AP Physics 1 Quiz: Angular Momentum And Angular Impulse
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Angular Momentum And Angular ImpulseQuestion 1 of 20

A rotor's angular momentum changes by +0.40kg⋅m2/s+0.40 \text{kg}\cdot\text{m}^2/\text{s} when a constant torque acts for 0.80s0.80 \text{s}. What torque magnitude acted?

0.50N\cdotpm0.50 \text{N·m}
0.32N\cdotpm0.32 \text{N·m}
1.2N\cdotpm1.2 \text{N·m}
0.40N\cdotpm\cdotps0.40 \text{N·m·s}
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AP Physics 1 Quiz

AP Physics 1 Quiz: Angular Momentum And Angular Impulse

Practice Angular Momentum And Angular 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.

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.

All questions

Question 1

A rotor's angular momentum changes by +0.40kg⋅m2/s+0.40 \text{kg}\cdot\text{m}^2/\text{s} when a constant torque acts for 0.80s0.80 \text{s}. What torque magnitude acted?

  1. 0.50N\cdotpm0.50 \text{N·m} (correct answer)
  2. 0.32N\cdotpm0.32 \text{N·m}
  3. 1.2N\cdotpm1.2 \text{N·m}
  4. 0.40N\cdotpm\cdotps0.40 \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/Δtτ = 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=τΔtΔL = τΔt, ensuring proper algebraic isolation.

Question 2

A fan initially has L=0L=0. A constant net torque of 0.90 N⋅m0.90\ \text{N}\cdot\text{m} acts for 1.0 s1.0\ \text{s}. What is the fan's angular momentum afterward?

  1. 0.90 kg\cdotpm2/s0.90\ \text{kg·m}^2/\text{s} (correct answer)
  2. 0.90 N\cdotpm0.90\ \text{N·m}
  3. 1.0 kg\cdotpm2/s1.0\ \text{kg·m}^2/\text{s}
  4. 0 kg\cdotpm2/s0\ \text{kg·m}^2/\text{s}
Explanation: This question probes how angular momentum evolves from zero under constant torque. Starting from rest (L=0), angular impulse from torque over time builds up the final angular momentum. The relationship is linear: more time or torque means higher final L. Since initial is zero, final L equals the impulse τΔt. A distractor like 0.90 N·m might ignore time, a misconception of confusing torque units with momentum units. As a strategy, verify units: angular momentum is kg·m²/s, requiring torque (N·m) times seconds.

Question 3

A wheel's motor applies a constant torque of 4.0 N⋅m4.0\ \text{N}\cdot\text{m} for 0.50 s0.50\ \text{s}. What angular impulse is delivered to the wheel?

  1. 8.0 kg\cdotpm2/s8.0\ \text{kg·m}^2/\text{s}
  2. 2.0 kg\cdotpm2/s2.0\ \text{kg·m}^2/\text{s} (correct answer)
  3. 4.0 kg\cdotpm2/s4.0\ \text{kg·m}^2/\text{s}
  4. 0.50 kg\cdotpm2/s0.50\ \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

A uniform disk rotates about a fixed axle. A constant torque of 0.60 N⋅m0.60\ \text{N}\cdot\text{m} acts for 2.0 s2.0\ \text{s}, then stops. What is the magnitude of the disk's change in angular momentum?

  1. 0.30 kg⋅m2/s0.30\ kg\cdot m^2/s
  2. 1.2 kg⋅m2/s1.2\ kg\cdot m^2/s (correct answer)
  3. 0.60 kg⋅m2/s0.60\ kg\cdot m^2/s
  4. 2.0 kg⋅m2/s2.0\ 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.

Question 5

A rotor experiences a constant opposing torque of magnitude 0.80 N⋅m0.80\ \text{N}\cdot\text{m} for 3.0 s3.0\ \text{s}. What is the magnitude of the angular impulse?

  1. 2.4 kg\cdotpm2/s2.4\ \text{kg·m}^2/\text{s} (correct answer)
  2. 0.27 kg\cdotpm2/s0.27\ \text{kg·m}^2/\text{s}
  3. 0.80 kg\cdotpm2/s0.80\ \text{kg·m}^2/\text{s}
  4. 3.0 kg\cdotpm2/s3.0\ \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 6

A wheel's angular momentum is 1.6 kg⋅m2/s1.6\ \text{kg}\cdot\text{m}^2/\text{s}, then a constant opposing torque of 0.40 N⋅m0.40\ \text{N}\cdot\text{m} acts for 2.0 s2.0\ \text{s}. What is the final angular momentum magnitude?

  1. 0.80 kg\cdotpm2/s0.80\ \text{kg·m}^2/\text{s} (correct answer)
  2. 2.4 kg\cdotpm2/s2.4\ \text{kg·m}^2/\text{s}
  3. 1.2 kg\cdotpm2/s1.2\ \text{kg·m}^2/\text{s}
  4. 0.40 kg\cdotpm2/s0.40\ \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 7

A rotor starts with Li=0L_i=0. A constant torque of 1.5 N⋅m1.5\,\text{N}\cdot\text{m} is applied for 4.0 s4.0\,\text{s}. What is LfL_f?

  1. 6.0 kg\cdotpm2/s6.0\,\text{kg·m}^2/\text{s} (correct answer)
  2. 1.5 kg\cdotpm2/s1.5\,\text{kg·m}^2/\text{s}
  3. 4.0 kg\cdotpm2/s4.0\,\text{kg·m}^2/\text{s}
  4. 0 kg\cdotpm2/s0\,\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 8

A spinning platform has Li=10 kg⋅m2/sL_i=10\,\text{kg}\cdot\text{m}^2/\text{s}. A constant torque of −1.0 N⋅m-1.0\,\text{N}\cdot\text{m} acts for 6.0 s6.0\,\text{s}. What is ΔL\Delta L?

  1. −6.0 kg\cdotpm2/s-6.0\,\text{kg·m}^2/\text{s} (correct answer)
  2. −1.0 kg\cdotpm2/s-1.0\,\text{kg·m}^2/\text{s}
  3. −60 kg\cdotpm2/s-60\,\text{kg·m}^2/\text{s}
  4. 6.0 kg\cdotpm2/s6.0\,\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 9

A wheel's angular momentum changes by 1.8 kg⋅m2/s1.8\ \text{kg}\cdot\text{m}^2/\text{s} when a constant torque acts. If the torque is 0.60 N⋅m0.60\ \text{N}\cdot\text{m}, how long did it act?

  1. 0.33 s0.33\ \text{s}
  2. 1.2 s1.2\ \text{s}
  3. 3.0 s3.0\ \text{s} (correct answer)
  4. 2.4 s2.4\ \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 10

A uniform disk initially at rest experiences a constant torque of 0.60 N⋅m0.60\ \text{N}\cdot\text{m} for 2.0 s2.0\ \text{s}. What is the magnitude of the disk's change in angular momentum?

  1. 0.30 kg\cdotpm2/s0.30\ \text{kg·m}^2/\text{s}
  2. 1.2 kg\cdotpm2/s1.2\ \text{kg·m}^2/\text{s} (correct answer)
  3. 0.60 kg\cdotpm2/s0.60\ \text{kg·m}^2/\text{s}
  4. 2.0 kg\cdotpm2/s2.0\ \text{kg·m}^2/\text{s}
Explanation: This question assesses the skill of understanding angular momentum and angular impulse in rotational dynamics. Angular impulse is defined as the product of a constant torque and the time interval over which it acts, directly causing a change in angular momentum. Qualitatively, just as linear impulse changes linear momentum, angular impulse changes angular momentum, with the relationship ΔL = τ Δt for constant torque. Since the disk starts at rest, its initial angular momentum is zero, so the change in angular momentum equals the angular impulse applied. One distractor, such as choice A (0.30 kg·m²/s), might be selected if someone mistakenly divides torque by time instead of multiplying, reflecting a misconception about the impulse-momentum theorem. A transferable strategy is to analogize rotational quantities to their linear counterparts, remembering that torque plays the role of force in changing momentum over time.

Question 11

A rotor experiences a net torque of 1.5 N⋅m1.5\ \text{N}\cdot\text{m} for 4.0 s4.0\ \text{s}. What is the magnitude of the change in angular momentum?

  1. 6.0 kg\cdotpm2/s6.0\ \text{kg·m}^2/\text{s} (correct answer)
  2. 0.38 kg\cdotpm2/s0.38\ \text{kg·m}^2/\text{s}
  3. 1.5 kg\cdotpm2/s1.5\ \text{kg·m}^2/\text{s}
  4. 4.0 kg\cdotpm2/s4.0\ \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 12

A uniform disk spins about a fixed axle. A constant torque of 0.60 N⋅m0.60\,\text{N}\cdot\text{m} acts for 3.0 s3.0\,\text{s}. What is ΔL\Delta L?

  1. 0.20 kg\cdotpm2/s0.20\,\text{kg·m}^2/\text{s}
  2. 1.8 kg\cdotpm2/s1.8\,\text{kg·m}^2/\text{s} (correct answer)
  3. 0.60 kg\cdotpm2/s0.60\,\text{kg·m}^2/\text{s}
  4. 3.0 kg\cdotpm2/s3.0\,\text{kg·m}^2/\text{s}
Explanation: This problem tests understanding of angular impulse and its relationship to change in angular momentum. Angular impulse equals the product of torque and time: J = τΔt = (0.60 N·m)(3.0 s) = 1.8 N·m·s. By the angular impulse-momentum theorem, angular impulse equals the change in angular momentum: ΔL = J = 1.8 kg·m²/s. The units N·m·s and kg·m²/s are equivalent because 1 N = 1 kg·m/s². Choice A (0.20) incorrectly divides torque by time instead of multiplying, revealing confusion about the impulse formula. When given constant torque and time, multiply them to find the change in angular momentum.

Question 13

A turntable experiences a constant external torque of 0.20 N⋅m0.20\ \text{N}\cdot\text{m} for 5.0 s5.0\ \text{s}. What is ΔL\Delta L for the turntable?

  1. 1.0 kg\cdotpm2/s1.0\ \text{kg·m}^2/\text{s} (correct answer)
  2. 0.20 kg\cdotpm2/s0.20\ \text{kg·m}^2/\text{s}
  3. 5.0 kg\cdotpm2/s5.0\ \text{kg·m}^2/\text{s}
  4. 0.040 kg\cdotpm2/s0.040\ \text{kg·m}^2/\text{s}
Explanation: This problem tests the concept of change in angular momentum due to angular impulse. The change in angular momentum, ΔL, is equal to the angular impulse, which for constant torque is τ multiplied by the time duration. This relationship holds because torque is the rate of change of angular momentum, so integrating over time gives the total change. For the turntable, the external torque directly produces this ΔL. Choice B (0.20 kg·m²/s) might be chosen if one confuses torque with impulse and omits the time factor, stemming from a misconception that torque alone changes momentum without time consideration. A useful strategy is to recall the rotational analog of Newton's second law, where τ = dL/dt, leading to ΔL = τ Δt for constant torque.

Question 14

A wheel on a low-friction axle experiences a constant torque of 4.0N⋅m4.0 \text{N}\cdot\text{m} for 0.50s0.50 \text{s}. What is the angular impulse delivered?

  1. 2.0N\cdotpm\cdotps2.0 \text{N·m·s} (correct answer)
  2. 8.0N\cdotpm\cdotps8.0 \text{N·m·s}
  3. 4.0N\cdotpm4.0 \text{N·m}
  4. 0.50N\cdotpm\cdotps0.50 \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 4.0N⋅m4.0 \text{N}\cdot\text{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 15

A wheel experiences a constant torque of 0.25N⋅m0.25 \text{N}\cdot\text{m} for 8.0s8.0 \text{s}. What is the angular impulse magnitude?

  1. 2.0N\cdotpm\cdotps2.0 \text{N·m·s} (correct answer)
  2. 0.031N\cdotpm\cdotps0.031 \text{N·m·s}
  3. 8.0N\cdotpm8.0 \text{N·m}
  4. 0.25kg\cdotpm2/s0.25 \text{kg·m}^2/\text{s}
Explanation: This question evaluates computing angular impulse from constant torque and duration. Angular impulse is τΔtτ Δt for constant torque, measuring total rotational effect. It relates qualitatively to how much spin is imparted, with units N⋅m⋅s\text{N}\cdot\text{m}\cdot\text{s}. The magnitude ignores direction here. A distractor like 8.0 N⋅m\text{N}\cdot\text{m} might use time alone, a misconception of swapping torque and time in the product. Strategically, recall impulse as area under torque-time graph; for constant, it's simple multiplication.

Question 16

A rotating rod has Li=7.0 kg⋅m2/sL_i = 7.0 \, \text{kg}\cdot\text{m}^2/\text{s}. A net torque of +0.50 N⋅m+0.50 \, \text{N}\cdot\text{m} acts for 2.0 s2.0 \, \text{s}. What is LfL_f?

  1. 6.0 kg\cdotpm2/s6.0 \, \text{kg·m}^2/\text{s}
  2. 8.0 kg\cdotpm2/s8.0 \, \text{kg·m}^2/\text{s} (correct answer)
  3. 7.5 kg\cdotpm2/s7.5 \, \text{kg·m}^2/\text{s}
  4. 9.0 kg\cdotpm2/s9.0 \, \text{kg·m}^2/\text{s}
Explanation: This question tests applying positive torque to increase angular momentum. The angular impulse J=τΔt=(+0.50 N⋅m)(2.0 s)=+1.0 kg⋅m2/sJ = \tau \Delta t = (+0.50 \, \text{N}\cdot\text{m})(2.0 \, \text{s}) = +1.0 \, \text{kg}\cdot\text{m}^2/\text{s}. The final angular momentum is Lf=Li+ΔL=7.0+1.0=8.0 kg⋅m2/sL_f = L_i + \Delta L = 7.0 + 1.0 = 8.0 \, \text{kg}\cdot\text{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 17

A disk's angular momentum decreases by 3.0 kg⋅m2/s3.0\ \text{kg}\cdot\text{m}^2/\text{s} due to a constant opposing torque acting for 2.0 s2.0\ \text{s}. What is the torque magnitude?

  1. 6.0 N\cdotpm6.0\ \text{N·m}
  2. 1.5 N\cdotpm1.5\ \text{N·m} (correct answer)
  3. 3.0 N\cdotpm\cdotps3.0\ \text{N·m·s}
  4. 0.67 N\cdotpm0.67\ \text{N·m}
Explanation: This question assesses calculating torque magnitude from change in angular momentum and time. The decrease in angular momentum implies an opposing torque, where magnitude of ΔLΔL = τΔtτ Δt. Qualitatively, torque acts as the rate of change, so dividing ΔLΔL by time gives ττ. The opposing nature determines the sign, but magnitude is absolute. The distractor 6.0N⋅m6.0 \text{N}\cdot\text{m} could result from multiplying instead of dividing, a misconception of inverting the impulse formula. A transferable approach is to rearrange ΔLΔL = τΔtτ Δt for the unknown, checking consistency with units.

Question 18

A wheel experiences a constant torque of 0.40N⋅m0.40 \text{N}\cdot\text{m} for 0.75s0.75 \text{s}. What is the magnitude of ΔL\Delta L?

  1. 0.53 kg\cdotpm2/s0.53 \text{ kg·m}^2/\text{s}
  2. 0.30 kg\cdotpm2/s0.30 \text{ kg·m}^2/\text{s} (correct answer)
  3. 1.15 kg\cdotpm2/s1.15 \text{ kg·m}^2/\text{s}
  4. 0.40 kg\cdotpm2/s0.40 \text{ kg·m}^2/\text{s}
Explanation: This problem assesses computing change in angular momentum from torque and time. For constant torque, ΔL\Delta L equals τ\tau times Δt\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 kg⋅m2/s1.15 \text{ kg}\cdot\text{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 19

A wheel experiences a constant net torque of 4.0 N⋅m4.0\,\text{N}\cdot\text{m} for 0.50 s0.50\,\text{s}. What is the angular impulse?

  1. 8.0 kg\cdotpm2/s8.0\,\text{kg·m}^2/\text{s}
  2. 2.0 kg\cdotpm2/s2.0\,\text{kg·m}^2/\text{s} (correct answer)
  3. 4.0 kg\cdotpm2/s4.0\,\text{kg·m}^2/\text{s}
  4. 0.50 kg\cdotpm2/s0.50\,\text{kg·m}^2/\text{s}
Explanation: This question requires calculating angular impulse from constant torque and time duration. Angular impulse J=τΔtJ = \tau \Delta t, where τ\tau is the net torque and Δt\Delta t is the time interval. Substituting the given values: J=(4.0 N⋅m)(0.50 s)=2.0 N⋅m⋅s=2.0 kg⋅m2/sJ = (4.0 \, \text{N}\cdot\text{m})(0.50 \, \text{s}) = 2.0 \, \text{N}\cdot\text{m}\cdot\text{s} = 2.0 \, \text{kg}\cdot\text{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 20

A fan blade's axle experiences a constant torque of 1.5N⋅m1.5 \text{N}\cdot\text{m} for 0.20s0.20 \text{s}. What change in angular momentum results?

  1. 1.7kg\cdotpm2/s1.7 \text{kg·m}^2/\text{s}
  2. 0.20kg\cdotpm2/s0.20 \text{kg·m}^2/\text{s}
  3. 0.30kg\cdotpm2/s0.30 \text{kg·m}^2/\text{s} (correct answer)
  4. 7.5kg\cdotpm2/s7.5 \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 ΔL=τΔt\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 (7.5kg⋅m2/s7.5 \text{kg}\cdot\text{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.