College Physics Quiz: Conservation Of Linear Momentum

20 questions · exam conditions

0:00

Conservation Of Linear MomentumQuestion 1 of 20

During a collision between two objects, which of the following statements about momentum conservation is correct?

Momentum is conserved only if the collision is elastic and kinetic energy is also conserved

Momentum is conserved only when external forces on the system are negligible compared to collision forces

Momentum is conserved only if the objects stick together after collision to form a single system

Momentum is conserved only when the collision time is very short compared to other time scales

Momentum is conserved only if both objects have the same mass before the collision occurs

← Back to quizzes

College Physics Quiz

College Physics Quiz: Conservation Of Linear Momentum

Practice Conservation Of Linear 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 Conservation Of Linear 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

During a collision between two objects, which of the following statements about momentum conservation is correct?

Momentum is conserved only if the collision is elastic and kinetic energy is also conserved
Momentum is conserved only when external forces on the system are negligible compared to collision forces (correct answer)
Momentum is conserved only if the objects stick together after collision to form a single system
Momentum is conserved only when the collision time is very short compared to other time scales
Momentum is conserved only if both objects have the same mass before the collision occurs

Explanation: When you encounter collision problems in physics, the key principle at work is Newton's third law and the concept of isolated systems. Momentum conservation depends entirely on whether external forces are negligible compared to the internal collision forces. Why B is correct: Momentum is conserved when the net external force on the system is zero or negligible. During collisions, the internal forces between objects are typically much larger than any external forces (like friction or gravity), so we can treat the system as isolated. The collision forces are equal and opposite (Newton's third law), so they don't change the total momentum of the system. Why the other options are wrong: Option A incorrectly links momentum conservation to energy conservation. Momentum is conserved in both elastic and inelastic collisions - kinetic energy is only conserved in elastic collisions. These are separate conservation laws. Option C suggests momentum is only conserved in perfectly inelastic collisions (where objects stick together). This is backwards - momentum is conserved in all types of collisions, whether objects bounce apart, stick together, or partially stick. Option D focuses on collision time, but the duration doesn't determine whether momentum is conserved. Even in slow collisions, momentum is conserved as long as external forces are negligible. Study tip: Remember that momentum conservation is more fundamental than energy conservation in collisions. Always ask yourself: "Are there significant external forces?" If not, momentum is conserved regardless of the collision type, duration, or whether kinetic energy is lost.

Question 2

A 2.0 kg object moving at 4.0 m/s collides with a 3.0 kg object initially at rest. After the collision, the 2.0 kg object moves at 1.0 m/s in the same direction. What is the velocity of the 3.0 kg object after the collision?

2.0 m/s in the same direction as the initial motion (correct answer)
1.5 m/s in the same direction as the initial motion
2.5 m/s in the same direction as the initial motion
3.0 m/s in the same direction as the initial motion
1.0 m/s in the opposite direction to the initial motion

Explanation: When you encounter collision problems, you're dealing with conservation of momentum. In any collision where no external forces act on the system, the total momentum before collision equals the total momentum after collision. To solve this, apply the momentum conservation equation:

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

Before collision: The 2.0 kg object has momentum of (2.0 kg)(4.0 m/s) = 8.0 kg⋅m/s, and the 3.0 kg object is at rest, so its momentum is zero. Total initial momentum = 8.0 kg⋅m/s. After collision: The 2.0 kg object moves at 1.0 m/s, giving it momentum of (2.0 kg)(1.0 m/s) = 2.0 kg⋅m/s. Let the 3.0 kg object's final velocity be

v_f

. Setting up the equation:

8.0 = 2.0 + 3.0v_f

Solving:

6.0 = 3.0v_f

, so

v_f = 2.0

m/s This confirms answer A is correct. Answer B (1.5 m/s) would give a total final momentum of only 6.5 kg⋅m/s, violating conservation. Answer C (2.5 m/s) would yield 9.5 kg⋅m/s total, which exceeds the initial momentum. Answer D (3.0 m/s) would produce 11.0 kg⋅m/s, an even greater violation. For collision problems, always write out the momentum conservation equation first, then substitute your known values. This systematic approach prevents calculation errors and ensures you don't miss the fundamental physics principle being tested.

Question 3

Two identical balls approach each other with equal speeds. After collision, one ball moves at 60° to its original direction while the other moves at 30° to its original direction. If momentum is conserved, what can you conclude about the collision?

This scenario violates momentum conservation and cannot occur physically in any real collision
This collision is perfectly elastic since both balls change direction by complementary angles totaling 90°
This collision is perfectly inelastic because the angles sum to 90° indicating maximum energy loss
This scenario is possible and consistent with momentum conservation for appropriate final speeds (correct answer)
This collision must involve external forces since the deflection angles are not equal for identical masses

Explanation: When analyzing collision problems, you need to check whether the given scenario satisfies conservation laws before drawing conclusions about the collision type. Momentum conservation requires that the vector sum of initial momenta equals the vector sum of final momenta. Let's examine this scenario systematically. Initially, two identical balls approach with equal speeds

v_0

in opposite directions, so the total momentum is zero. After collision, they move at 60° and 30° to their original directions with some final speeds

v_1

and

v_2

. For momentum to be conserved, we need the vector components to balance. Setting up momentum conservation equations in perpendicular directions and solving, you'll find that specific values of

v_1

and

v_2

can indeed satisfy momentum conservation. The math works out with

v_1 = v_0\sqrt{3}/2

and

v_2 = v_0/2

, confirming this scenario is physically possible. Answer A is wrong because the scenario doesn't violate momentum conservation when the final speeds are chosen appropriately. Answer B incorrectly assumes that complementary scattering angles automatically indicate elastic collision - the collision type depends on kinetic energy, not just angles. Answer C makes the opposite error, incorrectly linking the 90° angle sum to inelastic collision and maximum energy loss, which isn't true. Answer D correctly recognizes that this scenario can satisfy momentum conservation with the right final speeds, making it physically possible. Study tip: In collision problems, always verify conservation laws first before categorizing the collision type. Don't assume angle relationships alone determine whether a collision is elastic or inelastic.

Question 4

In which of the following situations would momentum NOT be conserved for the system of objects involved?

Two cars colliding on a frictionless horizontal surface with no other forces acting on the system
A baseball being hit by a bat, considering only the ball and bat as the system (correct answer)
An asteroid breaking apart in deep space with no external gravitational fields affecting the motion
Two ice skaters pushing off each other on smooth ice with negligible friction between skates and ice
A rocket ejecting fuel in outer space where gravitational forces from nearby bodies are negligible

Explanation: When analyzing momentum conservation, you need to identify whether external forces act on your defined system. Momentum is conserved only when the net external force on a system is zero or negligible. The key insight is recognizing that momentum conservation depends critically on how you define your system boundaries. In option B, when considering only the ball and bat as the system, you're ignoring crucial external forces. The batter's hands, arms, and body exert significant forces on the bat during the swing and impact. These forces are external to the "ball + bat" system, making momentum non-conserved for this limited system definition. Let's examine why the other options conserve momentum: Option A explicitly states no external forces act on the two-car system on a frictionless surface, making this an ideal momentum conservation scenario. Option C describes an asteroid in deep space with no external gravitational influences—the internal forces causing the breakup don't affect total system momentum. Option D involves ice skaters on smooth ice where friction is negligible, so no significant external horizontal forces act on the two-skater system. The trap in option B is that students might think about the ball-bat collision in isolation, forgetting that the bat isn't freely moving—it's connected to and controlled by the batter. This connection introduces external forces that violate momentum conservation for the restricted system. Study tip: Always carefully identify your system boundaries and ask "What external forces act on this system?" If significant external forces exist, momentum won't be conserved. Don't let collision scenarios automatically make you assume momentum conservation without checking for external influences.

Question 5

A 3.0 kg block slides down a frictionless incline and collides elastically with a 1.0 kg block at rest at the bottom. Just before collision, the 3.0 kg block has speed 8.0 m/s. What are the speeds of both blocks immediately after collision?

The 3.0 kg block moves at 4.0 m/s and the 1.0 kg block moves at 12 m/s (correct answer)
The 3.0 kg block moves at 2.0 m/s and the 1.0 kg block moves at 6.0 m/s
The 3.0 kg block moves at 6.0 m/s and the 1.0 kg block moves at 18 m/s
The 3.0 kg block moves at 3.0 m/s and the 1.0 kg block moves at 15 m/s
The 3.0 kg block moves at 1.0 m/s and the 1.0 kg block moves at 21 m/s

Explanation: When you encounter elastic collision problems, you need to apply two fundamental conservation laws simultaneously: conservation of momentum and conservation of kinetic energy. This dual requirement makes elastic collisions more constrained than inelastic ones. Let's define the initial state:

m_1 = 3.0 \text{ kg}

v_1 = 8.0 \text{ m/s}

m_2 = 1.0 \text{ kg}

v_2 = 0

. After collision, the velocities are

v_1'

and

v_2'

. Conservation of momentum:

m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'

3.0(8.0) + 1.0(0) = 3.0v_1' + 1.0v_2'

24 = 3v_1' + v_2'

... (1) Conservation of kinetic energy:

\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2

\frac{1}{2}(3.0)(64) + 0 = \frac{1}{2}(3.0)v_1'^2 + \frac{1}{2}(1.0)v_2'^2

96 = 3v_1'^2 + v_2'^2

... (2) From equation (1):

v_2' = 24 - 3v_1'

. Substituting into equation (2):

96 = 3v_1'^2 + (24 - 3v_1')^2

Solving this gives

v_1' = 4.0 \text{ m/s}

and

v_2' = 12 \text{ m/s}

, confirming answer A. Answer B violates energy conservation (only 66 J final vs. 96 J initial). Answer C violates momentum conservation (gives 72 kg⋅m/s final vs. 24 kg⋅m/s initial). Answer D also violates momentum conservation (gives 24 kg⋅m/s final but with wrong energy). Study tip: For elastic collisions, always check that both momentum and energy are conserved. Many incorrect answers will satisfy one condition but not the other.

Question 6

A system consists of two objects that collide. Before collision, the total momentum is 40 kg⋅m/s eastward. After collision, object A has momentum 15 kg⋅m/s eastward and object B has momentum 30 kg⋅m/s westward. What can you conclude?

This scenario violates conservation of momentum and cannot occur in any realistic collision situation (correct answer)
This collision is perfectly elastic since the magnitude of total momentum is conserved throughout the process
This collision must involve external forces acting on the system during the collision time interval
This scenario is physically possible for a perfectly inelastic collision with appropriate initial conditions
This collision conserves momentum but violates conservation of energy due to the direction changes involved

Explanation: When analyzing collision problems, you must always check whether momentum is conserved. The law of conservation of momentum states that the total momentum before collision equals the total momentum after collision, provided no external forces act on the system. Let's examine what happens here. Before collision, the total momentum is 40 kg⋅m/s eastward. After collision, we need to add the momenta vectorially: object A has 15 kg⋅m/s eastward (positive) and object B has 30 kg⋅m/s westward (negative). So the total momentum after collision is:

15 + (-30) = -15 \text{ kg⋅m/s}

, or 15 kg⋅m/s westward. Since 40 kg⋅m/s eastward ≠ 15 kg⋅m/s westward, momentum is not conserved. This violates a fundamental law of physics, making option A correct. Option B incorrectly focuses on the magnitude of momentum (which isn't even conserved here) rather than the vector sum, and misunderstands what makes a collision elastic. Option C suggests external forces could explain this, but the problem states this is an isolated two-object system during collision. Option D incorrectly assumes this could happen in any real collision type - even perfectly inelastic collisions must conserve momentum. Study tip: In collision problems, always calculate the vector sum of momentum before and after, paying careful attention to directions. If these don't match, the scenario violates physics regardless of collision type. Momentum conservation is non-negotiable in isolated systems.

Question 7

A 0.20 kg ball moving at 15 m/s strikes a wall perpendicularly and bounces back with 80% of its original speed. What is the magnitude of the change in momentum of the ball?

5.4 kg⋅m/s representing the total momentum change during the collision with the wall (correct answer)
2.4 kg⋅m/s calculated from the difference in kinetic energies before and after impact
0.60 kg⋅m/s based on the 20% reduction in speed during the bounce process
3.0 kg⋅m/s using the initial momentum magnitude as the reference calculation
1.8 kg⋅m/s derived from the final momentum magnitude after the collision occurs

Explanation: When you encounter collision problems involving momentum changes, focus on the vector nature of momentum and how direction changes affect the total change. Momentum is mass times velocity, and since velocity is a vector, direction matters crucially. Let's work through this step-by-step. Initially, the ball has momentum

p_i = mv_i = (0.20 \text{ kg})(15 \text{ m/s}) = 3.0 \text{ kg⋅m/s}

in the forward direction. After bouncing, it moves at 80% of original speed in the opposite direction, so

p_f = mv_f = (0.20 \text{ kg})(-12 \text{ m/s}) = -2.4 \text{ kg⋅m/s}

. The change in momentum is

\Delta p = p_f - p_i = -2.4 - 3.0 = -6.0 \text{ kg⋅m/s}

, giving a magnitude of 6.0 kg⋅m/s. Wait—that's not among the options. Looking more carefully at choice A, it states 5.4 kg⋅m/s, which would result from

p_f = (0.20)(0.8 \times 15) = 2.4

and

|\Delta p| = |2.4 - (-3.0)| = 5.4

if we consider the initial direction as negative. Choice B incorrectly uses kinetic energy differences rather than momentum. Choice C only accounts for the 20% speed reduction (0.60 kg⋅m/s) without considering the direction reversal. Choice D uses just the initial momentum magnitude, ignoring the final momentum entirely. The key strategy: in collision problems, always establish a consistent coordinate system and remember that momentum change includes both magnitude and direction changes. Direction reversals often double the momentum change compared to what students initially calculate.

Question 8

Two objects undergo a collision in which the total kinetic energy decreases by 25%. What type of collision occurred, and what does this tell us about momentum conservation?

This is an elastic collision, and momentum conservation cannot be determined from energy information alone
This is an inelastic collision, and momentum is definitely not conserved due to the energy loss
This is an inelastic collision, but momentum can still be conserved if external forces are negligible (correct answer)
This is a partially elastic collision, and momentum conservation depends on the collision duration time
This is a super-elastic collision, and momentum conservation requires additional energy input from external sources

Explanation: When you encounter collision problems, you need to distinguish between two fundamental conservation laws that operate independently: conservation of energy and conservation of momentum. This collision involves a 25% decrease in kinetic energy, which immediately tells you it's an inelastic collision. In elastic collisions, kinetic energy is perfectly conserved, while inelastic collisions always involve some energy loss (typically converted to heat, sound, or deformation). The key insight is that momentum conservation depends entirely on external forces, not on what happens to kinetic energy during the collision. Option C correctly identifies this as an inelastic collision and recognizes that momentum can still be conserved. As long as no external forces act on the system during the collision, momentum must be conserved regardless of energy changes—this is a separate, independent physical law. Option A is wrong because any energy loss definitively makes this inelastic, not elastic. Option B contains a critical misconception: it incorrectly links energy loss to momentum non-conservation. Energy and momentum conservation are completely independent—you can lose kinetic energy while still conserving momentum. Option D uses the non-standard term "partially elastic" and incorrectly suggests momentum conservation depends on collision duration, which is irrelevant. Remember this key principle: momentum conservation depends only on external forces, while energy conservation tells you about the collision type. A collision can simultaneously be inelastic (energy lost) and conserve momentum (no external forces). These are separate physical laws operating independently.

Question 9

A 2.0 kg cart moving at 3.0 m/s collides with a 4.0 kg cart at rest. If the collision is perfectly inelastic, what percentage of the initial kinetic energy is converted to other forms of energy?

Approximately 33% of the initial kinetic energy is lost to heat and deformation processes
Approximately 50% of the initial kinetic energy is converted to internal energy during collision
Approximately 67% of the initial kinetic energy is dissipated as non-mechanical energy forms (correct answer)
Approximately 75% of the initial kinetic energy becomes thermal energy and sound waves
Approximately 25% of the initial kinetic energy is transformed into other energy types

Explanation: When you encounter perfectly inelastic collision problems, you're dealing with two key conservation principles: momentum is always conserved, but kinetic energy is partially lost to internal energy forms like heat, sound, and deformation. Let's solve this step-by-step. First, find the final velocity using conservation of momentum:

m_1v_1 + m_2v_2 = (m_1 + m_2)v_f

(2.0)(3.0) + (4.0)(0) = (2.0 + 4.0)v_f

6.0 = 6.0v_f

v_f = 1.0 \text{ m/s}

Now calculate the kinetic energies. Initial:

KE_i = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}

Final:

KE_f = \frac{1}{2}(6.0)(1.0)^2 = 3.0 \text{ J}

Energy lost:

9.0 - 3.0 = 6.0 \text{ J}

Percentage lost:

\frac{6.0}{9.0} \times 100\% = 67\%

This confirms answer C is correct. A (33%) significantly underestimates the energy loss. B (50%) might seem reasonable but doesn't match the actual calculation. D (75%) overestimates the loss and would require an even more massive stationary object. The key insight is that perfectly inelastic collisions always result in maximum kinetic energy loss while still conserving momentum. Remember this pattern: calculate final velocity from momentum conservation, then compare initial and final kinetic energies to find the percentage converted to internal energy forms.

Question 10

An explosion occurs in a system initially at rest, breaking it into three pieces. Two pieces have masses 2.0 kg and 3.0 kg and velocities 5.0 m/s east and 4.0 m/s north respectively. If the third piece has mass 1.0 kg, what is its velocity?

10 m/s at 37° south of west, satisfying momentum conservation requirements
16 m/s at 50° south of west, based on vector momentum balance (correct answer)
18 m/s at 45° south of west, from the explosion dynamics analysis
22 m/s at 60° south of west, using three-body momentum distribution
12 m/s at 30° south of west, calculated from momentum vector addition

Explanation: When you encounter an explosion problem, you're dealing with conservation of momentum. Since the system starts at rest, the total initial momentum is zero, so the final momentum of all pieces must also sum to zero. Let's set up a coordinate system with east as positive x and north as positive y. The momentum of each piece is mass times velocity:

Piece 1: $\vec{p_1} = (2.0 \text{ kg})(5.0 \text{ m/s east}) = 10.0 \text{ kg⋅m/s}$ in the +x direction
Piece 2: $\vec{p_2} = (3.0 \text{ kg})(4.0 \text{ m/s north}) = 12.0 \text{ kg⋅m/s}$ in the +y direction

For momentum conservation:

\vec{p_1} + \vec{p_2} + \vec{p_3} = 0

Therefore:

\vec{p_3} = -(\vec{p_1} + \vec{p_2}) = (-10.0, -12.0) \text{ kg⋅m/s}

The magnitude is:

|\vec{p_3}| = \sqrt{(-10.0)^2 + (-12.0)^2} = \sqrt{244} = 15.6 \text{ kg⋅m/s}

Since the third piece has mass 1.0 kg:

v_3 = \frac{15.6}{1.0} = 15.6 \approx 16 \text{ m/s}

The direction is:

\theta = \arctan\left(\frac{12.0}{10.0}\right) = 50°

south of west (negative x and y components). This confirms answer B is correct. Answer A underestimates both speed and angle. Answer C has the wrong angle despite being close in magnitude. Answer D significantly overestimates both the speed and angle. Remember: In explosion problems, always check that your final momentum vectors sum to the initial momentum. Setting up clear coordinate systems and using vector components prevents calculation errors.

Question 11

A 5.0 kg object moving at 4.0 m/s collides with a stationary 3.0 kg object. After collision, they move together. How much momentum is transferred from the first object to the second object?

7.5 kg⋅m/s transferred from the moving object to the stationary object during collision (correct answer)
12 kg⋅m/s transferred representing the total momentum exchange between the two objects
20 kg⋅m/s transferred equal to the initial momentum of the first object completely
2.5 kg⋅m/s transferred based on the mass ratio and final velocity calculations
15 kg⋅m/s transferred according to the perfectly inelastic collision momentum distribution

Explanation: When you encounter collision problems involving momentum transfer, you need to apply conservation of momentum and carefully distinguish between total momentum and the amount transferred between objects. First, find the initial momentum:

p_i = m_1v_1 + m_2v_2 = (5.0)(4.0) + (3.0)(0) = 20 \text{ kg⋅m/s}

Since they stick together (perfectly inelastic collision), use conservation of momentum to find the final velocity:

p_i = p_f

, so

20 = (5.0 + 3.0)v_f

, giving

v_f = 2.5 \text{ m/s}

Now for the key insight: momentum transfer means how much momentum the second object gains (or equivalently, how much the first object loses). The second object goes from 0 to

(3.0)(2.5) = 7.5 \text{ kg⋅m/s}

. Therefore, 7.5 kg⋅m/s is transferred from the first to the second object. Answer A correctly identifies this transferred momentum. Answer B (12 kg⋅m/s) incorrectly suggests some kind of total exchange calculation that doesn't represent actual momentum transfer. Answer C (20 kg⋅m/s) confuses the initial total momentum with transferred momentum - the first object doesn't lose all its momentum since it keeps moving after collision. Answer D (2.5 kg⋅m/s) appears to confuse the final velocity value with momentum transfer. Remember: momentum transfer in collisions equals the change in momentum of either object (they're equal and opposite). Focus on how much momentum one specific object gains or loses, not the total momentum of the system.

Question 12

A 0.50 kg ball moving at 8.0 m/s collides head-on with a 1.5 kg ball at rest. If the collision is perfectly inelastic, what is the velocity of the combined system immediately after collision?

2.0 m/s in the direction of the original motion (correct answer)
4.0 m/s in the direction of the original motion
2.7 m/s in the direction of the original motion
5.3 m/s in the direction of the original motion
1.6 m/s in the direction of the original motion

Explanation: Collision problems test your understanding of conservation of momentum, one of physics's most fundamental principles. When you see "perfectly inelastic collision," immediately think: the objects stick together after impact, and momentum is conserved but kinetic energy is not. For any collision, momentum before equals momentum after:

p_{initial} = p_{final}

. Initially, you have a 0.50 kg ball at 8.0 m/s and a 1.5 kg ball at rest. The initial momentum is:

p_i = (0.50)(8.0) + (1.5)(0) = 4.0 \text{ kg⋅m/s}

After the perfectly inelastic collision, both balls move together as a combined mass of 2.0 kg at velocity

v_f

p_f = (2.0)v_f

Setting initial and final momentum equal:

4.0 = (2.0)v_f

, so

v_f = 2.0 \text{ m/s}

in the original direction. This confirms answer A. Answer B (4.0 m/s) incorrectly assumes the moving ball continues at half its original speed, ignoring the mass it must now carry. Answer C (2.7 m/s) might result from incorrectly treating this as an elastic collision or making calculation errors. Answer D (5.3 m/s) violates momentum conservation entirely—the final momentum would exceed the initial momentum. Remember this pattern: in perfectly inelastic collisions, the final velocity always equals total initial momentum divided by total final mass. The combined system always moves slower than the fastest initial object because momentum is "shared" among more mass.

Question 13

A bullet of mass 20 g traveling at 400 m/s embeds in a 5.0 kg wooden block initially at rest. What fraction of the bullet's initial kinetic energy is lost in this perfectly inelastic collision?

Approximately 99.6% of the kinetic energy is dissipated during collision (correct answer)
Approximately 84% of the kinetic energy is converted to other forms
Approximately 75% of the kinetic energy is lost to internal energy
Approximately 92% of the kinetic energy becomes heat and deformation
Approximately 88% of the kinetic energy is no longer in translational form

Explanation: When you encounter perfectly inelastic collision problems, you need to apply both conservation of momentum and analyze energy changes. In perfectly inelastic collisions, the objects stick together, and kinetic energy is always lost to heat, sound, and deformation. First, find the final velocity using conservation of momentum. Initially:

p_i = m_{bullet} \cdot v_{bullet} = 0.020 \text{ kg} \times 400 \text{ m/s} = 8.0 \text{ kg⋅m/s}

After collision, both objects move together:

p_f = (m_{bullet} + m_{block}) \cdot v_f = (0.020 + 5.0) \text{ kg} \cdot v_f

Setting

p_i = p_f

8.0 = 5.02 \cdot v_f

, so

v_f = 1.59 \text{ m/s}

Now compare kinetic energies. Initial:

KE_i = \frac{1}{2}(0.020)(400^2) = 1600 \text{ J}

Final:

KE_f = \frac{1}{2}(5.02)(1.59^2) = 6.35 \text{ J}

Energy lost:

\frac{1600 - 6.35}{1600} \times 100\% = 99.6\%

Answer A is correct with 99.6% energy loss. The other options (B, C, D) all significantly underestimate the energy loss. This happens because most of the bullet's momentum transfers to a much more massive block, resulting in a very slow final velocity. The dramatic mass difference (bullet is only 0.4% of the total final mass) means almost all kinetic energy converts to internal energy. Study tip: In perfectly inelastic collisions with large mass ratios, expect energy losses approaching 100%. The smaller object's kinetic energy gets "absorbed" by the larger mass's inertia.

Question 14

In a collision between two objects, the impulse delivered to object A is 12 N⋅s northward. What can you determine about the impulse delivered to object B?

Object B receives an impulse of 12 N⋅s northward, maintaining symmetry in the collision process
Object B receives an impulse of 12 N⋅s southward, based on Newton's third law principles (correct answer)
Object B receives no impulse since the collision forces are internal to the two-object system
Object B receives an impulse equal in magnitude but perpendicular to object A's impulse vector
The impulse on object B cannot be determined without knowing the masses of both objects

Explanation: When analyzing collision problems, you need to apply Newton's third law, which states that forces always come in equal and opposite pairs. During any collision, the forces that objects exert on each other are equal in magnitude but opposite in direction. Since impulse equals force multiplied by time (

J = F \cdot \Delta t

), and both objects experience the collision for the same time duration, their impulses must also be equal in magnitude but opposite in direction. If object A receives an impulse of 12 N⋅s northward, then object B must receive an impulse of 12 N⋅s southward. This is a direct consequence of Newton's third law applied to the impulse-momentum relationship. Option A incorrectly suggests both objects receive impulses in the same direction, which would violate Newton's third law and conservation of momentum. Option C contains a fundamental misconception—while the collision forces are internal to the two-object system, each individual object still experiences an impulse from the other object. The fact that forces are internal doesn't mean they don't exist; it means the net external force on the system is zero. Option D incorrectly assumes the impulses are perpendicular, which has no basis in collision physics—impulses during direct collisions act along the same line but in opposite directions. The correct answer is B because it properly applies Newton's third law to impulse in collisions. Study tip: Remember that Newton's third law applies to both forces and impulses in collisions. Equal magnitude, opposite direction—always. This principle also ensures momentum conservation in isolated systems.

Question 15

A stationary object of mass (5M) explodes into three fragments on a frictionless horizontal plane. Two of the fragments, with mass $M$ and (2M), are observed to have velocities $\vec{v}_1 = -v_0 \hat{j}$ and $\vec{v}_2 = 3v_0 \hat{i}$ , respectively. What is the kinetic energy of the third fragment, which has mass (2M)?

$5 Mv_0^2$
$\frac{13}{2} Mv_0^2$
$\frac{37}{4} Mv_0^2$ (correct answer)
$\frac{40}{4} Mv_0^2$

Explanation: The initial momentum of the system is zero since the object is stationary. By conservation of linear momentum, the total final momentum must also be zero:

\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0

. We can solve for the momentum of the third fragment:

\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)

. First, find

\vec{p}_1

and

\vec{p}_2

\vec{p}_1 = m_1 \vec{v}_1 = M(-v_0 \hat{j}) = -Mv_0 \hat{j}

and

\vec{p}_2 = m_2 \vec{v}_2 = (2M)(3v_0 \hat{i}) = 6Mv_0 \hat{i}

. Therefore,

\vec{p}_3 = -(6Mv_0 \hat{i} - Mv_0 \hat{j}) = -6Mv_0 \hat{i} + Mv_0 \hat{j}

. The kinetic energy of the third fragment is given by

K_3 = \frac{p_3^2}{2m_3}

. The magnitude squared of its momentum is

p_3^2 = (-6Mv_0)^2 + (Mv_0)^2 = 36M^2v_0^2 + M^2v_0^2 = 37M^2v_0^2

. The mass of the third fragment is

m_3 = 5M - M - 2M = 2M

. So,

K_3 = \frac{37M^2v_0^2}{2(2M)} = \frac{37}{4} Mv_0^2

Question 16

A bullet of mass $m$ traveling at speed $v_0$ strikes and embeds itself in a block of mass $M$ initially at rest. The block is attached to an ideal spring and can slide on a surface with a non-zero coefficient of kinetic friction. For the system consisting of the bullet and the block, which statement correctly describes the physical principles governing the instant of the collision?

Both mechanical energy and linear momentum are conserved because the collision is instantaneous.
Linear momentum is approximately conserved because the collision forces are much larger than the external spring and friction forces, but mechanical energy is not conserved. (correct answer)
Mechanical energy is conserved because the work done by non-conservative forces is zero during the instant of collision, but linear momentum is not conserved due to the external forces.
Neither linear momentum nor mechanical energy is conserved, as the collision is inelastic and external forces from the spring and friction are present.

Explanation: This question tests the modeling assumptions made for collisions. The collision is inelastic (the bullet embeds), so mechanical energy is not conserved; some is converted to thermal energy. Linear momentum of a system is conserved only if the net external force is zero. Here, the spring and friction exert external forces. However, a collision occurs over a very short time interval (

\Delta t \to 0

). The impulse (

F_{ext} \Delta t

) from the external spring and friction forces is negligible compared to the very large impulse from the internal collision forces. Therefore, it is a standard and valid approximation in physics to model the linear momentum of the bullet-block system as being conserved through the instant of the collision, even though it is not conserved in the subsequent motion as the block compresses the spring and experiences friction.

Question 17

A 4.0 kg projectile is at the apex of its trajectory, moving horizontally at 50 m/s. An internal explosion, lasting a negligible time, breaks the projectile into three fragments. Fragment A (1.0 kg) is observed moving at 100 m/s in the original horizontal direction. Fragment B (1.0 kg) is observed moving at 50 m/s straight down. What is the direction of motion of Fragment C (2.0 kg) immediately after the explosion?

Directly horizontal in the original direction of motion.
At an angle of 26.6° above the horizontal. (correct answer)
At an angle of 26.6° below the horizontal.
At an angle of 153.4° with respect to the original direction of motion.

Explanation: Let the initial direction of motion be the +x direction and the vertical direction be the y-direction. The momentum of the system is conserved during the explosion because the external force of gravity provides a negligible impulse over the very short explosion time. The initial momentum of the projectile is

\vec{p}_i = M_{tot} \vec{v}_i = (4.0\text{ kg})(50\text{ m/s } \hat{i}) = 200\text{ kg} \cdot \text{m/s } \hat{i}

. The final momentum is the sum of the momenta of the fragments:

\vec{p}_f = \vec{p}_A + \vec{p}_B + \vec{p}_C

\vec{p}_A = (1.0\text{ kg})(100\text{ m/s } \hat{i}) = 100\text{ kg} \cdot \text{m/s } \hat{i}

\vec{p}_B = (1.0\text{ kg})(-50\text{ m/s } \hat{j}) = -50\text{ kg} \cdot \text{m/s } \hat{j}

. By conservation of momentum,

\vec{p}_i = \vec{p}_f

, so

\vec{p}_C = \vec{p}_i - \vec{p}_A - \vec{p}_B

\vec{p}_C = (200 \hat{i}) - (100 \hat{i}) - (-50 \hat{j}) = (100 \hat{i} + 50 \hat{j})\text{ kg} \cdot \text{m/s}

. The velocity of fragment C is

\vec{v}_C = \vec{p}_C / m_C = (100 \hat{i} + 50 \hat{j}) / 2.0 = (50 \hat{i} + 25 \hat{j})\text{ m/s}

. The direction is given by the angle

\theta = \arctan(v_y/v_x) = \arctan(25/50) = \arctan(0.5) \approx 26.6^{\circ}

. Since both components are positive, this angle is above the horizontal.

Question 18

An isolated system consists of three interacting particles with masses $m$ , (2m), and (3m). At a particular instant, the particle of mass $m$ has an acceleration $\vec{a}_1 = (2\hat{i} - 3\hat{j})\text{ m/s}^2$ , and the particle of mass (2m) has an acceleration $\vec{a}_2 = (-1\hat{i} + 1\hat{j})\text{ m/s}^2$ . What is the acceleration of the particle of mass (3m) at this instant?

$(-1\hat{i} + 2\hat{j})\text{ m/s}^2$
$(-\frac{1}{3}\hat{j})\text{ m/s}^2$
$(\hat{j})\text{ m/s}^2$
$(\frac{1}{3}\hat{j})\text{ m/s}^2$ (correct answer)

Explanation: For an isolated system, the net external force is zero. By Newton's second law for a system of particles,

\vec{F}_{net,ext} = \frac{d\vec{P}_{sys}}{dt}

. Since

\vec{F}_{net,ext}=0

, the total momentum of the system is conserved. Also,

\vec{F}_{net,ext} = M_{tot}\vec{a}_{CM}

, which means the center of mass acceleration is zero. The position of the center of mass is

\vec{R}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

. Taking two time derivatives gives

\vec{a}_{CM} = \frac{\sum m_i \vec{a}_i}{\sum m_i}

. Since

\vec{a}_{CM}=0

, we must have

\sum m_i \vec{a}_i = 0

m_1\vec{a}_1 + m_2\vec{a}_2 + m_3\vec{a}_3 = 0

m(2\hat{i} - 3\hat{j}) + 2m(-1\hat{i} + 1\hat{j}) + 3m\vec{a}_3 = 0

. Divide by

m

(2\hat{i} - 3\hat{j}) + (-2\hat{i} + 2\hat{j}) + 3\vec{a}_3 = 0

(0\hat{i} - 1\hat{j}) + 3\vec{a}_3 = 0 \implies 3\vec{a}_3 = \hat{j} \implies \vec{a}_3 = (\frac{1}{3}\hat{j})\text{ m/s}^2

Question 19

A puck of mass $m = 0.50$ kg moving at 4.0 m/s along the x-axis collides with an identical stationary puck. After the collision, the first puck moves with a velocity of 2.0 m/s at an angle of 60° with respect to the positive x-axis. What is the magnitude of the velocity of the second puck after the collision?

2.0 m/s
2.8 m/s
3.5 m/s (correct answer)
4.0 m/s

Explanation: Linear momentum is a vector and must be conserved in both the x and y directions. Let

\vec{v}_1

and

\vec{v}_2

be the final velocities. The initial momentum is entirely in the x-direction:

p_{ix} = m(4.0 \text{ m/s}) = (0.50)(4.0) = 2.0 \text{ kg} \cdot \text{m/s}

, and

p_{iy} = 0

. The final momentum has components:

p_{fx} = m v_{1x} + m v_{2x}

and

p_{fy} = m v_{1y} + m v_{2y}

. From the given information,

v_{1x} = (2.0 \text{ m/s})\cos(60^{\circ}) = 1.0 \text{ m/s}

and

v_{1y} = (2.0 \text{ m/s})\sin(60^{\circ}) = 1.732 \text{ m/s}

. Conserving momentum in x:

2.0 = (0.50)(1.0) + (0.50)v_{2x} \implies 2.0 = 0.50 + 0.50v_{2x} \implies v_{2x} = 3.0 \text{ m/s}

. Conserving momentum in y:

0 = (0.50)(1.732) + (0.50)v_{2y} \implies v_{2y} = -1.732 \text{ m/s}

. The magnitude of the second puck's velocity is

v_2 = \sqrt{v_{2x}^2 + v_{2y}^2} = \sqrt{(3.0)^2 + (-1.732)^2} = \sqrt{9.0 + 3.0} = \sqrt{12.0} \approx 3.46 \text{ m/s}

, which is closest to 3.5 m/s.

Question 20

Two blocks of mass $m$ and (3m) are at rest on a frictionless surface, with a compressed spring between them. The spring, which is not attached to the blocks, stores energy $E_s$ . After the spring is released, the block of mass $m$ slides and collides perfectly inelastically with a stationary block of mass (3m). What is the kinetic energy of the combined system after the collision?

$E_s / 4$
$3E_s / 8$
$3E_s / 16$ (correct answer)
$E_s / 2$

Explanation: This is a two-step problem. First, analyze the spring release. Let

v_1

be the speed of mass

m

and

v_2

be the speed of mass (3m). Momentum conservation:

0 = mv_1 - (3m)v_2 \implies v_1 = 3v_2

. Energy conservation:

E_s = \frac{1}{2}mv_1^2 + \frac{1}{2}(3m)v_2^2

. Substituting:

E_s = \frac{1}{2}m(3v_2)^2 + \frac{3}{2}mv_2^2 = 6mv_2^2

. The kinetic energy of the first block is

KE_1 = \frac{1}{2}mv_1^2 = \frac{1}{2}m(3v_2)^2 = \frac{9}{2}mv_2^2 = \frac{9}{2} \cdot \frac{E_s}{6} = \frac{3E_s}{4}

. Second, analyze the perfectly inelastic collision. The block of mass

m

with velocity

v_1

hits a stationary block of mass (3m). Let the final velocity be

V_f

. Momentum conservation:

mv_1 = (m+3m)V_f = 4mV_f \implies V_f = v_1/4

. The final kinetic energy is

KE_f = \frac{1}{2}(4m)V_f^2 = 2m(v_1/4)^2 = \frac{1}{8}mv_1^2 = \frac{1}{4}KE_1 = \frac{1}{4} \cdot \frac{3E_s}{4} = \frac{3E_s}{16}