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. $+1.2\ \text{N·s}$ (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 2

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 3

1. $1.5\ \text{N·s}$ to the left (correct answer)
2. $2.5\ \text{N}$ to the left
3. $1.5\ \text{N·s}$ to the right
4. $0.24\ \text{N·s}$ 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 4

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 5

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 6

1. $+1.0\,\text{N}\cdot\text{s}$ (to the right)
2. $-4.0\,\text{kg}\cdot\text{m/s}$ (to the left)
3. $-1.0\,\text{kg}\cdot\text{m/s}$ (to the left) (correct answer)
4. $-2.0\,\text{kg}\cdot\text{m/s}$ (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 7

1. $-4.0\,\text{kg}\!\cdot\!\text{m/s}$ (leftward)
2. $+1.0\,\text{N}\!\cdot\!\text{s}$ (rightward)
3. $-1.0\,\text{N}\!\cdot\!\text{s}$ (leftward) (correct answer)
4. $-4.0\,\text{N}$ (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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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.