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

A 2.0 kg object moving at 8.0 m/s collides with a wall and bounces back at 6.0 m/s. If the collision lasts 0.15 s, what is the magnitude of the average force exerted by the wall on the object?

13 N
27 N
93 N
187 N
280 N
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College Physics Quiz

College Physics Quiz: Change In Momentum And Impulse

Practice Change In Momentum And Impulse in College Physics 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 College Physics.

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Question 1

A 2.0 kg object moving at 8.0 m/s collides with a wall and bounces back at 6.0 m/s. If the collision lasts 0.15 s, what is the magnitude of the average force exerted by the wall on the object?

  1. 13 N
  2. 27 N
  3. 93 N
  4. 187 N (correct answer)
  5. 280 N
Explanation: This collision problem tests your understanding of impulse and momentum change, particularly when an object reverses direction. When you see a collision with a "bounce back," pay special attention to the signs of velocity - they're crucial for getting the right answer. To find the average force, you'll use the impulse-momentum theorem: Favg=ΔpΔtF_{avg} = \frac{\Delta p}{\Delta t}. First, establish a coordinate system. Let's say the initial motion toward the wall is positive (+8.0 m/s), so bouncing back is negative (-6.0 m/s). The momentum change is: Δp=m(vf−vi)=2.0 kg(−6.0−8.0) m/s=−28.0 kg⋅m/s\Delta p = m(v_f - v_i) = 2.0 \text{ kg}(-6.0 - 8.0) \text{ m/s} = -28.0 \text{ kg⋅m/s} Therefore: Favg=−28.0 kg⋅m/s0.15 s=−187 NF_{avg} = \frac{-28.0 \text{ kg⋅m/s}}{0.15 \text{ s}} = -187 \text{ N} The magnitude is 187 N, making D correct. Option A (13 N) likely comes from incorrectly using just the final speed divided by time: 6.00.15×2.0=80\frac{6.0}{0.15} \times 2.0 = 80, then making calculation errors. Option B (27 N) probably results from dividing the momentum change by the wrong time or making arithmetic mistakes. Option C (93 N) might come from using only half the actual velocity change, perhaps by not properly accounting for the direction reversal. The key insight: when an object bounces back, the velocity change isn't just the difference in speeds - it's the sum of the magnitudes because the object completely reverses direction. Always carefully track the signs in collision problems.

Question 2

A 0.75 kg ball moving horizontally at 12 m/s hits a vertical wall and bounces back horizontally at 8.0 m/s. The contact time with the wall is 0.040 s. What is the magnitude of the average force exerted by the wall on the ball?

  1. 75 N
  2. 150 N
  3. 225 N
  4. 300 N
  5. 375 N (correct answer)
Explanation: This problem tests your understanding of impulse and momentum in collision scenarios. When an object collides with a surface and bounces back, you need to carefully consider the direction of velocities and apply the impulse-momentum theorem. First, establish a coordinate system. Let's say the initial velocity is positive (+12 m/s) and the final velocity after bouncing is negative (-8.0 m/s) since it's in the opposite direction. The change in momentum is: Δp=m(vf−vi)=0.75 kg×(−8.0−12) m/s=0.75×(−20)=−15 kg⋅m/s\Delta p = m(v_f - v_i) = 0.75 \text{ kg} \times (-8.0 - 12) \text{ m/s} = 0.75 \times (-20) = -15 \text{ kg⋅m/s} Using the impulse-momentum theorem, FavgΔt=ΔpF_{avg} \Delta t = \Delta p, so: Favg=ΔpΔt=−15 kg⋅m/s0.040 s=−375 NF_{avg} = \frac{\Delta p}{\Delta t} = \frac{-15 \text{ kg⋅m/s}}{0.040 \text{ s}} = -375 \text{ N} The magnitude is 375 N. Since this isn't among the given options A-D, the correct answer must be E (not shown but implied to be 375 N). Choice A (75 N) likely comes from incorrectly using only the final velocity. Choice B (150 N) might result from using just the initial velocity or making a factor-of-2 error. Choice C (225 N) could come from calculation errors in the momentum change. Choice D (300 N) is close but suggests a slight computational mistake or rounding error. Remember: in collision problems, always pay attention to the directions of velocities before and after impact. The change in momentum includes both the initial momentum that must be stopped AND the momentum gained in the opposite direction.

Question 3

A 3.0 kg object experiences a constant net force for 2.5 s, during which its momentum changes from 18 kg⋅m/s eastward to 12 kg⋅m/s westward. What is the average net force acting on the object?

  1. 2.4 N westward
  2. 7.2 N westward
  3. 12 N westward (correct answer)
  4. 18 N westward
  5. 30 N westward
Explanation: When you encounter momentum and force problems, remember that these quantities are vectors—direction matters just as much as magnitude. This problem tests your understanding of the impulse-momentum theorem and proper vector calculations. To find the average net force, you need to calculate the change in momentum and apply Favg=ΔpΔtF_{avg} = \frac{\Delta p}{\Delta t}. First, establish a coordinate system. Let's call eastward positive and westward negative. The initial momentum is +18 kg⋅m/s+18 \text{ kg⋅m/s} and the final momentum is −12 kg⋅m/s-12 \text{ kg⋅m/s}. The change in momentum is: Δp=pf−pi=(−12)−(+18)=−30 kg⋅m/s\Delta p = p_f - p_i = (-12) - (+18) = -30 \text{ kg⋅m/s} Therefore: Favg=−30 kg⋅m/s2.5 s=−12 NF_{avg} = \frac{-30 \text{ kg⋅m/s}}{2.5 \text{ s}} = -12 \text{ N} The negative sign indicates the force points westward, giving us 12 N westward. This confirms answer C is correct. Looking at the wrong answers: A) 2.4 N westward likely comes from dividing the final momentum by time instead of using the change in momentum. B) 7.2 N westward might result from taking the difference between the magnitudes (18 - 12 = 6) and making an arithmetic error. D) 18 N westward probably comes from dividing just the initial momentum by time, ignoring the final state entirely. Study tip: In momentum problems, always define your positive direction clearly and remember that momentum change equals final minus initial—don't just subtract magnitudes. The direction of the net force will be in the direction of the momentum change.

Question 4

A 1.5 kg ball is thrown straight up with an initial speed of 12 m/s. When it reaches its maximum height, what is the magnitude of the impulse that gravity has delivered to the ball since it was thrown?

  1. 0 N⋅s
  2. 9.0 N⋅s
  3. 18 N⋅s (correct answer)
  4. 36 N⋅s
  5. 72 N⋅s
Explanation: This question tests your understanding of impulse, which is the change in momentum caused by a force acting over time. When you see projectile motion problems asking about impulse, focus on how forces change the object's momentum throughout its trajectory. To find the impulse gravity delivers, you need to calculate the change in the ball's momentum from launch to maximum height. Initially, the ball has momentum pi=mvi=(1.5 kg)(12 m/s)=18 kg⋅m/sp_i = mv_i = (1.5 \text{ kg})(12 \text{ m/s}) = 18 \text{ kg⋅m/s} upward. At maximum height, the ball momentarily stops, so pf=0p_f = 0. The change in momentum is Δp=pf−pi=0−18=−18 kg⋅m/s\Delta p = p_f - p_i = 0 - 18 = -18 \text{ kg⋅m/s}. Since impulse equals change in momentum, gravity delivered an impulse of 18 N⋅s downward. The magnitude is 18 N⋅s. Option A (0 N⋅s) incorrectly assumes no impulse occurs because the ball returns to its starting point eventually, but impulse depends on momentum change, not position change. Option B (9.0 N⋅s) might come from incorrectly using half the initial momentum, perhaps confusing this with kinetic energy relationships. Option D (36 N⋅s) likely results from calculating the total momentum change for the entire up-and-down trip, but the question only asks about the upward journey to maximum height. Remember that impulse always equals change in momentum (J=ΔpJ = \Delta p), regardless of the forces involved. Don't get distracted by the projectile motion context—focus on the momentum change between the two specified moments.

Question 5

A hockey puck slides across ice with negligible friction. A constant horizontal force of 12 N is applied to the puck for 3.0 s, during which time the puck's velocity changes from 5.0 m/s to 14 m/s in the same direction. What is the mass of the puck?

  1. 0.75 kg
  2. 1.3 kg
  3. 4.0 kg (correct answer)
  4. 5.3 kg
  5. 12 kg
Explanation: This problem tests Newton's second law and kinematics together. When you see a question involving force, time, and velocity changes, you need to connect force to acceleration, then use the motion data to find the unknown quantity. Start by finding the puck's acceleration using kinematics. With initial velocity vi=5.0 m/sv_i = 5.0 \text{ m/s}, final velocity vf=14 m/sv_f = 14 \text{ m/s}, and time t=3.0 st = 3.0 \text{ s}: a=vf−vit=14−5.03.0=3.0 m/s2a = \frac{v_f - v_i}{t} = \frac{14 - 5.0}{3.0} = 3.0 \text{ m/s}^2 Now apply Newton's second law: F=maF = ma. Solving for mass: m=Fa=12 N3.0 m/s2=4.0 kgm = \frac{F}{a} = \frac{12 \text{ N}}{3.0 \text{ m/s}^2} = 4.0 \text{ kg} The answer is C) 4.0 kg. Looking at the wrong answers: A) 0.75 kg would result if you incorrectly used a=Fma = \frac{F}{m} but made calculation errors or used wrong values. B) 1.3 kg might come from dividing the velocity change by the force directly (912≈0.75\frac{9}{12} \approx 0.75, then making additional errors). D) 5.3 kg could result from using the final velocity instead of acceleration, like 12+143.0≈8.7\frac{12 + 14}{3.0} \approx 8.7, then making computational mistakes. Study tip: For force-motion problems, always identify what you're given and what you need. If you have force and need mass, you must find acceleration first using kinematics, then apply F=maF = ma. Don't try to shortcut directly from velocities to mass.

Question 6

A 1200 kg car traveling at 20 m/s applies its brakes and comes to a complete stop in 8.0 s. Assuming the braking force is constant, what is the magnitude of the impulse delivered to the car during braking?

  1. 1500 N⋅s
  2. 3000 N⋅s
  3. 9600 N⋅s
  4. 24000 N⋅s (correct answer)
  5. 192000 N⋅s
Explanation: This problem tests your understanding of impulse, which connects force, time, and momentum change. When you see braking scenarios, think about how impulse relates to the change in momentum. The impulse-momentum theorem states that impulse equals the change in momentum: J=Δp=m(vf−vi)J = \Delta p = m(v_f - v_i). Here, the car's initial momentum is pi=(1200 kg)(20 m/s)=24,000 kg⋅m/sp_i = (1200 \text{ kg})(20 \text{ m/s}) = 24,000 \text{ kg⋅m/s}, and its final momentum is zero since it stops completely. Therefore, the change in momentum is Δp=0−24,000=−24,000 kg⋅m/s\Delta p = 0 - 24,000 = -24,000 \text{ kg⋅m/s}. The magnitude of the impulse is 24,000 N⋅s24,000 \text{ N⋅s}, making D correct. Let's examine why the other answers are wrong. Choice A (1500 N⋅s) might come from incorrectly using the time alone without properly calculating momentum change. Choice B (3000 N⋅s) could result from dividing the momentum by the time instead of recognizing that impulse equals the total momentum change. Choice C (9600 N⋅s) might arise from calculation errors, possibly confusing kinetic energy concepts with momentum. Remember that impulse problems often provide extra information (like the 8.0 s braking time) that you don't actually need when using the impulse-momentum theorem directly. While you could calculate the braking force first (F=3000 NF = 3000 \text{ N}) then multiply by time, it's more efficient to recognize that impulse simply equals the momentum change. Focus on identifying what quantity is actually being asked for and choose the most direct path to the solution.

Question 7

A 0.25 kg object initially at rest is acted upon by a net force that varies with time. If the object's velocity after 4.0 s is 16 m/s, what was the total impulse delivered to the object?

  1. 1.0 N⋅s
  2. 4.0 N⋅s (correct answer)
  3. 16 N⋅s
  4. 32 N⋅s
  5. 64 N⋅s
Explanation: This question tests the impulse-momentum theorem, one of the most fundamental relationships in mechanics. When you see a problem involving forces acting over time and resulting velocity changes, think impulse and momentum. The impulse-momentum theorem states that impulse equals the change in momentum: J=Δp=m(vf−vi)J = \Delta p = m(v_f - v_i). Since the object starts from rest, vi=0v_i = 0, so the impulse is simply J=mvf=(0.25 kg)(16 m/s)=4.0 N⋅sJ = mv_f = (0.25 \text{ kg})(16 \text{ m/s}) = 4.0 \text{ N⋅s}. This confirms answer B is correct. Let's examine why the other choices are wrong. Choice A (1.0 N⋅s) appears to come from incorrectly using just the mass and time: 0.25×4=1.00.25 \times 4 = 1.0. This completely ignores the velocity and misapplies the given values. Choice C (16 N⋅s) likely results from using only the final velocity while forgetting to multiply by mass—a common error when students rush through momentum calculations. Choice D (32 N⋅s) seems to come from multiplying the final velocity by the time interval: 16×2=3216 \times 2 = 32 (possibly using half the time) or some other incorrect combination of the given numbers. Remember: impulse problems often provide extra information that isn't directly needed for the solution. Here, the varying force and 4-second time interval are red herrings. The impulse-momentum theorem gives you a direct path to the answer using only mass and velocity change. Always identify what you actually need before diving into calculations.

Question 8

A tennis ball of mass 60 g moving at 30 m/s is hit by a racket. After the hit, the ball moves in the opposite direction at 40 m/s. If the contact time between ball and racket is 5.0 ms, what is the average force exerted by the racket on the ball?

  1. 240 N
  2. 480 N
  3. 840 N (correct answer)
  4. 1200 N
  5. 2100 N
Explanation: This is a classic impulse-momentum problem that tests your understanding of how forces relate to changes in motion. When you see collision problems involving contact time, think impulse: the change in momentum equals the impulse (force × time). Start by identifying the momentum change. The ball's initial momentum is pi=mvi=(0.060 kg)(30 m/s)=1.8 kg⋅m/sp_i = mv_i = (0.060 \text{ kg})(30 \text{ m/s}) = 1.8 \text{ kg⋅m/s} in one direction. After the hit, it moves opposite at 40 m/s, so pf=(0.060 kg)(−40 m/s)=−2.4 kg⋅m/sp_f = (0.060 \text{ kg})(-40 \text{ m/s}) = -2.4 \text{ kg⋅m/s} (negative because it's opposite). The change in momentum is Δp=pf−pi=−2.4−1.8=−4.2 kg⋅m/s\Delta p = p_f - p_i = -2.4 - 1.8 = -4.2 \text{ kg⋅m/s}. Using the impulse-momentum theorem: FΔt=ΔpF \Delta t = \Delta p, so F=ΔpΔt=−4.2 kg⋅m/s0.005 s=−840 NF = \frac{\Delta p}{\Delta t} = \frac{-4.2 \text{ kg⋅m/s}}{0.005 \text{ s}} = -840 \text{ N}. The magnitude is 840 N, which is answer C. Choice A (240 N) likely comes from forgetting the direction change and only using the final velocity. Choice B (480 N) might result from incorrectly calculating the momentum change as just the difference in speeds (70 m/s) rather than properly accounting for the direction reversal. Choice D (1200 N) could come from computational errors in the momentum calculation. Study tip: Always establish a coordinate system first and treat velocity as a vector. In collision problems, "opposite direction" means you must add the magnitudes when calculating momentum change, not subtract them.

Question 9

A 4.0 kg object moving at 5.0 m/s in the positive x-direction collides with a wall. The collision lasts 0.20 s, and the object rebounds with a velocity of 3.0 m/s in the negative x-direction. During the collision, what was the change in the object's momentum?

  1. 8.0 kg⋅m/s in the negative x-direction
  2. 20 kg⋅m/s in the negative x-direction
  3. 32 kg⋅m/s in the negative x-direction (correct answer)
  4. 2.0 kg⋅m/s in the positive x-direction
  5. 12 kg⋅m/s in the positive x-direction
Explanation: When you encounter collision problems, you're dealing with momentum change, which requires careful attention to both magnitude and direction. Momentum is a vector quantity, so you must account for the sign conventions throughout your calculation. To find the change in momentum, use Δp=pf−pi=mvf−mvi\Delta p = p_f - p_i = m v_f - m v_i. Here, the object initially moves at +5.0 m/s (positive x-direction) and rebounds at -3.0 m/s (negative x-direction). Initial momentum: pi=(4.0 kg)(+5.0 m/s)=+20 kg⋅m/sp_i = (4.0 \text{ kg})(+5.0 \text{ m/s}) = +20 \text{ kg⋅m/s} Final momentum: pf=(4.0 kg)(−3.0 m/s)=−12 kg⋅m/sp_f = (4.0 \text{ kg})(-3.0 \text{ m/s}) = -12 \text{ kg⋅m/s} Change in momentum: Δp=−12−(+20)=−32 kg⋅m/s\Delta p = -12 - (+20) = -32 \text{ kg⋅m/s} The negative sign indicates the change is in the negative x-direction, making C correct. Choice A gives 8.0 kg⋅m/s in the negative x-direction, which incorrectly subtracts the speeds (5.0 - 3.0 = 2.0) then multiplies by mass, ignoring proper vector subtraction. Choice B gives 20 kg⋅m/s in the negative x-direction, which is just the magnitude of the initial momentum—this represents calculating only the final momentum change needed to stop the object, not the full rebound. Choice D gives 2.0 kg⋅m/s in the positive x-direction, combining the wrong magnitude from choice A with the wrong direction. Remember: momentum change in collisions often involves sign changes. Always subtract initial from final momentum as vectors, and the collision duration given here is irrelevant to momentum change calculations.

Question 10

A 0.40 kg soccer ball is kicked and experiences an average force of 80 N for 0.15 s during the kick. If the ball was initially at rest, what is the magnitude of its momentum immediately after the kick?

  1. 4.8 kg⋅m/s
  2. 12 kg⋅m/s (correct answer)
  3. 30 kg⋅m/s
  4. 48 kg⋅m/s
  5. 120 kg⋅m/s
Explanation: When you encounter a problem involving forces acting over time, you're dealing with impulse and momentum. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum: J=F⋅Δt=ΔpJ = F \cdot \Delta t = \Delta p. Since the soccer ball starts from rest, its initial momentum is zero. The impulse delivered to the ball equals its final momentum. Calculate the impulse: J=F⋅Δt=80 N×0.15 s=12 N⋅sJ = F \cdot \Delta t = 80 \text{ N} \times 0.15 \text{ s} = 12 \text{ N⋅s}. Since 1 N⋅s=1 kg⋅m/s1 \text{ N⋅s} = 1 \text{ kg⋅m/s}, the final momentum is 12 kg⋅m/s. Let's examine why the other answers are incorrect: A) 4.8 kg⋅m/s results from incorrectly multiplying the mass by the time: 0.40×12=4.80.40 \times 12 = 4.8. This completely ignores the force and misapplies the momentum formula. C) 30 kg⋅m/s comes from dividing force by mass and multiplying by time: 800.40×0.15=30\frac{80}{0.40} \times 0.15 = 30. This calculates acceleration times time (which gives velocity), but then forgets to multiply by mass to get momentum. D) 48 kg⋅m/s results from multiplying force by mass: 80×0.40=3280 \times 0.40 = 32, then perhaps adding some calculation error. This ignores time entirely and misuses the impulse formula. Study tip: When you see force, time, and momentum in the same problem, immediately think impulse-momentum theorem. Always check your units—momentum must have units of kg⋅m/s, and impulse has units of N⋅s, which are equivalent.

Question 11

Two identical 2.0 kg blocks are connected by a light string. Initially, both blocks are at rest. A constant force of 24 N is applied to one block for 3.0 s in the horizontal direction. What is the magnitude of the impulse delivered to the system of two blocks?

  1. 24 N⋅s
  2. 36 N⋅s
  3. 48 N⋅s
  4. 72 N⋅s (correct answer)
  5. 144 N⋅s
Explanation: When you encounter problems involving forces and time, think about impulse-momentum concepts. Impulse is defined as the change in momentum of a system, and it also equals the net external force multiplied by the time interval. The key insight here is to treat the two connected blocks as a single system. Since the blocks are connected by a string, they move together as one unit with a combined mass of 4.0 kg. When analyzing impulse on a system, you only consider external forces - the internal tension forces between the blocks cancel out. The impulse delivered to the system equals the external force times the time interval: J=F⋅t=24 N×3.0 s=72 N⋅sJ = F \cdot t = 24 \text{ N} \times 3.0 \text{ s} = 72 \text{ N⋅s} You can verify this using the impulse-momentum theorem. The system starts at rest (initial momentum = 0). The acceleration is a=24 N4.0 kg=6.0 m/s2a = \frac{24 \text{ N}}{4.0 \text{ kg}} = 6.0 \text{ m/s}^2. After 3.0 s, the velocity is v=6.0×3.0=18 m/sv = 6.0 \times 3.0 = 18 \text{ m/s}. The final momentum is 4.0 kg×18 m/s=72 kg⋅m/s4.0 \text{ kg} \times 18 \text{ m/s} = 72 \text{ kg⋅m/s}, confirming our answer. Choice A (24 N⋅s) incorrectly uses only 1 second instead of 3 seconds. Choice B (36 N⋅s) might result from using just one block's mass incorrectly. Choice C (48 N⋅s) appears to use 2 seconds instead of 3. Remember: for impulse problems involving connected objects, treat them as a single system and focus on external forces only. Internal forces within the system don't contribute to the net impulse.

Question 12

A 1.8 kg object is subject to a net force that causes its velocity to change from 4.0 m/s eastward to 7.0 m/s eastward in 2.0 s. What is the impulse delivered to the object?

  1. 2.7 N⋅s eastward
  2. 5.4 N⋅s eastward (correct answer)
  3. 7.2 N⋅s eastward
  4. 12.6 N⋅s eastward
  5. 19.8 N⋅s eastward
Explanation: When you encounter problems involving forces acting over time and changes in velocity, you're dealing with impulse and momentum concepts. Impulse represents the change in momentum of an object and can be calculated directly from the velocity change. The impulse-momentum theorem states that impulse equals the change in momentum: J=Δp=m(vf−vi)J = \Delta p = m(v_f - v_i). Here, you have a 1.8 kg object whose velocity changes from 4.0 m/s to 7.0 m/s eastward. The impulse is: J=1.8 kg×(7.0−4.0) m/s=1.8×3.0=5.4 N⋅s eastwardJ = 1.8 \text{ kg} \times (7.0 - 4.0) \text{ m/s} = 1.8 \times 3.0 = 5.4 \text{ N⋅s eastward}. This confirms answer B is correct. Looking at the wrong answers: A (2.7 N⋅s) likely results from using only half the mass or half the velocity change—perhaps from averaging incorrectly or making an arithmetic error. C (7.2 N⋅s) could come from multiplying the mass by the final velocity rather than the velocity change, using 1.8 × 4.0 instead of 1.8 × 3.0. D (12.6 N⋅s) appears to result from adding the initial and final velocities instead of finding their difference: 1.8 × (4.0 + 7.0) = 19.8, though the exact path to 12.6 might involve other calculation errors. Remember that impulse depends only on the change in momentum, not the time duration or the specific force values. When given initial velocity, final velocity, and mass, you can find impulse directly without needing to calculate acceleration or force first.

Question 13

A hockey puck slides across the ice with an initial velocity. The coefficient of kinetic friction between the puck and ice is 0.15, and the puck has a mass of 0.25 kg.

If the puck slides for 8.0 s before coming to rest, what was the magnitude of the impulse delivered by friction during this time?

  1. 2.9 N⋅s (correct answer)
  2. 5.8 N⋅s
  3. 12 N⋅s
  4. 18 N⋅s
  5. 29 N⋅s
Explanation: When you encounter problems involving friction and motion over time, think about impulse and momentum. The impulse-momentum theorem states that impulse equals the change in momentum, which often provides the most direct path to the answer. First, let's find the friction force. The kinetic friction force is fk=μkmg=0.15×0.25×9.8=0.37 Nf_k = \mu_k mg = 0.15 \times 0.25 \times 9.8 = 0.37 \text{ N}. Since friction opposes motion, this force acts for the entire 8.0 seconds until the puck stops. Using the impulse-momentum theorem, the impulse equals the change in momentum. Since the puck comes to rest, its final momentum is zero, so the impulse equals the magnitude of the initial momentum. We can find this using J=f×t=0.37×8.0=2.96 N⋅sJ = f \times t = 0.37 \times 8.0 = 2.96 \text{ N⋅s}, which rounds to 2.9 N⋅s. Answer A (2.9 N⋅s) is correct. Answer B (5.8 N⋅s) likely comes from doubling the correct answer, perhaps from a sign error or misconception about direction. Answer C (12 N⋅s) might result from using the wrong coefficient of friction (like 0.6 instead of 0.15) or calculation errors. Answer D (18 N⋅s) is too large and suggests a fundamental error in approach, possibly confusing impulse with another quantity entirely. Remember: impulse problems often have multiple solution paths (force × time or change in momentum), but both should give the same answer. When friction is involved, always calculate the friction force first, then apply it over the given time interval.

Question 14

A baseball of mass 0.15 kg is thrown horizontally at 25 m/s. It is caught by a player who brings it to rest in 0.050 s. During the catch, what is the average force exerted on the ball by the player's glove?

  1. 37.5 N in the direction of the initial motion
  2. 75 N opposite to the direction of initial motion (correct answer)
  3. 75 N in the direction of the initial motion
  4. 150 N opposite to the direction of initial motion
  5. 375 N opposite to the direction of initial motion
Explanation: When you encounter problems involving forces and motion changes, think about impulse and momentum. The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum. Let's work through this systematically. The baseball's initial momentum is pi=mvi=(0.15 kg)(25 m/s)=3.75 kg⋅m/sp_i = mv_i = (0.15 \text{ kg})(25 \text{ m/s}) = 3.75 \text{ kg⋅m/s} in the horizontal direction. Its final momentum is pf=0p_f = 0 since it comes to rest. The change in momentum is Δp=pf−pi=0−3.75=−3.75 kg⋅m/s\Delta p = p_f - p_i = 0 - 3.75 = -3.75 \text{ kg⋅m/s}. Using the impulse-momentum theorem: FavgΔt=ΔpF_{avg} \Delta t = \Delta p, so Favg=ΔpΔt=−3.75 kg⋅m/s0.050 s=−75 NF_{avg} = \frac{\Delta p}{\Delta t} = \frac{-3.75 \text{ kg⋅m/s}}{0.050 \text{ s}} = -75 \text{ N}. The negative sign indicates the force opposes the initial motion direction, which makes physical sense—the glove must push backward to slow the ball down. Choice A gives the wrong magnitude (37.5 N instead of 75 N) and wrong direction. Choice C has the correct magnitude but wrong direction—the force can't be in the same direction as the initial motion if it's slowing the ball down. Choice D has the correct direction but wrong magnitude (150 N), likely from incorrectly doubling somewhere in the calculation. Remember: when an object slows down, the net force always opposes its motion direction. Always check that your force direction makes physical sense with what's happening in the problem.

Question 15

A force varies with time according to F(t)=6tF(t) = 6t N, where tt is in seconds. This force acts on a 2.0 kg object initially at rest. What is the impulse delivered to the object during the first 4.0 seconds?

  1. 24 N⋅s
  2. 48 N⋅s (correct answer)
  3. 72 N⋅s
  4. 96 N⋅s
  5. 144 N⋅s
Explanation: When you see a time-varying force acting on an object, you're dealing with impulse-momentum problems. Impulse is defined as the change in momentum, but it can also be calculated as the integral of force over time: J=∫F(t) dtJ = \int F(t) \, dt. Given F(t)=6tF(t) = 6t N acting for 4.0 seconds, you need to integrate this force function over the time interval. The impulse is: J=∫046t dt=6∫04t dt=6[t22]04=6(162−0)=6×8=48 N⋅sJ = \int_0^4 6t \, dt = 6 \int_0^4 t \, dt = 6 \left[\frac{t^2}{2}\right]_0^4 = 6 \left(\frac{16}{2} - 0\right) = 6 \times 8 = 48 \text{ N⋅s} This confirms answer B is correct. Looking at the wrong answers: A (24 N⋅s) represents a common error where students might calculate 6×4=246 \times 4 = 24, treating the force as constant at F=6tF = 6t when t=1t = 1 second, or making an integration mistake. C (72 N⋅s) could result from incorrectly calculating 6×42÷2+246 \times 4^2 \div 2 + 24, mixing up the integration process. D (96 N⋅s) likely comes from calculating 6×42=966 \times 4^2 = 96, forgetting to divide by 2 during the power rule integration. Remember: when force varies with time, always integrate to find impulse. Don't be tempted to use average values or plug in specific time points. The integral accounts for how the force changes continuously over the entire time interval.

Question 16

A 0.60 kg basketball bounces off the floor. Just before hitting the floor, its velocity is 8.0 m/s downward. Just after leaving the floor, its velocity is 6.0 m/s upward. If the ball is in contact with the floor for 0.025 s, what is the average force exerted by the floor on the ball during the bounce?

  1. 48 N upward
  2. 336 N upward
  3. 342 N upward (correct answer)
  4. 384 N upward
  5. 390 N upward
Explanation: When you encounter collision problems involving momentum and forces, you're dealing with Newton's second law in its impulse-momentum form: Favg⋅Δt=ΔpF_{avg} \cdot \Delta t = \Delta p, where the impulse equals the change in momentum. First, establish your coordinate system. Let's define upward as positive. The ball's initial velocity is -8.0 m/s (downward) and final velocity is +6.0 m/s (upward). The momentum change is: Δp=m(vf−vi)=0.60 kg(6.0−(−8.0))=0.60×14.0=8.4 kg⋅m/s\Delta p = m(v_f - v_i) = 0.60 \text{ kg}(6.0 - (-8.0)) = 0.60 \times 14.0 = 8.4 \text{ kg⋅m/s} Now you need to consider all forces acting on the ball during contact. The floor exerts force FfloorF_{floor} upward, while gravity exerts mg=0.60×9.8=5.88mg = 0.60 \times 9.8 = 5.88 N downward. The net force creates the momentum change: Fnet=ΔpΔt=8.40.025=336F_{net} = \frac{\Delta p}{\Delta t} = \frac{8.4}{0.025} = 336 N upward. Since Fnet=Ffloor−mgF_{net} = F_{floor} - mg, we get: Ffloor=Fnet+mg=336+5.88=342F_{floor} = F_{net} + mg = 336 + 5.88 = 342 N upward. Choice A (48 N) likely comes from incorrectly calculating the momentum change. Choice B (336 N) is the net force, but the question asks for the floor's force specifically—a common trap where students forget to account for gravity. Choice D (384 N) might result from calculation errors in the momentum change. Remember: in collision problems, always identify all forces acting on the object. The force you're asked to find might not be the same as the net force causing the momentum change.

Question 17

A 2.5 kg block slides down a frictionless incline. At the bottom, it has a momentum of magnitude 15 kg⋅m/s. A spring at the bottom exerts a varying force on the block, bringing it to rest in 0.30 s. What is the average force exerted by the spring on the block?

  1. 12.5 N up the incline
  2. 25 N up the incline
  3. 37.5 N up the incline
  4. 50 N up the incline (correct answer)
  5. 125 N up the incline
Explanation: When you encounter problems involving forces and momentum changes, think about the impulse-momentum theorem, which connects the change in momentum to the average force applied over time. Here, you need to find the average force the spring exerts on the block. The block starts with momentum of 15 kg⋅m/s down the incline and ends at rest (momentum = 0). The change in momentum is Δp=0−15=−15\Delta p = 0 - 15 = -15 kg⋅m/s (negative because momentum decreases). Using the impulse-momentum theorem: Favg×Δt=ΔpF_{avg} \times \Delta t = \Delta p Solving for average force: Favg=ΔpΔt=−15 kg⋅m/s0.30 s=−50F_{avg} = \frac{\Delta p}{\Delta t} = \frac{-15 \text{ kg⋅m/s}}{0.30 \text{ s}} = -50 N The negative sign indicates the force opposes the block's motion, so it acts up the incline. The magnitude is 50 N up the incline. Answer A (12.5 N) likely comes from incorrectly using just the block's mass (2.5 kg) divided by time, ignoring momentum entirely. Answer B (25 N) might result from using the wrong time value or making an arithmetic error. Answer C (37.5 N) could come from incorrectly calculating the momentum change or mixing up the given values. The correct answer is D: 50 N up the incline. Study tip: For momentum problems, always identify the initial and final momentum clearly, then use Favg=ΔpΔtF_{avg} = \frac{\Delta p}{\Delta t}. Pay attention to direction—forces that slow objects down oppose the motion.

Question 18

A 0.50 kg ball drops vertically and hits the ground with a speed of 12 m/s. It bounces back upward with a speed of 8.0 m/s. If the contact time with the ground is 0.020 s, what is the impulse delivered to the ball by the ground?

  1. 2.0 N⋅s upward
  2. 4.0 N⋅s upward
  3. 10 N⋅s upward (correct answer)
  4. 20 N⋅s upward
  5. 100 N⋅s upward
Explanation: When you see a collision or bounce problem, you're dealing with impulse and momentum. Impulse equals the change in momentum, and it's crucial to carefully track the direction of velocities before and after impact. Let's establish our coordinate system with upward as positive. Initially, the ball moves downward at 12 m/s, so vi=−12v_i = -12 m/s. After bouncing, it moves upward at 8.0 m/s, so vf=+8.0v_f = +8.0 m/s. The impulse delivered by the ground equals the change in momentum: J=Δp=m(vf−vi)=0.50 kg×[8.0−(−12)] m/s=0.50×20=10J = \Delta p = m(v_f - v_i) = 0.50 \text{ kg} \times [8.0 - (-12)] \text{ m/s} = 0.50 \times 20 = 10 N⋅s upward. Looking at the wrong answers: Answer A (2.0 N⋅s) likely comes from incorrectly using just the final velocity: 0.50×8.0−0.50×12=−2.00.50 \times 8.0 - 0.50 \times 12 = -2.0 N⋅s, then taking the magnitude. Answer B (4.0 N⋅s) might result from using 0.50×(12−8)=2.00.50 \times (12 - 8) = 2.0 N⋅s and doubling it somehow. Answer D (20 N⋅s) comes from forgetting to multiply by the mass: just using the velocity change of 20 m/s directly. The key insight is that impulse depends on the total velocity change, not just the final velocity. When an object reverses direction, you must add the magnitudes of the initial and final speeds to get the total change. Always define your coordinate system clearly and stick to it throughout the problem.

Question 19

A constant force of 15 N acts on a 3.0 kg object for 4.0 s. If the object was initially moving at 2.0 m/s in the direction opposite to the force, what is the object's final momentum?

  1. 6.0 kg⋅m/s in the direction opposite to the force
  2. 14 kg⋅m/s in the direction of the force
  3. 54 kg⋅m/s in the direction of the force (correct answer)
  4. 60 kg⋅m/s in the direction of the force
  5. 66 kg⋅m/s in the direction of the force
Explanation: When you encounter a problem involving force acting over time, you're dealing with the impulse-momentum theorem: impulse equals change in momentum. This connects force, time, and motion in a single framework. Let's work through this systematically. The impulse delivered by the force is J=Ft=15 N×4.0 s=60 N⋅sJ = Ft = 15 \text{ N} \times 4.0 \text{ s} = 60 \text{ N⋅s}. Since the object's initial velocity is opposite to the force direction, we'll call the force direction positive and the initial velocity negative. The initial momentum is pi=mvi=3.0 kg×(−2.0 m/s)=−6.0 kg⋅m/sp_i = mv_i = 3.0 \text{ kg} \times (-2.0 \text{ m/s}) = -6.0 \text{ kg⋅m/s}. Using the impulse-momentum theorem: J=Δp=pf−piJ = \Delta p = p_f - p_i, so pf=pi+J=−6.0+60=54 kg⋅m/sp_f = p_i + J = -6.0 + 60 = 54 \text{ kg⋅m/s} in the direction of the force. Now for the wrong answers: Choice A (6.0 kg⋅m/s opposite to force) likely comes from confusing initial momentum with final momentum. Choice B (14 kg⋅m/s in force direction) appears to result from incorrectly subtracting initial momentum from impulse rather than adding. Choice D (60 kg⋅m/s in force direction) represents a common error of forgetting about initial momentum entirely—this would only be correct if the object started from rest. Study tip: In impulse-momentum problems, always establish a clear coordinate system first. Choose one direction as positive, then consistently apply signs to all velocities, momenta, and forces. This prevents sign errors that lead to wrong answers.

Question 20

A 2.0 kg object moving at 8.0 m/s to the right collides with a wall and bounces back at 6.0 m/s to the left. If the collision lasts 0.15 s, what is the magnitude of the average force exerted by the wall on the object?

  1. 93 N
  2. 187 N (correct answer)
  3. 27 N
  4. 13 N
Explanation: The impulse-momentum theorem states that FavgΔt=ΔpF_{avg} \Delta t = \Delta p. Taking rightward as positive, the initial momentum is pi=(2.0 kg)(8.0 m/s)=16 kg⋅m/sp_i = (2.0 \text{ kg})(8.0 \text{ m/s}) = 16 \text{ kg⋅m/s}. The final momentum is pf=(2.0 kg)(−6.0 m/s)=−12 kg⋅m/sp_f = (2.0 \text{ kg})(-6.0 \text{ m/s}) = -12 \text{ kg⋅m/s}. The change in momentum is Δp=pf−pi=−12−16=−28 kg⋅m/s\Delta p = p_f - p_i = -12 - 16 = -28 \text{ kg⋅m/s}. Therefore, Favg=ΔpΔt=−28 kg⋅m/s0.15 s=−187 NF_{avg} = \frac{\Delta p}{\Delta t} = \frac{-28 \text{ kg⋅m/s}}{0.15 \text{ s}} = -187 \text{ N}. The magnitude is 187 N. Choice A uses only the final velocity change. Choice C incorrectly adds the speeds instead of considering the direction change. Choice D uses half the correct momentum change.