3.0 kg·m/s to the right, irrespective of initial motion. Choice A is a distractor as it wrongly assigns a leftward direction, perhaps confusing with initial velocity. Consistently use Δp⃗ = J⃗ to handle direction correctly in momentum problems.

Question 8

1. $8\,\text{kg}\!\cdot\!\text{m/s}$
2. $6\,\text{kg}\!\cdot\!\text{m/s}$ (correct answer)
3. $0.094\,\text{kg}\!\cdot\!\text{m/s}$
4. $0.75\,\text{kg}\!\cdot\!\text{m/s}$

Explanation: This question asks for the magnitude of momentum change given force and time. Using the impulse-momentum theorem, the magnitude of impulse is |J| = |F|·Δt = (8 N)(0.75 s) = 6 N·s. Since impulse equals change in momentum, |Δp| = 6 kg·m/s. The problem states the force acts in the direction of motion, so both impulse and momentum change are positive. Choice A incorrectly divides force by time, while choice D gives just the time value. When calculating momentum change from constant force, multiply force magnitude by time duration to get impulse magnitude.

Question 9

1. $+2.4\,\text{kg}\!\cdot\!\text{m/s}$ (rightward)
2. $-1.6\,\text{kg}\!\cdot\!\text{m/s}$ (leftward)
3. $-2.4\,\text{kg}\!\cdot\!\text{m/s}$ (leftward) (correct answer)
4. $-2.4\,\text{N}$ (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 10

1. $+1.2\ \text{N·s}$ (to the right)
2. $-1.2\ \text{N·s}$ (to the left) (correct answer)
3. $-3.0\ \text{N·s}$ (to the left)
4. $-0.75\ \text{N·s}$ (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 11

1. $1.2\,\text{N\cdot s}$ (correct answer)
2. $6\,\text{N}$
3. $0.20\,\text{s}$
4. $30\,\text{N\cdot s}$

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 $6 \, \text{N}$ for $0.20 \, \text{s}$ results in an impulse magnitude of $1.2 \, \text{N·s}$, 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 $F \, \Delta t$ for constant forces to determine momentum changes accurately.

Question 12

1. $-25\ \text{N·s}$ (left)
2. $-1.0\ \text{N·s}$ (left) (correct answer)
3. $+1.0\ \text{N·s}$ (right)
4. $-4.0\ \text{N·s}$ (left)

Explanation: This problem tests understanding of change in momentum and impulse. Using the impulse-momentum theorem, Δp = J = F·Δt = (5 N)(0.20 s) = 1.0 N·s. Since the force is to the left, Δp = -1.0 N·s (left), where negative indicates leftward direction. The mass of the cart doesn't affect the impulse calculation when force and time are given. Choice A incorrectly multiplies force by 5 instead of the given time. To find change in momentum from force and time, always use Δp = F·Δt, regardless of the object's mass.

Question 13

1. $12\ \text{kg·m/s}$ to the right
2. $3.0\ \text{kg·m/s}$ to the right (correct answer)
3. $6.0\ \text{kg·m/s}$ to the right
4. $3.0\ \text{kg·m/s}$ 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). Given F = 6 N to the right and Δt = 0.50 s, we calculate J = (6 N)(0.50 s) = 3.0 N·s to the right. Therefore, Δp = 3.0 kg·m/s to the right (since N·s = kg·m/s). Choice A incorrectly multiplies force by mass instead of time. When finding change in momentum from a constant force, always use Δp = F·Δt, not F·m.

Question 14

1. $0.60\,\text{s}$ (correct answer)
2. $15\,\text{s}$
3. $1.7\,\text{s}$
4. $0.40\,\text{s}$

Explanation: This problem requires solving for time given force and momentum change. Using J = F·Δt = Δp, we can solve for time: Δt = Δp/F. The momentum change is -3.0 kg·m/s (negative for leftward), and the force is -5.0 N (also leftward). Therefore, Δt = (-3.0 kg·m/s)/(-5.0 N) = 0.60 s. The negative signs cancel because both quantities point left. Choice C incorrectly divides 5.0 by 3.0, while choice B multiplies instead of dividing. When finding time from force and momentum change, divide momentum change by force, being careful with signs.

Question 15

1. $1.2\ \text{kg·m/s}$ forward (correct answer)
2. $12\ \text{kg·m/s}$ forward
3. $0.30\ \text{kg·m/s}$ forward
4. $1.2\ \text{kg·m/s}$ 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 16

1. $2.0\,\text{N}\!\cdot\!\text{s}$ downward
2. $10\,\text{N}$ upward
3. $0.20\,\text{N}\!\cdot\!\text{s}$ upward
4. $2.0\,\text{N}\!\cdot\!\text{s}$ upward (correct answer)

Explanation: This question asks for impulse given force and time duration. Impulse is calculated as J = F·Δt = (10 N upward)(0.20 s) = 2.0 N·s upward. The impulse has the same direction as the applied force and represents the momentum change the object will experience. By the impulse-momentum theorem, this 2.0 N·s impulse will produce a 2.0 kg·m/s change in momentum. Choice B incorrectly gives just the force value without time multiplication. To find impulse, always multiply the force vector by the time interval, maintaining proper direction.

Question 17

1. $5.0\,\text{N}\!\cdot\!\text{s}$ to the right (correct answer)
2. $50\,\text{N}$ to the right
3. $0.10\,\text{N}\!\cdot\!\text{s}$ to the right
4. $5.0\,\text{N}\!\cdot\!\text{s}$ to the left

Explanation: This question directly asks for impulse calculation. Impulse is defined as the product of force and the time interval over which it acts: J = F·Δt. With a rightward force of 50 N acting for 0.10 s, the impulse is J = (50 N)(0.10 s) = 5.0 N·s to the right. The impulse-momentum theorem tells us this impulse will equal the ball's change in momentum. Choice B incorrectly gives just the force value without multiplying by time. To find impulse, always multiply the force magnitude by the time duration and maintain the force's direction.

Question 18

1. $+1.2\,\text{N}\!\cdot\!\text{s}$ (to the right)
2. $-7.5\,\text{kg}\!\cdot\!\text{m/s}$ (to the left)
3. $-1.2\,\text{N}\!\cdot\!\text{s}$ (to the left) (correct answer)
4. $-3.0\,\text{N}$ (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 19

1. $1.2\,\text{N\!\cdot\!s}$ upward (correct answer)
2. $6\,\text{N}$ upward
3. $30\,\text{N\!\cdot\!s}$ upward
4. $1.2\,\text{N\!\cdot\!s}$ 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 20

1. Left, because the cart was moving left initially
2. Right, because impulse and $\Delta \vec p$ 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.