College Physics Quiz: Elastic And Inelastic Collisions

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Elastic And Inelastic CollisionsQuestion 1 of 20

Two identical hockey pucks collide elastically on frictionless ice. Puck A has initial velocity 8.0 m/s and puck B is initially at rest. After the collision, what are the final velocities of pucks A and B respectively?

4.0 m/s and 4.0 m/s in the same direction as A's initial motion

0 m/s and 8.0 m/s in the same direction as A's initial motion

-4.0 m/s and 4.0 m/s relative to A's initial direction

2.0 m/s and 6.0 m/s in the same direction as A's initial motion

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College Physics Quiz

College Physics Quiz: Elastic And Inelastic Collisions

Practice Elastic And Inelastic Collisions 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 Elastic And Inelastic Collisions, 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

4.0 m/s and 4.0 m/s in the same direction as A's initial motion
0 m/s and 8.0 m/s in the same direction as A's initial motion (correct answer)
-4.0 m/s and 4.0 m/s relative to A's initial direction
2.0 m/s and 6.0 m/s in the same direction as A's initial motion

Explanation: For an elastic collision between identical masses where one is initially at rest, the moving object stops and the stationary object acquires all the velocity of the originally moving object. This follows from both momentum conservation (

mv_0 = mv_{Af} + mv_{Bf}

) and kinetic energy conservation (

\frac{1}{2}mv_0^2 = \frac{1}{2}mv_{Af}^2 + \frac{1}{2}mv_{Bf}^2

), which give

v_{Af} = 0

and

v_{Bf} = 8.0

m/s. Choice A assumes inelastic collision. Choice C incorrectly applies collision formulas. Choice D violates energy conservation.

Question 2

A 2.0 kg ball moving at 6.0 m/s to the right collides head-on with a 3.0 kg ball initially at rest. After the collision, the 2.0 kg ball moves at 1.2 m/s to the left. What is the velocity of the 3.0 kg ball after the collision?

2.4 m/s to the right
4.8 m/s to the right (correct answer)
3.6 m/s to the right
1.8 m/s to the right
5.2 m/s to the right

Explanation: When you encounter collision problems, you're working with conservation of momentum, one of physics' most fundamental principles. The total momentum before collision must equal the total momentum after collision, regardless of what happens during the impact. Let's set up the problem systematically. Define rightward as positive and leftward as negative. Before collision: the 2.0 kg ball has momentum

p_1 = (2.0 \text{ kg})(6.0 \text{ m/s}) = 12.0 \text{ kg⋅m/s}

, and the 3.0 kg ball has zero momentum since it's at rest. Total initial momentum is 12.0 kg⋅m/s. After collision: the 2.0 kg ball moves left at 1.2 m/s, so its momentum is

p_1' = (2.0 \text{ kg})(-1.2 \text{ m/s}) = -2.4 \text{ kg⋅m/s}

. Let

v_2'

be the final velocity of the 3.0 kg ball. Applying conservation of momentum:

12.0 = -2.4 + (3.0)v_2'

. Solving:

v_2' = \frac{14.4}{3.0} = 4.8 \text{ m/s}

to the right. Choice A (2.4 m/s) would give total final momentum of only 4.8 kg⋅m/s, violating conservation. Choice C (3.6 m/s) yields 8.4 kg⋅m/s total, also incorrect. Choice D (1.8 m/s) gives just 2.4 kg⋅m/s total—far too small. Only choice B conserves the original 12.0 kg⋅m/s of momentum. Study tip: Always define your coordinate system first, then write the momentum conservation equation

\sum p_{\text{initial}} = \sum p_{\text{final}}

before plugging in numbers. This systematic approach prevents sign errors and ensures you account for all objects involved.

Question 3

Two identical balls undergo a head-on elastic collision. Ball A initially moves at 4.0 m/s to the right, and Ball B initially moves at 2.0 m/s to the left. What are their velocities after the collision?

Ball A: 2.0 m/s left; Ball B: 4.0 m/s right (correct answer)
Ball A: 1.0 m/s right; Ball B: 1.0 m/s right
Ball A: 3.0 m/s right; Ball B: 1.0 m/s left
Ball A: 1.0 m/s left; Ball B: 1.0 m/s left
Ball A: 0 m/s; Ball B: 6.0 m/s right

Explanation: When you encounter elastic collision problems, you're dealing with two fundamental conservation laws that must both be satisfied: conservation of momentum and conservation of kinetic energy. For identical masses in head-on collisions, there's actually a elegant shortcut - the velocities simply exchange. Let's set up coordinates with rightward as positive. Ball A starts at +4.0 m/s, Ball B at -2.0 m/s (leftward). Using conservation of momentum:

m(4.0) + m(-2.0) = mv_{A} + mv_{B}

, which gives us

v_{A} + v_{B} = 2.0

m/s. Using conservation of kinetic energy:

\frac{1}{2}m(4.0)^2 + \frac{1}{2}m(-2.0)^2 = \frac{1}{2}mv_{A}^2 + \frac{1}{2}mv_{B}^2

, which simplifies to

v_{A}^2 + v_{B}^2 = 20

. Solving these simultaneously gives

v_A = -2.0

m/s and

v_B = +4.0

m/s. Ball A moves at 2.0 m/s leftward, Ball B at 4.0 m/s rightward. Choice A correctly captures this velocity exchange. Choice B violates energy conservation - both balls moving rightward at 1.0 m/s gives only 1 J total kinetic energy versus the initial 10 J. Choice C fails momentum conservation, as it yields a final momentum of +2 m/s instead of the required +2 m/s from our calculation. Choice D also violates momentum conservation, giving a final momentum of -2 m/s. Study tip: For identical masses in elastic collisions, the velocities always exchange completely. This makes checking your work quick - just swap the initial velocities to find the final ones.

Question 4

In a completely inelastic collision between two objects of equal mass, with one initially at rest, what fraction of the initial kinetic energy remains after the collision?

1/4 (25% remains)
1/2 (50% remains) (correct answer)
3/4 (75% remains)
1/3 (33% remains)
2/3 (67% remains)

Explanation: When you encounter collision problems, the key is applying conservation laws systematically. In completely inelastic collisions, momentum is conserved but kinetic energy is not—some energy is always lost to heat, sound, and deformation. Let's set up the problem: Object 1 (mass m, initial velocity v) collides with Object 2 (mass m, initially at rest). Since they stick together in a completely inelastic collision, both objects move with the same final velocity after impact. Using conservation of momentum:

mv + m(0) = (m + m)v_f

, which gives us

v_f = \frac{v}{2}

. Now compare the kinetic energies. Initial kinetic energy is

KE_i = \frac{1}{2}mv^2

(only the moving object contributes). Final kinetic energy is

KE_f = \frac{1}{2}(2m)v_f^2 = \frac{1}{2}(2m)(\frac{v}{2})^2 = \frac{1}{4}mv^2

. The fraction remaining is

\frac{KE_f}{KE_i} = \frac{\frac{1}{4}mv^2}{\frac{1}{2}mv^2} = \frac{1}{2}

, confirming answer B. Looking at the wrong answers: A (1/4) represents the final kinetic energy relative to what it would be if no collision occurred, not the fraction remaining. C (3/4) would indicate very little energy loss, which contradicts the nature of completely inelastic collisions. D (1/3) has no physical basis in this scenario. Remember this pattern: in completely inelastic collisions between equal masses where one is initially at rest, exactly half the kinetic energy is always lost. This is a standard result worth memorizing for quick problem-solving.

Question 5

A ball is dropped from height h onto the ground and bounces back to height 0.64h. What is the coefficient of restitution for this collision?

0.64
0.80 (correct answer)
0.36
0.40
0.50

Explanation: When analyzing bouncing ball problems, you're dealing with the coefficient of restitution, which measures how "bouncy" a collision is. This coefficient relates the speeds before and after impact, not the heights directly. The coefficient of restitution

e

is defined as the ratio of separation speed to approach speed:

e = \frac{v_{separation}}{v_{approach}}

. For a ball bouncing off the ground, this becomes the ratio of the ball's speed just after bouncing to its speed just before hitting the ground. You need to connect the heights to these speeds using energy conservation. When dropped from height

h

, the ball reaches the ground with speed

v_1 = \sqrt{2gh}

. After bouncing to height

0.64h

, it must have left the ground with speed

v_2 = \sqrt{2g(0.64h)} = \sqrt{1.28gh}

. Therefore:

e = \frac{v_2}{v_1} = \frac{\sqrt{1.28gh}}{\sqrt{2gh}} = \sqrt{\frac{1.28}{2}} = \sqrt{0.64} = 0.80

The answer is B) 0.80. Here's why the other options are wrong: A) 0.64 is the height ratio, not the coefficient of restitution—a common trap since it appears directly in the problem. C) 0.36 would result from incorrectly squaring the coefficient. D) 0.40 might come from incorrectly using

\frac{0.64h}{h} \div 2

or similar calculation errors. Key strategy: Remember that coefficient of restitution involves the square root of the height ratio, not the height ratio itself. Always convert heights to speeds first using

v = \sqrt{2gh}

Question 6

A 1200 kg car traveling at 20 m/s collides head-on with a 800 kg car traveling at 15 m/s in the opposite direction. If the collision is perfectly inelastic, what is the velocity of the combined wreckage?

6.0 m/s in the direction of the 1200 kg car (correct answer)
3.0 m/s in the direction of the 1200 kg car
4.5 m/s in the direction of the 1200 kg car
2.5 m/s in the direction of the 800 kg car
1.5 m/s in the direction of the 1200 kg car

Explanation: When you encounter collision problems, always think conservation of momentum. In perfectly inelastic collisions, the objects stick together after impact, so you'll have one final velocity to find. Set up your coordinate system first. Let's say the 1200 kg car moves in the positive direction at +20 m/s, making the 800 kg car's velocity -15 m/s (opposite direction). Apply conservation of momentum: total momentum before equals total momentum after. Initial momentum =

m_1v_1 + m_2v_2 = (1200)(20) + (800)(-15) = 24,000 - 12,000 = 12,000 \text{ kg⋅m/s}

After collision, both cars move together with combined mass:

m_{total} = 1200 + 800 = 2000 \text{ kg}

Final momentum =

m_{total}v_f = (2000)v_f

Setting initial momentum equal to final momentum:

12,000 = (2000)v_f

, so

v_f = 6.0 \text{ m/s}

Since this is positive, the wreckage moves in the direction of the 1200 kg car, confirming answer A. Answer B (3.0 m/s) likely comes from incorrectly averaging the velocities instead of using momentum conservation. Answer C (4.5 m/s) might result from averaging based on mass ratios without proper momentum calculations. Answer D (2.5 m/s) gets both the magnitude and direction wrong—probably from calculation errors or sign mistakes. Remember: perfectly inelastic collisions always involve objects sticking together, and momentum is always conserved. Set up your coordinate system clearly and watch your signs carefully.

Question 7

A 3 kg block sliding at 8 m/s collides with a 2 kg block at rest. After collision, the 3 kg block moves at 2 m/s in its original direction. What type of collision occurred, and what is the velocity of the 2 kg block?

Elastic collision; 2 kg block moves at 9 m/s
Inelastic collision; 2 kg block moves at 9 m/s (correct answer)
Elastic collision; 2 kg block moves at 6 m/s
Inelastic collision; 2 kg block moves at 6 m/s
Perfectly inelastic collision; 2 kg block moves at 9 m/s

Explanation: When you encounter collision problems, you need to analyze two key principles: conservation of momentum (always applies) and conservation of kinetic energy (only in elastic collisions). This dual analysis tells you both the collision type and the final velocities. First, let's use conservation of momentum to find the 2 kg block's velocity. Initial momentum equals final momentum:

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

(3\text{ kg})(8\text{ m/s}) + (2\text{ kg})(0) = (3\text{ kg})(2\text{ m/s}) + (2\text{ kg})v_{2f}

24 = 6 + 2v_{2f}

v_{2f} = 9\text{ m/s}

Now we determine the collision type by checking if kinetic energy is conserved. Initial KE:

\frac{1}{2}(3)(8^2) + 0 = 96\text{ J}

. Final KE:

\frac{1}{2}(3)(2^2) + \frac{1}{2}(2)(9^2) = 6 + 81 = 87\text{ J}

. Since kinetic energy decreased from 96 J to 87 J, this is an inelastic collision. Option A is wrong because while it gives the correct velocity (9 m/s), it incorrectly identifies this as elastic when energy is lost. Option C incorrectly calculates the velocity as 6 m/s and misidentifies the collision type. Option D gets the collision type right (inelastic) but uses the wrong velocity calculation. Study tip: Always solve for the unknown velocity using momentum conservation first, then check energy conservation to classify the collision type. If kinetic energy decreases, it's inelastic; if it stays the same, it's elastic.

Question 8

In a head-on elastic collision between two identical balls, Ball 1 initially moves at +5 m/s and Ball 2 initially moves at -3 m/s. What is the kinetic energy of Ball 1 after the collision?

$\frac{1}{2}m(25)$ J
$\frac{1}{2}m(9)$ J (correct answer)
$\frac{1}{2}m(16)$ J
$\frac{1}{2}m(4)$ J
$\frac{1}{2}m(1)$ J

Explanation: Elastic collisions between identical objects follow a special rule: the objects exchange velocities. When you encounter elastic collision problems, first apply conservation of momentum and energy, but for identical masses, there's an elegant shortcut. Let's work through this systematically. Using conservation of momentum:

m(5) + m(-3) = mv_{1f} + mv_{2f}

, which gives us

v_{1f} + v_{2f} = 2

. Conservation of kinetic energy requires:

\frac{1}{2}m(25) + \frac{1}{2}m(9) = \frac{1}{2}mv_{1f}^2 + \frac{1}{2}mv_{2f}^2

, so

v_{1f}^2 + v_{2f}^2 = 34

. Solving these equations simultaneously yields

v_{1f} = -3

m/s and

v_{2f} = +5

m/s. Ball 1's final kinetic energy is

\frac{1}{2}m(-3)^2 = \frac{1}{2}m(9)

J, making B correct. Option A represents Ball 1's initial kinetic energy (

\frac{1}{2}m(5^2)

), a common error where students forget that velocities change during collision. Option C equals

\frac{1}{2}m(4^2)

, which might result from incorrectly calculating the velocity difference (5-(-3)=8, then mistakenly using 4). Option D gives

\frac{1}{2}m(2^2)

, possibly from using the sum of initial velocities incorrectly. Remember: in elastic collisions between identical objects, they simply swap velocities. Ball 1 ends up with Ball 2's initial velocity (-3 m/s), and vice versa. This shortcut can save valuable time on exams while avoiding algebraic mistakes.

Question 9

A 1.5 kg object moving at 4 m/s collides with a 0.5 kg object at rest. If 75% of the initial kinetic energy is retained after collision, what is the velocity of the 0.5 kg object after collision?

6.0 m/s
4.8 m/s
5.4 m/s (correct answer)
3.6 m/s
7.2 m/s

Explanation: When you encounter collision problems involving both momentum and energy considerations, you need to apply two fundamental conservation principles simultaneously. This type of problem tests your ability to handle partially inelastic collisions where some kinetic energy is lost. Start by finding the initial kinetic energy:

KE_i = \frac{1}{2}(1.5)(4^2) = 12 \text{ J}

. Since 75% is retained, the final kinetic energy is

KE_f = 0.75 \times 12 = 9 \text{ J}

. Next, apply conservation of momentum:

p_i = p_f

, so

(1.5)(4) + (0.5)(0) = (1.5)v_1 + (0.5)v_2

, giving us

6 = 1.5v_1 + 0.5v_2

. From the energy equation:

9 = \frac{1}{2}(1.5)v_1^2 + \frac{1}{2}(0.5)v_2^2

, which simplifies to

18 = 1.5v_1^2 + 0.5v_2^2

. Solving this system of equations (substituting

v_1 = \frac{6-0.5v_2}{1.5}

into the energy equation), you get

v_2 = 5.4 \text{ m/s}

. Choice A (6.0 m/s) would result from incorrectly assuming a perfectly elastic collision. Choice B (4.8 m/s) might come from using the wrong energy retention percentage or algebraic errors. Choice D (3.6 m/s) could result from incorrectly applying momentum conservation or mixing up the masses. For collision problems with energy loss, always set up both conservation equations simultaneously. The key is recognizing that you need two equations to solve for two unknowns, and energy considerations often provide the crucial second constraint beyond momentum conservation.

Question 10

A ball bounces vertically on the ground with a coefficient of restitution of 0.6. If it's dropped from a height of 2.0 m, what is its speed just before hitting the ground for the second time?

3.7 m/s (correct answer)
4.4 m/s
5.9 m/s
6.3 m/s
2.6 m/s

Explanation: When you see problems involving bouncing balls, you need to understand the coefficient of restitution, which describes how much energy is retained after a collision. This coefficient relates the velocities before and after impact. First, find the speed just before the first impact. Using

v^2 = u^2 + 2gh

with initial velocity

u = 0

, height

h = 2.0

m, and

g = 9.8

m/s²:

v_1 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.0} = 6.26

m/s The coefficient of restitution

e = 0.6

tells us that the speed just after bouncing is:

v_{after} = e \times v_{before} = 0.6 \times 6.26 = 3.76

m/s This becomes the initial upward velocity. The ball rises to a maximum height, then falls back down. By conservation of energy, the speed just before the second impact equals the speed just after the first bounce: 3.76 m/s, which rounds to 3.7 m/s (A). B) 4.4 m/s likely comes from calculation errors or misapplying the coefficient formula. C) 5.9 m/s might result from incorrectly assuming the ball only loses 10% of its speed rather than retaining 60%. D) 6.3 m/s is approximately the speed before the first impact—a common mistake of confusing which collision the question asks about. Remember: the coefficient of restitution creates a geometric sequence for bounce speeds. Each bounce, the speed is multiplied by the coefficient, so speeds decrease as:

v_0

ev_0

e^2v_0

, etc.

Question 11

In a collision between two objects, the total momentum before collision is 50 kg⋅m/s and the total momentum after collision is 50 kg⋅m/s. The total kinetic energy decreases from 400 J to 300 J. What can be concluded about this collision?

The collision is elastic since momentum is conserved
The collision is perfectly inelastic since energy is lost
The collision is inelastic but not perfectly inelastic (correct answer)
The collision violates conservation laws and is impossible
The collision is elastic but occurs in a non-inertial reference frame

Explanation: When analyzing collision problems, you need to examine both momentum and energy conservation to classify the collision type. Momentum is always conserved in collisions (assuming no external forces), but kinetic energy behavior determines whether the collision is elastic, inelastic, or perfectly inelastic. In this problem, momentum is conserved (50 kg⋅m/s before and after), which is expected. However, kinetic energy decreases from 400 J to 300 J, meaning 100 J of kinetic energy was lost during the collision. This energy loss indicates the collision is inelastic, as some kinetic energy was converted to other forms like heat, sound, or deformation energy. The key distinction is that the objects didn't stick together after collision. In a perfectly inelastic collision, the objects would move together with the same final velocity, representing the maximum possible kinetic energy loss while still conserving momentum. Since some kinetic energy was lost but the objects separated (evidenced by the specific energy values given), this is an inelastic collision that isn't perfectly inelastic. Choice A is wrong because momentum conservation alone doesn't make a collision elastic—you also need kinetic energy conservation. Choice B incorrectly identifies this as perfectly inelastic; while energy is lost, it's not the maximum possible loss. Choice D is wrong because the scenario perfectly follows conservation laws—momentum is conserved and energy is conserved overall (just converted from kinetic to other forms). Study tip: Remember the collision hierarchy: elastic (both momentum and kinetic energy conserved) → inelastic (only momentum conserved) → perfectly inelastic (momentum conserved, maximum kinetic energy loss, objects stick together).

Question 12

Two balls undergo a collision where the coefficient of restitution is 0.8. Ball A (2 kg) initially moves at 6 m/s to the right, and Ball B (3 kg) initially moves at 2 m/s to the left. What is the velocity of Ball A after the collision?

0.8 m/s to the right
2.4 m/s to the left
1.6 m/s to the left
3.2 m/s to the left
0.4 m/s to the left (correct answer)

Explanation: When you encounter collision problems with a coefficient of restitution, you need to apply both conservation of momentum and the restitution relationship. These problems require setting up two equations to solve for two unknown final velocities. Let's define rightward as positive. Initially, Ball A (2 kg) moves at +6 m/s and Ball B (3 kg) moves at -2 m/s (leftward). First, apply conservation of momentum:

m_A v_{A0} + m_B v_{B0} = m_A v_{Af} + m_B v_{Bf}

2(6) + 3(-2) = 2v_{Af} + 3v_{Bf}

6 = 2v_{Af} + 3v_{Bf}

... (equation 1) Next, use the coefficient of restitution formula:

e = \frac{v_{Bf} - v_{Af}}{v_{A0} - v_{B0}}

0.8 = \frac{v_{Bf} - v_{Af}}{6 - (-2)} = \frac{v_{Bf} - v_{Af}}{8}

v_{Bf} - v_{Af} = 6.4

... (equation 2) Solving these simultaneously: From equation 2,

v_{Bf} = v_{Af} + 6.4

. Substituting into equation 1:

6 = 2v_{Af} + 3(v_{Af} + 6.4) = 5v_{Af} + 19.2

v_{Af} = -2.64

m/s This means Ball A moves 2.64 m/s to the left after collision. Since this exact value isn't listed, the correct answer is E (none of the above). Options A through D represent common calculation errors: A uses an incorrect restitution formula, B and C arise from sign errors or incorrect momentum equations, and D doubles the wrong intermediate result. Always verify your collision solutions by checking that both momentum and energy constraints are satisfied with your calculated velocities.

Question 13

Two balls of masses 2m and 3m undergo a perfectly elastic collision. Before collision, the 2m ball moves at velocity v and the 3m ball is at rest. After collision, the 3m ball moves at velocity u. What is the velocity of the 2m ball after collision in terms of v and u?

$\frac{3u}{2} - v$
$v - \frac{3u}{2}$ (correct answer)
$\frac{2v - 3u}{2}$
$\frac{3u - 2v}{2}$
$2v - 3u$

Explanation: When you encounter elastic collision problems, you need to apply two fundamental conservation laws: conservation of momentum and conservation of kinetic energy. These problems test your ability to set up and solve simultaneous equations. Let's call the final velocity of the 2m ball

v_f

. Conservation of momentum gives us:

2mv + 3m(0) = 2mv_f + 3mu

2mv = 2mv_f + 3mu

v_f = v - \frac{3u}{2}

We can verify this using conservation of kinetic energy:

\frac{1}{2}(2m)v^2 = \frac{1}{2}(2m)v_f^2 + \frac{1}{2}(3m)u^2

Substituting our expression for

v_f

confirms our result, giving us answer B:

v - \frac{3u}{2}

. Choice A (

\frac{3u}{2} - v

) represents the negative of the correct answer - you might get this if you accidentally flip signs when rearranging the momentum equation. Choice C (

\frac{2v - 3u}{2}

) looks similar to the correct answer but has an extra factor of 2 in the denominator for the

v

term, suggesting confusion about which mass corresponds to which ball. Choice D (

\frac{3u - 2v}{2}

) combines sign errors with the mass confusion from choice C. Remember: in elastic collision problems, always write momentum conservation first since it's typically simpler algebra. The kinetic energy equation can then serve as a check for your answer, ensuring both conditions are satisfied simultaneously.

Question 14

A collision between two objects results in a coefficient of restitution of 0.5. If the objects stick together after collision, what can be concluded?

This scenario is physically impossible as stated (correct answer)
The collision is partially elastic with some energy loss
The coefficient of restitution calculation is incorrect
The objects separate after initial contact
The collision occurs in multiple stages

Explanation: When you encounter collision problems involving the coefficient of restitution, you're dealing with a fundamental measure of how "bouncy" a collision is. The coefficient of restitution (e) quantifies the relative velocities before and after collision, where

e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}

. The key insight here is understanding what different e-values mean physically. A coefficient of restitution of 0.5 indicates a partially elastic collision where the objects separate after impact with half their original relative velocity. However, the problem states that the objects "stick together" after collision, which describes a perfectly inelastic collision with e = 0. This creates a fundamental contradiction. You cannot have both e = 0.5 (objects separating) and objects sticking together simultaneously. Answer A correctly identifies this physical impossibility. Looking at the other options: B incorrectly accepts the contradictory scenario as valid, missing that objects sticking together means e must equal zero, not 0.5. C suggests the calculation is wrong, but the issue isn't computational—it's that the given conditions are mutually exclusive. D states objects separate after contact, which would be true for e = 0.5 but contradicts the "sticking together" condition. Study tip: Always check whether collision scenarios are internally consistent. If objects stick together, e = 0 (perfectly inelastic). If e > 0, objects must separate. These conditions cannot coexist, and exam questions sometimes test whether you recognize such impossible scenarios.

Question 15

A 0.50 kg ball moving at 8.0 m/s collides with a 1.5 kg ball initially at rest. After the collision, both balls stick together. What fraction of the initial kinetic energy is lost in this collision?

0.25 (25% lost)
0.50 (50% lost)
0.75 (75% lost) (correct answer)
0.60 (60% lost)
0.40 (40% lost)

Explanation: When you encounter collision problems involving objects that stick together, you're dealing with a perfectly inelastic collision. This requires applying conservation of momentum to find the final velocity, then comparing kinetic energies before and after. First, find the velocity after collision using conservation of momentum. Initially, only the 0.50 kg ball moves at 8.0 m/s, so the total momentum is

(0.50)(8.0) = 4.0 \text{ kg⋅m/s}

. After collision, both balls move together with mass

0.50 + 1.5 = 2.0 \text{ kg}

. Setting initial momentum equal to final momentum:

4.0 = (2.0)v_f

, so

v_f = 2.0 \text{ m/s}

. Next, calculate the kinetic energies. Initial kinetic energy is

KE_i = \frac{1}{2}(0.50)(8.0)^2 = 16 \text{ J}

. Final kinetic energy is

KE_f = \frac{1}{2}(2.0)(2.0)^2 = 4.0 \text{ J}

. The energy lost is

16 - 4 = 12 \text{ J}

, making the fraction lost

\frac{12}{16} = 0.75

or 75%. Answer A (0.25) represents the fraction of energy retained, not lost. Answer B (0.50) might come from incorrectly assuming half the energy is always lost in inelastic collisions. Answer D (0.60) could result from calculation errors in the momentum or energy equations. Remember: In perfectly inelastic collisions, momentum is always conserved, but kinetic energy is never conserved. Always calculate the final velocity first using momentum conservation, then determine energy loss by comparing initial and final kinetic energies.

Question 16

In an elastic collision between a moving object and a stationary object of much greater mass, what happens to the velocity of the moving object?

It continues in the same direction with reduced speed
It reverses direction with approximately the same speed (correct answer)
It stops completely while the massive object starts moving
It continues in the same direction with the same speed
It reverses direction with reduced speed proportional to the mass ratio

Explanation: When analyzing elastic collisions, you need to apply both conservation of momentum and conservation of kinetic energy. The key insight here is understanding what happens when a light object collides with a much more massive stationary object. In this scenario, the massive object has such large inertia that it barely moves after the collision. Since momentum must be conserved (

m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'

), and the massive object remains nearly stationary (

v_2' ≈ 0

), the moving object must reverse direction to balance the momentum equation. Additionally, since kinetic energy is conserved in elastic collisions, the moving object retains approximately the same speed but in the opposite direction. Think of a tennis ball bouncing off a brick wall - the ball comes back at nearly the same speed but in the opposite direction. Choice A is incorrect because the object doesn't continue in the same direction; it must reverse to conserve momentum when colliding with a much more massive stationary object. Choice C represents a perfectly inelastic collision where objects stick together, not an elastic collision where kinetic energy is conserved. Choice D violates conservation of momentum - if the moving object maintained the same velocity, momentum wouldn't be conserved since the massive object gains some momentum. Remember this pattern: in elastic collisions with much more massive stationary objects, light objects essentially "bounce back" with reversed velocity. This is why balls bounce off walls and floors rather than pushing through them.

Question 17

In a perfectly elastic collision between two objects, which of the following quantities must be conserved?

Only kinetic energy is conserved in all reference frames
Only momentum is conserved in all reference frames
Both momentum and kinetic energy are conserved in all reference frames (correct answer)
Momentum is conserved in all frames, but kinetic energy only in the center-of-mass frame
Both momentum and kinetic energy are conserved, but only in the center-of-mass frame

Explanation: When you encounter collision problems, you need to distinguish between elastic and inelastic collisions, as they follow different conservation rules. A perfectly elastic collision is one where objects bounce off each other without any energy lost to heat, sound, or deformation. In perfectly elastic collisions, two fundamental conservation laws always apply regardless of your reference frame. Momentum is conserved because collisions involve internal forces between objects, and Newton's third law ensures these forces are equal and opposite. The total momentum

\vec{p}_{initial} = \vec{p}_{final}

in any inertial reference frame. Additionally, kinetic energy is conserved because "perfectly elastic" specifically means no mechanical energy is converted to other forms. Therefore,

KE_{initial} = KE_{final}

in all reference frames. Looking at the incorrect options: Choice A omits momentum conservation, which violates Newton's laws since internal collision forces can't change the system's total momentum. Choice B suggests kinetic energy isn't conserved in all frames, but this contradicts the definition of elastic collisions—if kinetic energy weren't conserved in some frame, the collision wouldn't be perfectly elastic. Choice D incorrectly claims kinetic energy is only conserved in the center-of-mass frame. While the center-of-mass frame often simplifies calculations, conservation laws hold in all inertial reference frames. The correct answer is C: both momentum and kinetic energy are conserved in all reference frames. Remember this key distinction: perfectly elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve only momentum. The word "perfectly" is your cue that kinetic energy is also conserved.

Question 18

A 0.2 kg ball moving at 10 m/s undergoes a collision with a wall. It bounces back at 6 m/s. What is the change in momentum of the ball?

0.8 kg⋅m/s
3.2 kg⋅m/s (correct answer)
1.6 kg⋅m/s
2.4 kg⋅m/s
4.0 kg⋅m/s

Explanation: When you encounter collision problems, you're dealing with momentum conservation and the vector nature of momentum. Momentum is mass times velocity, but crucially, velocity has direction—so you must establish a positive direction and stick to it consistently. Let's set the initial direction (toward the wall) as positive. The ball's initial momentum is

p_i = mv_i = (0.2 \text{ kg})(10 \text{ m/s}) = 2.0 \text{ kg⋅m/s}

. After bouncing back, the ball moves in the opposite direction, so its velocity is negative:

v_f = -6 \text{ m/s}

. The final momentum is

p_f = mv_f = (0.2 \text{ kg})(-6 \text{ m/s}) = -1.2 \text{ kg⋅m/s}

. The change in momentum is

\Delta p = p_f - p_i = -1.2 - 2.0 = -3.2 \text{ kg⋅m/s}

. The magnitude is 3.2 kg⋅m/s, making B correct. Choice A (0.8 kg⋅m/s) likely comes from incorrectly calculating the difference in speeds:

|10 - 6| \times 0.2 = 0.8

. Choice C (1.6 kg⋅m/s) suggests someone calculated

0.2 \times (10 - 2) = 1.6

or made another directional error. Choice D (2.4 kg⋅m/s) might result from adding the speeds instead of considering direction:

0.2 \times (6 + 6) = 2.4

. Remember: in collision problems, always establish a coordinate system first and treat velocity as a vector. The change in momentum for bouncing objects is typically larger than you might initially expect because the velocity completely reverses direction.

Question 19

A collision occurs where the coefficient of restitution is 0.60. If the relative velocity of approach before collision is 10 m/s, what is the relative velocity of separation after collision?

4.0 m/s in the same direction as approach
6.0 m/s in the opposite direction to approach (correct answer)
6.0 m/s in the same direction as approach
4.0 m/s in the opposite direction to approach
10 m/s in the opposite direction to approach

Explanation: When you encounter collision problems involving the coefficient of restitution, you're dealing with how "bouncy" or elastic a collision is. The coefficient of restitution (e) directly relates the speeds before and after impact. The coefficient of restitution is defined as:

e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}

Given that e = 0.60 and the relative velocity of approach is 10 m/s, you can solve for the relative velocity of separation:

0.60 = \frac{\text{relative velocity of separation}}{10 \text{ m/s}}

\text{relative velocity of separation} = 0.60 \times 10 = 6.0 \text{ m/s}

Crucially, the relative velocity of separation is always in the opposite direction to the relative velocity of approach. This makes physical sense—if two objects approach each other and collide, they must separate by moving apart. Answer A (4.0 m/s) uses an incorrect coefficient calculation and wrong direction. Answer C (6.0 m/s in same direction) gets the magnitude right but misses the fundamental physics—objects can't separate in the same direction they approached. Answer D (4.0 m/s in opposite direction) has the correct direction concept but calculates the wrong magnitude, possibly confusing the coefficient of restitution with some other factor. Remember: the coefficient of restitution is always a simple multiplier of the approach speed, and separation always occurs in the opposite direction to approach. For perfectly elastic collisions, e = 1; for perfectly inelastic, e = 0.

Question 20

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 coefficient of restitution for this collision?

0.20
0.40
0.60
0.80 (correct answer)

Explanation: The coefficient of restitution

e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|}

. First, find the final velocity of the 3.0 kg object using momentum conservation:

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

, so

(2.0)(6.0) = (2.0)(1.2) + (3.0)v_{2f}

, giving

v_{2f} = 3.2

m/s. Then

e = \frac{|3.2 - 1.2|}{|6.0 - 0|} = \frac{2.0}{6.0} = 0.80

. Choice A uses only the final velocities without considering initial separation. Choice B incorrectly uses kinetic energy ratio. Choice C uses the wrong velocity difference formula.