College Physics Quiz: Momentum
20 questions · exam conditions
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MomentumQuestion 1 of 20

A 0.50 kg ball moving at 8.0 m/s strikes a wall and rebounds elastically. The contact time with the wall is 0.020 s. What is the magnitude of the impulse delivered to the ball during the collision?

200 N⋅s
4.0 N⋅s
0.16 N⋅s
8.0 N⋅s
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College Physics Quiz

College Physics Quiz: Momentum

Practice Momentum in College Physics 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 Momentum, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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 0.50 kg ball moving at 8.0 m/s strikes a wall and rebounds elastically. The contact time with the wall is 0.020 s. What is the magnitude of the impulse delivered to the ball during the collision?

  1. 200 N⋅s
  2. 4.0 N⋅s
  3. 0.16 N⋅s
  4. 8.0 N⋅s (correct answer)
Explanation: When you encounter collision problems involving impulse, remember that impulse equals the change in momentum, regardless of the contact time or forces involved. For this elastic collision, you need to find the ball's momentum change. Initially, the ball moves toward the wall at 8.0 m/s, so its momentum is pi=(0.50 kg)(8.0 m/s)=4.0 kg⋅m/sp_i = (0.50 \text{ kg})(8.0 \text{ m/s}) = 4.0 \text{ kg⋅m/s}. Since the collision is elastic, the ball rebounds at the same speed but in the opposite direction, giving final momentum pf=(0.50 kg)(−8.0 m/s)=−4.0 kg⋅m/sp_f = (0.50 \text{ kg})(-8.0 \text{ m/s}) = -4.0 \text{ kg⋅m/s}. The impulse is J=Δp=pf−pi=−4.0−4.0=−8.0 N⋅sJ = \Delta p = p_f - p_i = -4.0 - 4.0 = -8.0 \text{ N⋅s}. The magnitude is 8.0 N⋅s, making D correct. Choice A (200 N⋅s) likely comes from incorrectly using F⋅tF \cdot t where students calculate force as F=ma=m⋅ΔvΔtF = ma = m \cdot \frac{\Delta v}{\Delta t}, then multiply by time again. Choice B (4.0 N⋅s) represents only half the momentum change—perhaps forgetting that elastic collisions involve both stopping and reversing direction. Choice C (0.16 N⋅s) might result from dividing instead of adding the momentum magnitudes. The key insight: impulse depends only on momentum change, not on contact time or force details. In elastic collisions, always account for the complete velocity reversal—the object doesn't just stop, it bounces back at the same speed.

Question 2

A rocket in space ejects 150 kg of fuel at a speed of 2000 m/s relative to the rocket. If the rocket's mass (excluding the ejected fuel) is 1800 kg, what is the magnitude of the rocket's momentum change?

  1. 3.0 × 10⁵ kg⋅m/s because momentum is conserved for the fuel-rocket system (correct answer)
  2. 1.67 × 10⁵ kg⋅m/s because this accounts for the reduced rocket mass after fuel ejection
  3. 4.5 × 10⁶ kg⋅m/s because both fuel and rocket momenta must be considered equally
  4. 2.5 × 10⁴ kg⋅m/s because only the mass ratio determines the momentum transfer
  5. 1.2 × 10⁵ kg⋅m/s because the fuel speed relative to space differs from rocket speed
Explanation: When you encounter rocket propulsion problems, you're dealing with conservation of momentum in an isolated system. The key insight is that the rocket and ejected fuel form a closed system where total momentum is conserved. Initially, both the rocket and fuel are at rest (assuming the rocket starts from rest), so the total momentum is zero. After fuel ejection, the momentum of the ejected fuel must equal the magnitude of the rocket's momentum change, but in opposite directions. The momentum of the ejected fuel is straightforward: pfuel=mfuel×vfuel=150 kg×2000 m/s=3.0×105 kg⋅m/sp_{fuel} = m_{fuel} \times v_{fuel} = 150 \text{ kg} \times 2000 \text{ m/s} = 3.0 \times 10^5 \text{ kg⋅m/s} Since momentum is conserved, the rocket must gain momentum of equal magnitude in the opposite direction. Answer A is correct because it properly applies conservation of momentum to find the momentum change magnitude: 3.0×105 kg⋅m/s3.0 \times 10^5 \text{ kg⋅m/s}. Answer B incorrectly attempts to account for mass reduction by dividing the fuel momentum by some factor, but momentum conservation doesn't work this way. Answer C makes the error of somehow adding or multiplying the momenta rather than recognizing they're equal and opposite. Answer D uses an incorrect calculation that doesn't follow from any valid physics principle for this scenario. Remember: in rocket problems, focus on momentum conservation for the entire system. The momentum gained by ejected mass equals the momentum change of the rocket, regardless of mass ratios or complex corrections.

Question 3

A 2.0 kg object moving at 6.0 m/s collides head-on with a 3.0 kg object initially at rest. After the collision, the 2.0 kg object moves at 1.2 m/s in its original direction. What is the momentum of the 3.0 kg object after the collision?

  1. 9.6 kg⋅m/s in the original direction of the 2.0 kg object (correct answer)
  2. 12.0 kg⋅m/s in the original direction of the 2.0 kg object
  3. 3.6 kg⋅m/s in the original direction of the 2.0 kg object
  4. 9.6 kg⋅m/s opposite to the original direction of the 2.0 kg object
  5. 14.4 kg⋅m/s in the original direction of the 2.0 kg object
Explanation: When you encounter collision problems, you're dealing with conservation of momentum—one of physics's most fundamental principles. In any collision where no external forces act, the total momentum before equals the total momentum after. Let's set up the problem systematically. Before collision: the 2.0 kg object has momentum p1=(2.0 kg)(6.0 m/s)=12.0 kg⋅m/sp_1 = (2.0 \text{ kg})(6.0 \text{ m/s}) = 12.0 \text{ kg⋅m/s}, and the 3.0 kg object at rest has zero momentum. Total initial momentum is 12.0 kg⋅m/s. After collision: the 2.0 kg object moves at 1.2 m/s in its original direction, so its momentum is p1′=(2.0 kg)(1.2 m/s)=2.4 kg⋅m/sp_1' = (2.0 \text{ kg})(1.2 \text{ m/s}) = 2.4 \text{ kg⋅m/s}. Using conservation of momentum: 12.0=2.4+p2′12.0 = 2.4 + p_2', where p2′p_2' is the 3.0 kg object's final momentum. Therefore, p2′=9.6 kg⋅m/sp_2' = 9.6 \text{ kg⋅m/s} in the original direction. Choice A gives the correct momentum magnitude and direction. Choice B (12.0 kg⋅m/s) incorrectly assumes the 3.0 kg object gained all the original momentum, ignoring that the 2.0 kg object still moves forward. Choice C (3.6 kg⋅m/s) represents a calculation error—perhaps subtracting incorrectly or confusing mass with momentum values. Choice D has the right magnitude but wrong direction; since both objects move forward after collision, momentum transfer occurs in the same direction. Remember: always define a positive direction first, then apply conservation of momentum algebraically. This prevents sign errors and helps you track whether objects move in the same or opposite directions after collision.

Question 4

Two ice skaters, each with mass 60 kg, are initially at rest and holding opposite ends of a rope. One skater pulls on the rope and they move toward each other. When they meet, the first skater has a speed of 2.0 m/s. What was the speed of the second skater just before they met?

  1. 2.0 m/s in the same direction as the first skater
  2. 2.0 m/s in the opposite direction from the first skater (correct answer)
  3. 1.0 m/s in the opposite direction from the first skater
  4. 4.0 m/s in the opposite direction from the first skater
  5. 0 m/s because the system started from rest
Explanation: When you encounter problems involving objects moving toward each other from rest, think conservation of momentum. Since no external forces act horizontally on this system of two skaters, the total momentum must remain constant throughout their motion. Initially, both skaters are at rest, so the total momentum is zero. By conservation of momentum, the total momentum must remain zero when they meet. This means: m1v1+m2v2=0m_1v_1 + m_2v_2 = 0 Since both skaters have the same mass (60 kg), we get: 60 kg×2.0 m/s+60 kg×v2=060 \text{ kg} \times 2.0 \text{ m/s} + 60 \text{ kg} \times v_2 = 0 Solving for v2v_2: v2=−2.0 m/sv_2 = -2.0 \text{ m/s} The negative sign indicates the second skater moves in the opposite direction from the first skater, with a speed of 2.0 m/s. This makes perfect sense—equal masses moving toward each other must have equal and opposite velocities to maintain zero total momentum. Answer A is wrong because both skaters can't move in the same direction and still approach each other. Answer C gives the wrong speed (1.0 m/s instead of 2.0 m/s)—this might result from incorrectly thinking unequal speeds are needed. Answer D has the right direction but wrong magnitude (4.0 m/s)—this could come from mistakenly doubling the first skater's speed. Study tip: In momentum conservation problems, always check that your final answer makes physical sense. Equal masses starting from rest and moving toward each other must have equal speeds in opposite directions.

Question 5

A hockey puck of mass 0.16 kg is struck by a stick, causing its velocity to change from 2.0 m/s eastward to 12.0 m/s westward in 0.02 s. What is the magnitude of the average force exerted on the puck during this time interval?

  1. 80 N because the impulse equals the change in momentum magnitude
  2. 112 N because both initial and final momentum magnitudes must be considered (correct answer)
  3. 32 N because only the speed increase from 2.0 to 12.0 m/s matters
  4. 224 N because the force must overcome both initial momentum and create final momentum
  5. 96 N because the time interval affects the momentum change calculation
Explanation: When you encounter problems involving forces and velocity changes, think about impulse and momentum. The impulse-momentum theorem states that impulse equals the change in momentum: FavgΔt=Δp=mΔvF_{avg} \Delta t = \Delta p = m\Delta v. The key insight here is properly handling the direction change. Since eastward and westward are opposite directions, treat them as positive and negative. Let's say east is positive: initial velocity is +2.0 m/s, final velocity is -12.0 m/s. The change in velocity is: Δv=vf−vi=(−12.0)−(+2.0)=−14.0 m/s\Delta v = v_f - v_i = (-12.0) - (+2.0) = -14.0 \text{ m/s} Using the impulse-momentum theorem: Favg=mΔvΔt=(0.16 kg)(−14.0 m/s)0.02 s=−112 NF_{avg} = \frac{m\Delta v}{\Delta t} = \frac{(0.16 \text{ kg})(-14.0 \text{ m/s})}{0.02 \text{ s}} = -112 \text{ N} The magnitude is 112 N, confirming answer B. Now for the wrong answers: A (80 N) incorrectly calculates the change in speed rather than velocity, using ∣12.0−2.0∣=10.0|12.0 - 2.0| = 10.0 m/s instead of the proper vector subtraction. C (32 N) makes the same conceptual error but with different arithmetic. D (224 N) double-counts by incorrectly adding the magnitudes of initial and final momentum, showing a misunderstanding of vector subtraction. Study tip: Always establish a coordinate system when dealing with direction changes. Vector subtraction for opposite directions means you add the magnitudes, but you must account for the signs in your calculation. Practice identifying when "change" means vector subtraction versus simple magnitude differences.

Question 6

A 75 kg person jumps horizontally from a 1200 kg boat initially at rest. If the person's velocity relative to the water is 3.0 m/s horizontally, what is the velocity of the boat relative to the water immediately after the person jumps?

  1. 0.19 m/s in the direction opposite to the person's motion (correct answer)
  2. 0.25 m/s in the same direction as the person's motion
  3. 0.19 m/s in the same direction as the person's motion
  4. 0.063 m/s in the direction opposite to the person's motion
  5. 3.0 m/s in the direction opposite to the person's motion
Explanation: When you encounter problems involving people jumping from boats, think conservation of momentum. Since no external horizontal forces act on the person-boat system, the total momentum before and after the jump must be equal. Initially, both the person and boat are at rest, so the total momentum is zero. After the jump, the momentum must still sum to zero: mperson⋅vperson+mboat⋅vboat=0m_{person} \cdot v_{person} + m_{boat} \cdot v_{boat} = 0 Substituting the given values: (75 kg)(3.0 m/s)+(1200 kg)(vboat)=0(75 \text{ kg})(3.0 \text{ m/s}) + (1200 \text{ kg})(v_{boat}) = 0 Solving for the boat's velocity: vboat=−(75)(3.0)1200=−2251200=−0.1875 m/sv_{boat} = -\frac{(75)(3.0)}{1200} = -\frac{225}{1200} = -0.1875 \text{ m/s} The negative sign indicates the boat moves opposite to the person's direction, and rounding gives 0.19 m/s. This confirms answer A is correct. Answer B incorrectly suggests the boat moves in the same direction as the person, violating momentum conservation. Answer C makes the same directional error as B while getting the magnitude right through coincidence. Answer D has the correct direction but uses an incorrect calculation—possibly dividing by the total mass (1275 kg) instead of just the boat's mass, giving 2251275≈0.18\frac{225}{1275} \approx 0.18, which rounds to their stated 0.063 m/s. Remember: in momentum conservation problems, when the initial momentum is zero, the final momenta must be equal in magnitude but opposite in direction. Always check that your answer makes physical sense—the lighter object should have higher speed than the heavier one.

Question 7

Two objects undergo a head-on elastic collision. Object 1 (mass = 2.0 kg, initial velocity = 5.0 m/s) collides with object 2 (mass = 3.0 kg, initial velocity = -2.0 m/s). What is the total momentum of the system after the collision?

  1. 4.0 kg⋅m/s because momentum is conserved in all collisions (correct answer)
  2. 16.0 kg⋅m/s because both objects contribute positively to the final momentum
  3. 10.0 kg⋅m/s because this uses only the momentum of the faster object
  4. 14.0 kg⋅m/s because elastic collisions increase the total system momentum
  5. 0 kg⋅m/s because elastic collisions conserve both momentum and kinetic energy
Explanation: When you encounter collision problems, the fundamental principle to remember is that momentum is always conserved in any collision, regardless of whether it's elastic or inelastic. This is a direct consequence of Newton's third law and applies to isolated systems where no external forces act. To find the total momentum after collision, you need to calculate the total momentum before collision. Momentum is a vector quantity, so direction matters. Taking rightward as positive: Object 1 has momentum p1=(2.0 kg)(5.0 m/s)=10.0 kg⋅m/sp_1 = (2.0 \text{ kg})(5.0 \text{ m/s}) = 10.0 \text{ kg⋅m/s}, and Object 2 has momentum p2=(3.0 kg)(−2.0 m/s)=−6.0 kg⋅m/sp_2 = (3.0 \text{ kg})(-2.0 \text{ m/s}) = -6.0 \text{ kg⋅m/s}. The total initial momentum is 10.0+(−6.0)=4.0 kg⋅m/s10.0 + (-6.0) = 4.0 \text{ kg⋅m/s}. Since momentum is conserved, the final momentum must also be 4.0 kg⋅m/s. Choice A is correct because it recognizes momentum conservation and provides the right value. Choice B incorrectly treats the negative velocity as positive, ignoring the vector nature of momentum. Choice C mistakenly assumes only one object's momentum matters, overlooking that both objects contribute to the system's total momentum. Choice D contains a fundamental misconception—elastic collisions conserve both momentum and kinetic energy, but they don't create additional momentum. Remember: momentum conservation is universal in collisions. Always treat momentum as a vector, carefully accounting for direction, and calculate the total initial momentum to find the total final momentum.

Question 8

A 1500 kg car moving at 20 m/s collides with a stationary 1000 kg car. After the collision, both cars move together. What fraction of the initial kinetic energy is lost in this completely inelastic collision?

  1. 0.40 because this represents the momentum transfer efficiency in inelastic collisions (correct answer)
  2. 0.60 because most kinetic energy is typically lost in completely inelastic collisions
  3. 0.25 because the mass ratio determines the energy loss fraction
  4. 0.75 because only a small fraction of energy is retained in such collisions
  5. 0.50 because equal energy is lost and retained in momentum conservation
Explanation: When you encounter completely inelastic collision problems, you need to apply both conservation of momentum and calculate kinetic energy before and after impact. In completely inelastic collisions, the objects stick together, maximizing energy loss while momentum remains conserved. First, find the final velocity using conservation of momentum: m1v1+m2v2=(m1+m2)vfm_1v_1 + m_2v_2 = (m_1 + m_2)v_f. With the moving car (1500 kg at 20 m/s) and stationary car (1000 kg), you get: 1500×20+1000×0=(1500+1000)vf1500 \times 20 + 1000 \times 0 = (1500 + 1000)v_f, so vf=12 m/sv_f = 12 \text{ m/s}. Next, compare kinetic energies. Initial: KEi=12(1500)(20)2=300,000 JKE_i = \frac{1}{2}(1500)(20)^2 = 300,000 \text{ J}. Final: KEf=12(2500)(12)2=180,000 JKE_f = \frac{1}{2}(2500)(12)^2 = 180,000 \text{ J}. The energy lost is 120,000 J, making the fraction lost 120,000300,000=0.40\frac{120,000}{300,000} = 0.40. Answer A correctly identifies this value and gives the right reasoning about momentum transfer efficiency. Answer B incorrectly assumes most energy is always lost in these collisions without calculating. Answer C wrongly suggests the mass ratio alone determines energy loss—while mass affects the outcome, you must calculate the actual energy change. Answer D makes an unfounded claim about typical energy retention without doing the physics. Remember: completely inelastic collision problems always require two steps—use momentum conservation to find final velocity, then compare kinetic energies before and after. Don't rely on rules of thumb about "typical" energy loss.

Question 9

A force of 120 N acts on a 3.0 kg object for 0.25 s, changing its velocity from 8.0 m/s to 18.0 m/s in the same direction. What is the momentum of the object after the force stops acting?

  1. 54 kg⋅m/s because this represents the final momentum state (correct answer)
  2. 30 kg⋅m/s because this accounts for the momentum change during force application
  3. 24 kg⋅m/s because this represents the initial momentum before force application
  4. 84 kg⋅m/s because both initial momentum and impulse contribute to final momentum
  5. 48 kg⋅m/s because this averages the initial and final momentum values
Explanation: This problem tests your understanding of momentum and impulse. When forces act on objects, you need to determine the final state after all interactions are complete. Momentum is simply mass times velocity: p=mvp = mv. After the force stops acting, the object has a final velocity of 18.0 m/s and mass of 3.0 kg, so its momentum is p=(3.0 kg)(18.0 m/s)=54 kg⋅m/sp = (3.0 \text{ kg})(18.0 \text{ m/s}) = 54 \text{ kg⋅m/s}. The question asks for the momentum "after the force stops acting," which means you want the final momentum state. Let's examine why the other answers miss the mark. Answer B (30 kg⋅m/s) represents the change in momentum during the force application: Δp=m(vf−vi)=3.0(18.0−8.0)=30 kg⋅m/s\Delta p = m(v_f - v_i) = 3.0(18.0 - 8.0) = 30 \text{ kg⋅m/s}. However, this is the impulse, not the final momentum. Answer C (24 kg⋅m/s) gives you the initial momentum: pi=(3.0)(8.0)=24 kg⋅m/sp_i = (3.0)(8.0) = 24 \text{ kg⋅m/s}, but again, this isn't what's being asked. Answer D (84 kg⋅m/s) incorrectly adds the initial momentum and impulse together, which double-counts the physics. The key insight is recognizing what the question is actually asking for. When you see "momentum after" or "final momentum," calculate mvmv using the final velocity. Don't get distracted by all the given information about forces and time intervals—sometimes the most direct approach using basic definitions is correct.

Question 10

Two identical 2.0 kg blocks slide toward each other on a frictionless surface. Block A moves at 5.0 m/s to the right, and block B moves at 3.0 m/s to the left. After an elastic collision, block A moves at 3.0 m/s to the left. What is the momentum of block B after the collision?

  1. 10.0 kg⋅m/s to the right because momentum conservation determines this value (correct answer)
  2. 6.0 kg⋅m/s to the left because this maintains the system's total momentum
  3. 4.0 kg⋅m/s to the right because elastic collisions preserve momentum magnitudes
  4. 16.0 kg⋅m/s to the right because both blocks contribute to the final momentum
  5. 2.0 kg⋅m/s to the right because momentum is shared equally in elastic collisions
Explanation: When you encounter collision problems, momentum conservation is your fundamental tool. The total momentum before collision must equal the total momentum after collision, regardless of the collision type. Let's establish our coordinate system with rightward as positive. Initially, block A has momentum pA,i=(2.0 kg)(5.0 m/s)=10.0 kg⋅m/sp_{A,i} = (2.0 \text{ kg})(5.0 \text{ m/s}) = 10.0 \text{ kg⋅m/s} and block B has momentum pB,i=(2.0 kg)(−3.0 m/s)=−6.0 kg⋅m/sp_{B,i} = (2.0 \text{ kg})(-3.0 \text{ m/s}) = -6.0 \text{ kg⋅m/s}. The total initial momentum is 10.0+(−6.0)=4.0 kg⋅m/s10.0 + (-6.0) = 4.0 \text{ kg⋅m/s}. After collision, block A moves at 3.0 m/s to the left, giving pA,f=(2.0)(−3.0)=−6.0 kg⋅m/sp_{A,f} = (2.0)(-3.0) = -6.0 \text{ kg⋅m/s}. Using conservation of momentum: ptotal,initial=ptotal,finalp_{total,initial} = p_{total,final}, so 4.0=−6.0+pB,f4.0 = -6.0 + p_{B,f}. Therefore, pB,f=10.0 kg⋅m/sp_{B,f} = 10.0 \text{ kg⋅m/s} (rightward). Answer A is correct because momentum conservation directly determines this value. Answer B incorrectly calculates 6.0 kg⋅m/s leftward, which would give a total final momentum of -12.0 kg⋅m/s, violating conservation. Answer C states 4.0 kg⋅m/s rightward based on the misconception that elastic collisions preserve momentum magnitudes rather than total momentum. Answer D incorrectly suggests 16.0 kg⋅m/s, apparently adding momentum magnitudes without considering directions. For collision problems, always start with momentum conservation—it applies to all collisions. Set up your coordinate system clearly, calculate initial total momentum, then use the given final information to find the unknown quantities.

Question 11

Two identical freight cars, each of mass 2.5 × 10⁴ kg, roll along a track. Car A moves at 4.0 m/s and car B moves at 1.0 m/s in the opposite direction. They couple together upon collision. What is the momentum of the coupled system immediately after collision?

  1. 7.5 × 10⁴ kg⋅m/s in the direction car A was originally moving (correct answer)
  2. 1.25 × 10⁵ kg⋅m/s in the direction car A was originally moving
  3. 2.5 × 10⁴ kg⋅m/s in the direction car A was originally moving
  4. 1.5 × 10⁵ kg⋅m/s because both cars contribute positively to total momentum
  5. 0 kg⋅m/s because the cars have equal mass and opposite directions
Explanation: When you encounter collision problems, you're dealing with conservation of momentum, one of physics's most fundamental principles. The key insight is that momentum is a vector quantity, so direction matters crucially. To find the total momentum after collision, you need the momentum of each car before collision. Car A has momentum pA=(2.5×104 kg)(4.0 m/s)=1.0×105 kg⋅m/sp_A = (2.5 × 10^4 \text{ kg})(4.0 \text{ m/s}) = 1.0 × 10^5 \text{ kg⋅m/s} in its direction of motion. Car B moves in the opposite direction, so its momentum is pB=(2.5×104 kg)(−1.0 m/s)=−2.5×104 kg⋅m/sp_B = (2.5 × 10^4 \text{ kg})(-1.0 \text{ m/s}) = -2.5 × 10^4 \text{ kg⋅m/s} (negative because it's opposite to car A's direction). The total momentum is ptotal=1.0×105−2.5×104=7.5×104 kg⋅m/sp_{total} = 1.0 × 10^5 - 2.5 × 10^4 = 7.5 × 10^4 \text{ kg⋅m/s} in car A's original direction. This confirms answer A is correct. Answer B (1.25×1051.25 × 10^5) incorrectly adds the momentum magnitudes without considering opposite directions. Answer C (2.5×1042.5 × 10^4) appears to represent just one car's momentum, missing the vector addition entirely. Answer D (1.5×1051.5 × 10^5) makes the critical error of treating both momenta as positive, ignoring that the cars move in opposite directions. Remember: in collision problems, always establish a coordinate system first and assign positive/negative signs based on direction. Momentum conservation means the vector sum before collision equals the vector sum after collision—direction is just as important as magnitude.

Question 12

A 0.15 kg baseball moving at 40 m/s is caught by a catcher whose glove moves 0.30 m while bringing the ball to rest. Assuming constant deceleration, what is the magnitude of the average momentum of the baseball during the catching process?

  1. 3.0 kg⋅m/s because this represents the average of initial and final momenta (correct answer)
  2. 6.0 kg⋅m/s because the initial momentum determines the average throughout the process
  3. 1.5 kg⋅m/s because the catching distance affects the momentum calculation
  4. 4.5 kg⋅m/s because the momentum decreases non-linearly during constant deceleration
  5. 0 kg⋅m/s because the final momentum is zero after the ball is caught
Explanation: When analyzing momentum during processes with constant acceleration, remember that momentum changes linearly with time, making the average momentum simply the arithmetic mean of initial and final values. Let's calculate the average momentum step by step. The initial momentum is pi=mvi=(0.15 kg)(40 m/s)=6.0 kg⋅m/sp_i = mv_i = (0.15 \text{ kg})(40 \text{ m/s}) = 6.0 \text{ kg⋅m/s}. The final momentum is pf=mvf=(0.15 kg)(0 m/s)=0 kg⋅m/sp_f = mv_f = (0.15 \text{ kg})(0 \text{ m/s}) = 0 \text{ kg⋅m/s} since the ball comes to rest. Under constant deceleration, momentum decreases linearly with time, so the average momentum is pavg=pi+pf2=6.0+02=3.0 kg⋅m/sp_{avg} = \frac{p_i + p_f}{2} = \frac{6.0 + 0}{2} = 3.0 \text{ kg⋅m/s}. Choice A is correct because it properly applies the linear relationship between momentum and time during constant acceleration. Choice B incorrectly assumes the initial momentum represents the average throughout the process, ignoring that momentum continuously decreases. Choice C suggests the catching distance affects momentum calculation, but distance only determines the acceleration magnitude—not the momentum averaging process. Choice D claims momentum decreases non-linearly, which contradicts the constant deceleration given in the problem. Remember that under constant acceleration, any quantity that changes linearly with time (like velocity and momentum) has an average equal to the arithmetic mean of initial and final values. This principle applies whether the object is speeding up or slowing down.

Question 13

A 0.20 kg projectile moving horizontally at 400 m/s embeds in a 4.8 kg wooden block initially at rest on a horizontal surface. What is the momentum of the block-projectile system immediately after impact?

  1. 80 kg⋅m/s in the original direction of the projectile (correct answer)
  2. 2000 kg⋅m/s because this accounts for the combined mass after collision
  3. 16 kg⋅m/s because the wooden block's large mass reduces the total momentum
  4. 96 kg⋅m/s because both objects contribute to the final system momentum
  5. 320 kg⋅m/s because the momentum distributes according to the mass ratio
Explanation: This problem tests conservation of momentum in collisions. When you encounter collision problems, remember that momentum is always conserved in the absence of external forces, regardless of whether the collision is elastic or inelastic. To find the system's momentum immediately after impact, you need to apply conservation of momentum. The total momentum before collision equals the total momentum after collision. Before impact, only the projectile is moving: pinitial=mprojectile×vprojectile=0.20 kg×400 m/s=80 kg⋅m/sp_{initial} = m_{projectile} \times v_{projectile} = 0.20 \text{ kg} \times 400 \text{ m/s} = 80 \text{ kg⋅m/s}. The wooden block contributes zero momentum since it's at rest. Since momentum is conserved, the block-projectile system must have 80 kg⋅m/s immediately after impact, making choice A correct. Choice B incorrectly multiplies the total mass (5.0 kg) by the projectile's initial velocity, which violates conservation of momentum. Choice C suggests the wooden block's mass somehow reduces total momentum, which is impossible since momentum is a conserved quantity. Choice D appears to add arbitrary contributions from both objects without proper calculation. The key insight is that momentum conservation doesn't depend on the masses involved or the type of collision. Whether the projectile bounces off or embeds in the block, the total momentum of the system remains constant at the moment of impact. Study tip: In collision problems, always start by calculating the initial momentum of the entire system. This value must equal the final momentum, regardless of how the objects interact during the collision.

Question 14

During a tennis match, a 0.057 kg ball traveling at 45 m/s is returned by a racket, leaving at 38 m/s in the opposite direction. If the contact time between ball and racket is 4.0 × 10⁻³ s, what is the average force exerted by the racket on the ball?

  1. 1200 N in the direction of the ball's final velocity (correct answer)
  2. 540 N in the direction of the ball's final velocity
  3. 100 N in the direction opposite to the ball's initial velocity
  4. 1200 N in the direction opposite to the ball's initial velocity
  5. 640 N in the direction of the ball's final velocity
Explanation: When you encounter collision or impact problems involving force and time, you're dealing with impulse-momentum concepts. The key insight is using the impulse-momentum theorem: the impulse (force × time) equals the change in momentum. First, establish your coordinate system. Let's say the initial direction is positive, so the ball starts at +45 m/s and ends at -38 m/s (opposite direction). The momentum change is: Δp=m(vf−vi)=0.057 kg(−38−45) m/s=0.057(−83)=−4.73 kg⋅m/s\Delta p = m(v_f - v_i) = 0.057 \text{ kg}(-38 - 45) \text{ m/s} = 0.057(-83) = -4.73 \text{ kg⋅m/s} Using the impulse-momentum theorem: F⋅Δt=ΔpF \cdot \Delta t = \Delta p, so F=ΔpΔt=−4.734.0×10−3=−1183 N≈−1200 NF = \frac{\Delta p}{\Delta t} = \frac{-4.73}{4.0 \times 10^{-3}} = -1183 \text{ N} \approx -1200 \text{ N} The negative sign means the force is opposite to our chosen positive direction (the initial velocity direction), which makes physical sense—the racket must push against the ball's original motion to reverse it. Option A correctly states 1200 N in the direction of the ball's final velocity, which is indeed opposite to the initial direction. Option B gives the wrong magnitude (540 N), likely from calculation errors. Option C has both wrong magnitude (100 N) and describes the direction confusingly. Option D gives the correct magnitude but incorrectly describes the direction as "opposite to initial velocity" rather than "same as final velocity"—these mean the same thing, but the wording in D is less precise than A. Remember: in collision problems, always define your coordinate system clearly and check that your force direction makes physical sense.

Question 15

A 0.25 kg ball is dropped from rest and hits the ground with a speed of 6.0 m/s. It bounces back upward with a speed of 4.0 m/s. What is the magnitude of the impulse delivered to the ball by the ground?

  1. 2.5 N⋅s because this represents the momentum change magnitude (correct answer)
  2. 0.5 N⋅s because only the speed reduction matters for impulse calculation
  3. 1.5 N⋅s because impulse equals the final momentum magnitude only
  4. 5.0 N⋅s because the ball changes direction completely during the collision
  5. 1.0 N⋅s because this is the difference between initial and final momentum magnitudes
Explanation: When you encounter collision problems involving impulse, focus on the impulse-momentum theorem: impulse equals the change in momentum, or J=Δp=m(vf−vi)J = \Delta p = m(v_f - v_i). The key insight here is properly defining your coordinate system and recognizing that velocity is a vector. Let's choose downward as positive. Initially, the ball moves downward at +6.0 m/s just before impact. After bouncing, it moves upward at -4.0 m/s (negative because it's opposite to our chosen positive direction). The momentum change is: Δp=m(vf−vi)=0.25 kg×(−4.0−(+6.0)) m/s=0.25×(−10.0)=−2.5 kg⋅m/s\Delta p = m(v_f - v_i) = 0.25 \text{ kg} \times (-4.0 - (+6.0)) \text{ m/s} = 0.25 \times (-10.0) = -2.5 \text{ kg⋅m/s} The magnitude of impulse is ∣J∣=∣Δp∣=2.5 N⋅s|J| = |\Delta p| = 2.5 \text{ N⋅s}. This represents the total momentum change magnitude, making A correct. B is wrong because impulse depends on the complete velocity change, not just speed reduction. The direction change is crucial. C incorrectly suggests impulse equals only final momentum magnitude (mvf=1.0 N⋅smv_f = 1.0 \text{ N⋅s}), ignoring the initial momentum. D gets the wrong numerical value—while it's true the ball changes direction completely, the calculation must account for both initial and final momentum values properly. Study tip: Always establish a clear coordinate system in collision problems and remember that impulse = change in momentum, not just final momentum. The direction change often doubles the momentum change compared to what students initially expect.

Question 16

A 60 kg astronaut floating in space throws a 2.0 kg tool at 15 m/s relative to herself. What is the magnitude of the astronaut's recoil momentum?

  1. 30 kg⋅m/s because momentum conservation requires equal and opposite momenta (correct answer)
  2. 900 kg⋅m/s because the astronaut's larger mass affects the momentum calculation
  3. 0.50 kg⋅m/s because the tool's small mass minimizes the recoil effect
  4. 450 kg⋅m/s because both the tool and astronaut momenta contribute to recoil
  5. 15 kg⋅m/s because the speed determines the momentum magnitude regardless of mass
Explanation: When you encounter problems involving objects separating in space, you're dealing with conservation of momentum in an isolated system. Since there are no external forces acting on the astronaut-tool system, the total momentum before and after the throw must be equal. Initially, both the astronaut and tool are at rest, so the total momentum is zero. After the throw, the momentum must still sum to zero, meaning the astronaut and tool must have equal and opposite momenta. The tool's momentum is: ptool=mtool×vtool=2.0 kg×15 m/s=30 kg⋅m/sp_{tool} = m_{tool} \times v_{tool} = 2.0 \text{ kg} \times 15 \text{ m/s} = 30 \text{ kg⋅m/s} Therefore, the astronaut must have a momentum of 30 kg⋅m/s in the opposite direction to maintain zero total momentum. Looking at the wrong answers: Choice B (900 kg⋅m/s) incorrectly multiplies the tool's momentum by the astronaut's mass, which has no basis in momentum conservation. Choice C (0.50 kg⋅m/s) appears to divide the tool's mass by its velocity, which is backwards and ignores the conservation principle entirely. Choice D (450 kg⋅m/s) seems to add or multiply various quantities without applying conservation laws correctly. The key insight is that momentum conservation requires the magnitudes to be equal, regardless of the different masses involved. The astronaut moves slower than the tool due to her larger mass, but the momentum magnitudes are identical. Study tip: In momentum conservation problems, always start by identifying the total momentum before the event—it must equal the total momentum after the event.

Question 17

A 2.0 kg object moving at 6.0 m/s collides with a 3.0 kg object initially at rest. After the collision, the 2.0 kg object moves at 2.0 m/s in the same direction as its initial motion. What is the momentum of the 3.0 kg object immediately after the collision?

  1. 8.0 kg⋅m/s in the same direction as the initial motion of the 2.0 kg object (correct answer)
  2. 6.0 kg⋅m/s in the same direction as the initial motion of the 2.0 kg object
  3. 4.0 kg⋅m/s in the opposite direction to the initial motion of the 2.0 kg object
  4. 12 kg⋅m/s in the same direction as the initial motion of the 2.0 kg object
Explanation: Using conservation of momentum: Initial momentum = Final momentum. Initial: pi=(2.0)(6.0)+(3.0)(0)=12p_i = (2.0)(6.0) + (3.0)(0) = 12 kg⋅m/s. Final: pf=(2.0)(2.0)+p3kg=4.0+p3kgp_f = (2.0)(2.0) + p_{3kg} = 4.0 + p_{3kg}. Therefore: 12=4.0+p3kg12 = 4.0 + p_{3kg}, so p3kg=8.0p_{3kg} = 8.0 kg⋅m/s in the positive direction. Choice B uses the initial velocity of the first object. Choice C assumes the second object moves backward. Choice D uses the total initial momentum.

Question 18

Two identical cars, each with mass mm, are involved in a collision. Car A travels east at speed vv, while car B travels north at the same speed vv. They stick together after collision. What is the magnitude of the momentum of the combined wreckage immediately after the collision?

  1. mv2mv\sqrt{2} (correct answer)
  2. 2mv2mv
  3. mvmv
  4. mv2\frac{mv}{\sqrt{2}}
Explanation: The initial momenta are perpendicular: Car A has momentum mvmv east, Car B has momentum mvmv north. Using vector addition, the total momentum has components px=mvp_x = mv and py=mvp_y = mv. The magnitude is ∣p⃗∣=(mv)2+(mv)2=mv2|\vec{p}| = \sqrt{(mv)^2 + (mv)^2} = mv\sqrt{2}. Choice B would be correct if momenta were parallel. Choice C ignores one car. Choice D incorrectly divides by 2\sqrt{2}.

Question 19

A railway car of mass 8000 kg moving at 4.0 m/s collides and couples with an identical stationary railway car. A 60 kg person standing in the moving car continues moving forward at 3.0 m/s relative to the ground immediately after collision. What was the person's velocity relative to the moving car before the collision?

  1. 7.0 m/s in the direction of the car's motion
  2. 3.0 m/s in the direction of the car's motion
  3. 4.0 m/s in the direction of the car's motion
  4. 1.0 m/s in the direction of the car's motion (correct answer)
Explanation: When you encounter collision problems involving people moving within vehicles, you need to carefully track reference frames and apply conservation of momentum to the entire system. First, let's find the velocity of the coupled cars after collision. Using conservation of momentum for the two cars: m1v1+m2v2=(m1+m2)vfm_1v_1 + m_2v_2 = (m_1 + m_2)v_f, where 8000×4.0+8000×0=16000×vf8000 \times 4.0 + 8000 \times 0 = 16000 \times v_f. This gives vf=2.0 m/sv_f = 2.0 \text{ m/s} for the coupled cars. Now we can find the person's initial velocity relative to the moving car. The person moves at 3.0 m/s relative to the ground after collision, while the coupled cars move at 2.0 m/s. So the person's velocity relative to the cars after collision is 3.0−2.0=1.0 m/s3.0 - 2.0 = 1.0 \text{ m/s}. Since no external forces act on the person during the brief collision, their velocity relative to the car remains constant. Therefore, the person was moving at 1.0 m/s relative to the moving car before collision. Answer A (7.0 m/s) incorrectly adds the car's initial speed to the person's final ground speed. Answer B (3.0 m/s) confuses the person's final ground velocity with their relative velocity. Answer C (4.0 m/s) assumes the person was at rest relative to the car initially, which contradicts the given information. Remember: in collision problems with multiple moving parts, always identify your reference frame first, then apply conservation principles systematically to each component of the system.

Question 20

A force F(t)=20tF(t) = 20t N (where tt is in seconds) acts on a 5.0 kg object initially at rest. What is the momentum of the object at t=3.0t = 3.0 s?

  1. 180 kg⋅m/s
  2. 60 kg⋅m/s
  3. 90 kg⋅m/s (correct answer)
  4. 300 kg⋅m/s
Explanation: When you encounter a time-varying force acting on an object, you need to connect force to momentum through the impulse-momentum theorem. This states that the impulse (integral of force over time) equals the change in momentum. Since the object starts at rest, its initial momentum is zero, so the final momentum equals the impulse delivered by the force. For F(t)=20tF(t) = 20t, you calculate the impulse from t=0t = 0 to t=3.0t = 3.0 s: J=∫03.0F(t) dt=∫03.020t dt=20⋅t22∣03.0=10t2∣03.0=10(9)−10(0)=90 N⋅sJ = \int_0^{3.0} F(t) \, dt = \int_0^{3.0} 20t \, dt = 20 \cdot \frac{t^2}{2}\Big|_0^{3.0} = 10t^2\Big|_0^{3.0} = 10(9) - 10(0) = 90 \text{ N⋅s} Since impulse equals change in momentum, and 1 N⋅s=1 kg⋅m/s1 \text{ N⋅s} = 1 \text{ kg⋅m/s}, the momentum at t=3.0t = 3.0 s is 90 kg⋅m/s. Looking at the wrong answers: (A) 180 kg⋅m/s likely comes from incorrectly doubling the result or mishandling the integration. (B) 60 kg⋅m/s suggests using F=20(3)=60F = 20(3) = 60 N and multiplying by 1 second instead of properly integrating. (D) 300 kg⋅m/s might result from multiplying the final force value F(3)=60F(3) = 60 N by the mass (5 kg), confusing momentum with some other quantity. Remember: for time-varying forces, always use the impulse-momentum theorem. Don't try shortcuts with average forces unless you're certain about the calculation—integration is your reliable path to the correct answer.