AP Physics 1 Quiz: Angular Momentum And Angular Impulse

What this quiz covers

This quiz focuses on Angular Momentum And Angular 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. $0.53 \text{ kg·m}^2/\text{s}$
2. $0.30 \text{ kg·m}^2/\text{s}$ (correct answer)
3. $1.15 \text{ kg·m}^2/\text{s}$
4. $0.40 \text{ kg·m}^2/\text{s}$

Explanation: This problem assesses computing change in angular momentum from torque and time. For constant torque, $\Delta L$ equals $\tau$ times $\Delta t$, embodying the angular impulse. This ties into how sustained torque accumulates change in rotational momentum over time. The wheel experiences this direct proportionality. Choice C ($1.15 \text{ kg·m}^2/\text{s}$) could be selected by adding torque and time instead of multiplying, revealing a misconception about the multiplicative nature of impulse. A key strategy is to memorize the rotational equivalents: force $\to$ torque, momentum $\to$ angular momentum, impulse $\to$ angular impulse.

Question 2

1. $0.33\ \text{s}$
2. $1.2\ \text{s}$
3. $3.0\ \text{s}$ (correct answer)
4. $2.4\ \text{s}$

Explanation: This scenario tests calculating time from change in angular momentum and torque. Rearranging ΔL = τ Δt gives Δt = ΔL / τ for constant torque. This reflects the duration needed for torque to effect the momentum change. The wheel's constant torque determines this time. Choice D (2.4 s) could result from dividing incorrectly, like 1.8 / 0.75, indicating a numerical misconception in division. A transferable approach is to check reasonability: larger ΔL or smaller τ should yield longer times, aiding error detection.

Question 3

1. $8.0\ \text{kg·m}^2/\text{s}$
2. $2.0\ \text{kg·m}^2/\text{s}$ (correct answer)
3. $4.0\ \text{kg·m}^2/\text{s}$
4. $0.50\ \text{kg·m}^2/\text{s}$

Explanation: This question evaluates knowledge of angular impulse in the context of rotational motion. Angular impulse results from a torque applied over a time interval and is calculated as τ Δt for constant torque. It represents the total change in angular momentum imparted to the system, similar to how force over time changes linear momentum. In this case, the motor delivers this impulse directly to the wheel. A common distractor like choice C (4.0 kg·m²/s) could arise from forgetting to multiply by time and just using the torque value, indicating a misconception of impulse as instantaneous rather than time-integrated. To approach such problems effectively, always verify units: angular impulse has units of kg·m²/s, matching angular momentum.

Question 4

1. $2.1\,\text{kg·m}^2/\text{s}$
2. $3.0\,\text{kg·m}^2/\text{s}$ (correct answer)
3. $2.5\,\text{kg·m}^2/\text{s}$
4. $1.2\,\text{kg·m}^2/\text{s}$

Explanation: This question requires calculating change in angular momentum from torque and time. Using the angular impulse-momentum theorem: $\Delta L = \tau \Delta t = (2.5 \, \text{N·m})(1.2 \, \text{s}) = 3.0 \, \text{kg·m}^2/\text{s}$. The change represents how much the angular momentum increases due to the applied torque. The calculation is straightforward multiplication of the two given values. Choice A (2.1) might result from calculation error or misreading the values, while C (2.5) incorrectly uses just the torque value. Always multiply torque by time to find the change in angular momentum.

Question 5

1. $8.0\,\text{kg·m}^2/\text{s}$
2. $2.0\,\text{kg·m}^2/\text{s}$ (correct answer)
3. $4.0\,\text{kg·m}^2/\text{s}$
4. $0.50\,\text{kg·m}^2/\text{s}$

Explanation: This question requires calculating angular impulse from constant torque and time duration. Angular impulse $J = \tau \Delta t$, where $\tau$ is the net torque and $\Delta t$ is the time interval. Substituting the given values: $J = (4.0 \, \text{N·m})(0.50 \, \text{s}) = 2.0 \, \text{N·m·s} = 2.0 \, \text{kg·m}^2/\text{s}$. The angular impulse represents the total angular effect of the torque over the time period. Choice A (8.0) incorrectly multiplies 4.0 by 2 instead of 0.50, suggesting a calculation error or misreading of the time value. To find angular impulse, always multiply the constant torque by the time duration.

Question 6

1. $-6.0\,\text{kg·m}^2/\text{s}$ (correct answer)
2. $-1.0\,\text{kg·m}^2/\text{s}$
3. $-60\,\text{kg·m}^2/\text{s}$
4. $6.0\,\text{kg·m}^2/\text{s}$

Explanation: This problem tests understanding of negative torque effects on angular momentum. The angular impulse J = τΔt = (-1.0 N·m)(6.0 s) = -6.0 kg·m²/s. Since angular impulse equals change in angular momentum, ΔL = -6.0 kg·m²/s. The negative value indicates the angular momentum decreases by this amount. Choice C (-60) incorrectly multiplies by an extra factor of 10, suggesting a decimal place error. When calculating change in angular momentum, multiply torque by time and preserve the sign to indicate direction.

Question 7

1. $0.16\,\text{N·m}$
2. $4.0\,\text{N·m}$ (correct answer)
3. $0.80\,\text{N·m}$
4. $1.0\,\text{N·m}$

Explanation: This question involves finding torque from angular impulse and time duration. The angular impulse J = τΔt, so τ = J/Δt = (0.80 kg·m²/s)/(0.20 s) = 4.0 N·m. The torque must be sufficient to produce the given angular impulse in the specified time. Shorter time requires larger torque for the same impulse. Choice A (0.16) incorrectly multiplies the two given values instead of dividing, revealing confusion about the relationship between impulse, torque, and time. When given angular impulse and time, divide impulse by time to find the constant torque.

Question 8

1. $6.0 \, \text{kg·m}^2/\text{s}$
2. $8.0 \, \text{kg·m}^2/\text{s}$ (correct answer)
3. $7.5 \, \text{kg·m}^2/\text{s}$
4. $9.0 \, \text{kg·m}^2/\text{s}$

Explanation: This question tests applying positive torque to increase angular momentum. The angular impulse $J = \tau \Delta t = (+0.50 \, \text{N·m})(2.0 \, \text{s}) = +1.0 \, \text{kg·m}^2/\text{s}$. The final angular momentum is $L_f = L_i + \Delta L = 7.0 + 1.0 = 8.0 \, \text{kg·m}^2/\text{s}$. The positive torque adds to the existing angular momentum in the same direction. Choice A (6.0) incorrectly subtracts instead of adding, treating the positive torque as negative. When torque and initial angular momentum have the same sign, add the angular impulse to find final angular momentum.

Question 9

1. $3.2\,\text{s}$
2. $0.50\,\text{s}$
3. $2.0\,\text{s}$ (correct answer)
4. $1.2\,\text{s}$

Explanation: This problem involves finding time duration from torque and angular momentum change. From ΔL = τΔt, we solve for time: Δt = ΔL/τ = (0.80 kg·m²/s)/(0.40 N·m) = 2.0 s. The time represents how long the torque must act to produce the given momentum change. Smaller torque requires more time for the same momentum change. Choice B (0.50) incorrectly multiplies the values instead of dividing, showing confusion about rearranging the impulse equation. To find time from momentum change and torque, divide the momentum change by the torque.

Question 10

1. $6.0\ \text{kg·m}^2/\text{s}$ (correct answer)
2. $0.38\ \text{kg·m}^2/\text{s}$
3. $1.5\ \text{kg·m}^2/\text{s}$
4. $4.0\ \text{kg·m}^2/\text{s}$

Explanation: This question examines the direct link between net torque, time, and magnitude of change in angular momentum. Net torque over time delivers angular impulse, which equals the absolute change in angular momentum. Qualitatively, longer torque application or stronger torque leads to greater momentum change. No initial conditions are needed since only the change is requested. The distractor 1.5 kg·m²/s may come from using torque alone, a misconception of equating torque to momentum change without time. Strategically, always multiply constant torque by time to find ΔL, mirroring linear dynamics.

Question 11

1. $0.50 \text{N·m}$ (correct answer)
2. $0.32 \text{N·m}$
3. $1.2 \text{N·m}$
4. $0.40 \text{N·m·s}$

Explanation: This question tests determining torque from angular momentum change and time. The positive change indicates torque direction aligns with increasing L, where $τ = ΔL / Δt$. Qualitatively, torque accelerates rotation, and its magnitude is the ratio of change to time. The constant nature simplifies to division. The distractor 0.40 N·m·s may confuse impulse with torque, a misconception of using ΔL directly as τ. Always solve for unknowns using $ΔL = τΔt$, ensuring proper algebraic isolation.

Question 12

1. $0.63 \, \text{kg·m}^2/\text{s}$
2. $6.5 \, \text{kg·m}^2/\text{s}$
3. $10 \, \text{kg·m}^2/\text{s}$ (correct answer)
4. $2.5 \, \text{kg·m}^2/\text{s}$

Explanation: This question involves determining change in angular momentum for a spinning platform. The net torque over time produces angular impulse, equaling $ΔL$. Qualitatively, this shows how external torques alter a system's rotational state. The constant net torque here results in a straightforward calculation. Distractor B ($6.5 \, \text{kg·m}^2/\text{s}$) might arise from using 2.5 * 2.6 or a miscalculation, pointing to an arithmetic misconception rather than conceptual error. For wider application, use the formula $ΔL = τ Δt$ as a checkpoint in more complex rotational problems involving moments of inertia.

Question 13

1. $0.10 \text{kg·m}^2/\text{s}$
2. $30 \text{kg·m}^2/\text{s}$
3. $3.0 \text{kg·m}^2/\text{s}$
4. $0.30 \text{kg·m}^2/\text{s}$ (correct answer)

Explanation: This question evaluates angular impulse delivered to a flywheel. Angular impulse is $τ Δt$ for constant torque, matching the change in angular momentum. Qualitatively, it quantifies the rotational 'kick' from the torque over a short time. The flywheel receives this impulse directly. Choice C ($3.0 \text{kg·m}^2/\text{s}$) may be chosen by using torque without time, showing a misconception that impulse equals torque, not its time integral. Generally, reinforce understanding by comparing to linear impulse problems, noting the parallel structures in calculations.

Question 14

1. $36\,\text{N·m}$
2. $9.0\,\text{N·m}$
3. $4.0\,\text{N·m}$ (correct answer)
4. $12\,\text{N·m}$

Explanation: This problem requires finding torque from given change in angular momentum and time. Using the angular impulse-momentum theorem: ΔL = τΔt, we can solve for torque: τ = ΔL/Δt = (12 kg·m²/s)/(3.0 s) = 4.0 N·m. The torque represents the rate of change of angular momentum. A larger torque would produce the same momentum change in less time. Choice A (36) incorrectly multiplies instead of dividing, showing confusion about rearranging the impulse equation. To find torque from momentum change and time, divide the momentum change by the time duration.

Question 15

1. $1.7 \text{kg·m}^2/\text{s}$
2. $0.20 \text{kg·m}^2/\text{s}$
3. $0.30 \text{kg·m}^2/\text{s}$ (correct answer)
4. $7.5 \text{kg·m}^2/\text{s}$

Explanation: This question focuses on calculating the change in angular momentum from a given torque and time. Angular momentum changes when a net torque is applied, with the magnitude of change given by $\Delta L = \tau \Delta t$ for constant torque. This qualitative link mirrors the linear case where impulse equals change in momentum. Here, the fan blade's axle torque causes the specified change. Distractor D ($7.5 \text{kg·m}^2/\text{s}$) could result from multiplying torque by time incorrectly, like using 1.5 * 5 instead of 0.20 s, showing a misconception in reading the time value accurately. For transferable skills, practice dimensional analysis to ensure calculations yield the correct units for angular momentum.

Question 16

1. $2.4\ \text{kg·m}^2/\text{s}$ (correct answer)
2. $0.27\ \text{kg·m}^2/\text{s}$
3. $0.80\ \text{kg·m}^2/\text{s}$
4. $3.0\ \text{kg·m}^2/\text{s}$

Explanation: This scenario examines angular impulse in an opposing torque context. The magnitude of angular impulse is τ Δt, regardless of direction, as it quantifies the total rotational impetus. Qualitatively, it equals the absolute change in angular momentum, opposing the rotor's motion in this case. The constant torque over time delivers this impulse. Choice C (0.80 kg·m²/s) may be picked by mistaking impulse for torque alone, reflecting a misconception that time is not a factor in impulse calculations. A broad strategy is to draw parallels between linear and angular dynamics, treating impulse as the 'push' over time in both realms.

Question 17

1. $6.0\,\text{kg·m}^2/\text{s}$ (correct answer)
2. $1.5\,\text{kg·m}^2/\text{s}$
3. $4.0\,\text{kg·m}^2/\text{s}$
4. $0\,\text{kg·m}^2/\text{s}$

Explanation: This question tests calculating final angular momentum when starting from rest with constant torque. The angular impulse J = τΔt = (1.5 N·m)(4.0 s) = 6.0 kg·m²/s. Since the rotor starts from rest (Li = 0), the final angular momentum equals the angular impulse: Lf = Li + ΔL = 0 + 6.0 = 6.0 kg·m²/s. Starting from rest means all the angular impulse becomes the final angular momentum. Choice B (1.5) incorrectly uses only the torque value without multiplying by time, confusing torque with angular momentum. Remember that angular momentum change equals torque multiplied by time, not just torque alone.

Question 18

1. $2.0 \text{N·m·s}$ (correct answer)
2. $8.0 \text{N·m·s}$
3. $4.0 \text{N·m}$
4. $0.50 \text{N·m·s}$

Explanation: This question evaluates understanding of angular impulse delivered by a constant torque over time. Angular impulse is the product of constant torque and the time interval, representing the total 'push' in the rotational sense. It quantifies how much the angular momentum changes, but here the question directly asks for the impulse itself. The low-friction axle implies negligible other torques, so the given torque is net. A distractor like $4.0 \text{N·m}$ could arise from mistaking torque for impulse, a misconception of overlooking the multiplication by time. Remember as a strategy that impulse always involves integrating force or torque over time, ensuring units include seconds.

Question 19

1. $0.80\ \text{kg·m}^2/\text{s}$ (correct answer)
2. $2.4\ \text{kg·m}^2/\text{s}$
3. $1.2\ \text{kg·m}^2/\text{s}$
4. $0.40\ \text{kg·m}^2/\text{s}$

Explanation: This question assesses finding final angular momentum after an opposing torque acts. Opposing torque reduces angular momentum, with change ΔL = -τΔt. Qualitatively, it slows rotation, and if sufficient, could reverse direction, but here it halves. Final L = initial - τΔt, assuming one-dimensional rotation. The distractor 2.4 kg·m²/s could come from adding instead of subtracting, a misconception of ignoring 'opposing' direction. Strategically, note directional words like 'opposing' to assign signs, treating as vectors along the axis.

Question 20

1. $0.30\ kg\cdot m^2/s$
2. $1.2\ kg\cdot m^2/s$ (correct answer)
3. $0.60\ kg\cdot m^2/s$
4. $2.0\ kg\cdot m^2/s$

Explanation: This question assesses the relationship between torque, time, and change in angular momentum for rotational systems. Angular impulse, which is torque multiplied by the time it acts, equals the change in angular momentum of the disk. Since the torque is constant, the change in angular momentum is directly proportional to both the torque magnitude and the duration. This relationship holds regardless of the disk's initial angular momentum or moment of inertia, as we're only finding the change. One distractor, such as 0.60 kg·m²/s, might be selected by confusing torque with angular impulse, a misconception of ignoring the time factor in impulse calculations. A transferable strategy is to always calculate angular impulse as τΔt when torque is constant, analogous to linear impulse FΔt.