Statics and Dynamics Quiz: Conservation Of Linear Momentum

7 questions · exam conditions

0:00

Conservation Of Linear MomentumQuestion 1 of 7

A projectile of mass $m$ traveling horizontally at speed $v_0$ embeds itself in a stationary block of mass $M$ resting on a frictionless table. The block-plus-projectile then slides off the edge of the table and undergoes projectile motion, landing a horizontal distance $d$ from the table's base. The table height is $h$ .

Which expression correctly gives the initial speed $v_0$ of the projectile in terms of $m$ , $M$ , $g$ , $h$ , and $d$ ?

v_0 = \frac{(m+M)}{m} \cdot d \cdot \sqrt{\frac{g}{2h}}

, obtained by combining conservation of momentum during impact with projectile kinematics after the block leaves the table.

v_0 = \frac{m}{(m+M)} \cdot d \cdot \sqrt{\frac{g}{2h}}

, obtained by applying conservation of momentum but inverting the mass ratio, treating the projectile as the post-collision system.

v_0 = \frac{(m+M)}{m} \cdot \sqrt{2gh}

, obtained by setting the projectile's initial kinetic energy equal to the gravitational potential energy at table height, then correcting for the mass ratio.

v_0 = \frac{m}{(m+M)} \cdot \sqrt{\frac{2h}{g}} \cdot d

, obtained by using the time of flight in the denominator rather than combining it with the horizontal velocity from momentum conservation.

← Back to quizzes

Statics and Dynamics Quiz

Statics and Dynamics Quiz: Conservation Of Linear Momentum

Practice Conservation Of Linear Momentum in Statics and Dynamics 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 Statics and Dynamics.

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

Which expression correctly gives the initial speed $v_0$ of the projectile in terms of $m$ , $M$ , $g$ , $h$ , and $d$ ?

$v_0 = \frac{(m+M)}{m} \cdot d \cdot \sqrt{\frac{g}{2h}}$ , obtained by combining conservation of momentum during impact with projectile kinematics after the block leaves the table. (correct answer)
$v_0 = \frac{m}{(m+M)} \cdot d \cdot \sqrt{\frac{g}{2h}}$ , obtained by applying conservation of momentum but inverting the mass ratio, treating the projectile as the post-collision system.
$v_0 = \frac{(m+M)}{m} \cdot \sqrt{2gh}$ , obtained by setting the projectile's initial kinetic energy equal to the gravitational potential energy at table height, then correcting for the mass ratio.
$v_0 = \frac{m}{(m+M)} \cdot \sqrt{\frac{2h}{g}} \cdot d$ , obtained by using the time of flight in the denominator rather than combining it with the horizontal velocity from momentum conservation.

Explanation: When a problem combines a collision with subsequent projectile motion, your instinct should be to split it into two independent phases: the impact (governed by momentum conservation) and the flight (governed by kinematics). Mixing the two phases — or misapplying either — is exactly where the distractors lead you astray. Phase 1 — Impact: The projectile embeds in the block, so momentum is conserved:

mv_0 = (m+M)V

, giving the post-collision speed

V = \frac{m}{m+M}v_0

. Phase 2 — Projectile flight: The block leaves the table horizontally at speed

V

. Falling height

h

gives a fall time via

h = \frac{1}{2}gt^2

, so

t = \sqrt{\frac{2h}{g}}

. The horizontal distance is then

d = V \cdot t = V\sqrt{\frac{2h}{g}}

, which means

V = \frac{d}{t} = d\sqrt{\frac{g}{2h}}

. Combining both phases: Substituting into the momentum result:

v_0 = \frac{m+M}{m} \cdot V = \frac{m+M}{m} \cdot d\sqrt{\frac{g}{2h}}

. That's exactly answer A. Answer B inverts the mass ratio — it solves for

V

rather than

v_0

, confusing which object's speed you want. Answer C uses energy conservation through the fall height to find a speed, but that conflates the kinematic phase with the collision phase and produces the wrong quantity entirely. Answer D places the time of flight in the denominator of an otherwise correct-looking expression, which dimensionally and physically gives the wrong result. Your study tip: always label which speed belongs to which phase. Write

v_0

(before collision) and

V

(after collision) explicitly — this single habit prevents the mass-ratio inversion error that traps most students.

Question 2

A 10 kg cart moves to the right at $4 \ \text{m/s}$ on a frictionless track. A 2 kg block sits on top of the cart with significant friction between them. The block is initially moving to the right at $10 \ \text{m/s}$ relative to the ground. Eventually, the block and cart reach a common final velocity.

What is the final common velocity of the block-cart system, and how does the impulse received by the cart from the block compare to the impulse received by the block from the cart?

Final velocity is $5 \ \text{m/s}$ to the right. The cart receives a greater impulse magnitude than the block because the cart is five times heavier and its momentum changes by a larger amount: $\Delta p_{\text{cart}} = 10(5-4) = 10 \ \text{N·s}$ versus $\Delta p_{\text{block}} = 2(5-10) = -10 \ \text{N·s}$ .
Final velocity is $5 \ \text{m/s}$ to the right. The impulse magnitudes are equal and the impulses are opposite in direction, because the friction force is an internal force pair and internal forces always cancel, producing zero net impulse on each individual object.
Final velocity is $\frac{(10)(4)+(2)(10)}{12} = 5 \ \text{m/s}$ to the right. The cart receives a larger forward impulse than the backward impulse the block receives, because friction acts over a greater sliding distance on the heavier cart, producing more work and therefore more impulse on that object.
Final velocity is $\frac{(10)(4)+(2)(10)}{12} = 5 \ \text{m/s}$ to the right. The impulse magnitudes are equal and opposite in direction — forward on the cart, backward on the block — because by Newton's third law the friction force the block exerts on the cart equals in magnitude the force the cart exerts on the block, and both forces act over the same time interval, so $J_{\text{cart}} = -J_{\text{block}}$ . (correct answer)

Explanation: When a question involves two objects interacting through a contact force, your first instincts should be conservation of momentum and Newton's third law — these two principles together fully determine both the final velocity and the impulse relationship. For the final velocity, apply conservation of momentum to the system. The track is frictionless, so no external horizontal forces act, meaning total momentum is conserved:

v_f = \frac{(10)(4) + (2)(10)}{10 + 2} = \frac{60}{12} = 5 \ \text{m/s}

to the right. Now for the impulse comparison: friction between the block and cart is an interaction pair. By Newton's third law, the force the cart exerts on the block equals in magnitude — and opposes in direction — the force the block exerts on the cart. Crucially, these forces act over the same time interval (from first contact until they reach common velocity). Since impulse is

J = F \cdot \Delta t

, equal and opposite forces over identical time intervals produce equal and opposite impulses. Answer D captures this correctly. Answer A makes the right numerical observation —

|\Delta p_{\text{cart}}| = 10 \ \text{N·s}

and

|\Delta p_{\text{block}}| = 10 \ \text{N·s}

— but then contradicts itself by claiming the magnitudes are unequal. They're actually identical, confirming Newton's third law. Answer B confuses individual and system analysis. Internal forces cancel for the system, but each individual object absolutely experiences a net impulse — that's precisely why their velocities change. Answer C incorrectly links impulse to sliding distance. Work and impulse are different quantities; distance determines work, but time determines impulse. Study tip: Whenever impulse is compared between two interacting objects, go straight to Newton's third law plus equal time intervals — the impulses must be equal and opposite, regardless of mass.

Question 3

A 1500 kg car traveling east at $20 \ \text{m/s}$ collides with a 1000 kg car traveling north at $15 \ \text{m/s}$ at an intersection. The two cars lock together upon impact (perfectly inelastic collision).

What is the magnitude of the combined wreckage's velocity immediately after the collision?

$10 \ \text{m/s}$ , found by computing the scalar sum of the two momenta and dividing by total mass, treating the collision as one-dimensional.
$\frac{\sqrt{(30000)^2+(15000)^2}}{2500} \ \text{m/s} = \sqrt{(12)^2+(6)^2} \ \text{m/s} \approx 13.4 \ \text{m/s}$ , found by applying vector addition of momenta separately in each direction then dividing by total mass. (correct answer)
$\frac{(1500)(20)+(1000)(15)}{2500} = 18 \ \text{m/s}$ , found by summing all momentum magnitudes as scalars and dividing by the combined mass.
$\frac{\sqrt{(1500)^2+(1000)^2}}{2500} \times 20 \ \text{m/s} \approx 7.2 \ \text{m/s}$ , found by taking the vector magnitude of the masses and multiplying by a reference speed.

Explanation: Whenever you see a 2D collision problem, the key is remembering that momentum is a vector — you must handle each direction independently before combining results. In a perfectly inelastic collision, total momentum is conserved. Here, the east-traveling car contributes momentum only in the x-direction, and the north-traveling car contributes momentum only in the y-direction. You calculate each component separately:

p_x = (1500)(20) = 30{,}000 \ \text{kg·m/s}

p_y = (1000)(15) = 15{,}000 \ \text{kg·m/s}

Dividing each by the combined mass (2500 kg) gives the velocity components:

v_x = 12 \ \text{m/s}, \quad v_y = 6 \ \text{m/s}

The magnitude is then found using the Pythagorean theorem:

v = \sqrt{12^2 + 6^2} = \sqrt{180} \approx 13.4 \ \text{m/s}

This confirms B is correct. A is wrong because it treats the two momenta as if they act along the same line — you can't simply add scalar magnitudes when the vectors point in perpendicular directions. C makes the same dimensional error: adding

30{,}000 + 15{,}000 = 45{,}000

and dividing by 2500 gives 18 m/s, which ignores the perpendicular geometry entirely. D is a nonsensical construction — taking a "vector magnitude of the masses" has no physical meaning; mass is a scalar, not a vector quantity. Study tip: Any time two objects collide at right angles, immediately set up x- and y-components separately, then use the Pythagorean theorem at the end. Treating perpendicular momenta as scalars is the single most common trap in 2D collision problems.

Question 4

A railroad car of mass $M = 2000 \ \text{kg}$ rolls without friction at $v_0 = 5 \ \text{m/s}$ to the right. Rain falls vertically at a rate such that the car accumulates mass at $\dot{m} = 10 \ \text{kg/s}$ . Assume the rain has zero horizontal velocity component before hitting the car.

Which of the following correctly describes the effect of the accumulating rain on the car's horizontal velocity over time, and which physical principle is most directly responsible?

The car decelerates because the rain exerts a downward normal force on the car, increasing friction with the track; conservation of energy governs since rain's kinetic energy converts to heat, slowing the car.
The car maintains constant velocity because the rain falls vertically and exerts no horizontal force on the car; conservation of horizontal momentum means neither the rain nor the car's horizontal motion is affected by the purely vertical rainfall.
The car decelerates because the horizontal momentum of the system (car + accumulated rain) must be conserved; as the total mass increases while horizontal momentum stays constant, the velocity must decrease according to $v(t) = \frac{M v_0}{M + \dot{m}t}$ . (correct answer)
The car decelerates because each raindrop exerts a backward horizontal impulse on the car equal to its vertical momentum; Newton's third law causes horizontal deceleration proportional to the rain's vertical momentum flux.

Explanation: When a problem involves a system whose mass is changing over time, your instinct should be to reach for conservation of momentum — not energy, not Newton's second law in its simple form. This is a variable-mass system, and that changes everything. Here's the core physics: the rain falls vertically, meaning it carries zero horizontal momentum. The car, however, carries horizontal momentum

p = Mv_0

. As rain accumulates, the total mass of the system grows, but no external horizontal force acts on it — so horizontal momentum is conserved. Since

p = (M + \dot{m}t)\,v(t)

must equal

Mv_0

, you can solve directly for the velocity:

v(t) = \frac{Mv_0}{M + \dot{m}t}

This is exactly what C states, making it correct. The car slows down not because a force pushes it backward, but because it must "share" its fixed horizontal momentum with an ever-growing mass. A is wrong on two counts: the track is frictionless (given), and conservation of energy doesn't govern this scenario — energy is not conserved in perfectly inelastic collisions like rain accumulating on the car. B is temptingly logical but misses the key insight. Yes, rain has no horizontal force, but it still dilutes the car's momentum by adding mass with zero horizontal velocity. Momentum conservation doesn't mean velocity is unchanged — it means

mv

is unchanged. D confuses directions entirely. Rain's vertical momentum creates no horizontal impulse on the car. Newton's third law operates vertically here, not horizontally. Study tip: Whenever mass is being added or shed from a moving object, immediately write down

p = mv

and ask what's conserved — that equation will almost always unlock the problem.

Question 5

Two ice skaters, A (mass $70 \ \text{kg}$ ) and B (mass $50 \ \text{kg}$ ), stand facing each other on frictionless ice. They push off from each other starting from rest. After the push, skater A moves at $3 \ \text{m/s}$ to the left.

A student claims: 'Because the push is an internal force to the two-skater system, the total kinetic energy after the push must be zero.' Which of the following best evaluates this claim?

The claim is correct; internal forces cannot change the total mechanical energy of an isolated system, so all kinetic energy gained by one skater is offset by an equal loss in the other skater's kinetic energy.
The claim is incorrect; while internal forces cannot change total linear momentum (which remains zero), they can convert stored potential energy (e.g., chemical energy in muscles) into kinetic energy, so the total kinetic energy after the push is positive and nonzero. (correct answer)
The claim is incorrect; internal forces violate Newton's third law in this context, allowing the system's momentum to change and enabling net kinetic energy to be produced from nothing without any energy source.
The claim is correct; because total momentum is conserved and equals zero, and kinetic energy is directly proportional to momentum, the total kinetic energy must also be zero after the push.

Explanation: Whenever you see a question mixing momentum conservation with energy, recognize that these are separate conservation laws governed by different conditions. Momentum is conserved when no external forces act on the system. Energy can still change form — even within an isolated system — if an internal energy source (like chemical energy in muscles) is present. Here, the two skaters start from rest, so total momentum is zero. By conservation of momentum, it remains zero after the push: skater A moves left at

3 \ \text{m/s}

, so skater B must move right. Using momentum conservation:

v_B = \frac{70 \times 3}{50} = 4.2 \ \text{m/s}

. The total kinetic energy after the push is

\frac{1}{2}(70)(3)^2 + \frac{1}{2}(50)(4.2)^2 = 315 + 441 = 756 \ \text{J}

, which is clearly nonzero. This energy came from the skaters' muscles (stored chemical energy), converted to kinetic energy through the internal push. That's why B is correct: internal forces can conserve momentum while still transforming stored energy into kinetic energy. A contains a subtle but critical error — it wrongly claims internal forces cannot change mechanical energy. They absolutely can when a stored energy source exists; they just can't change total momentum. C is entirely wrong on two counts: internal forces do not violate Newton's third law, and momentum does not change. D confuses momentum with kinetic energy. Zero total momentum does not imply zero total kinetic energy — momentum is a vector (it can cancel), while kinetic energy is always a positive scalar (it cannot). Your strategy: always treat momentum and energy as independent bookkeeping systems. Zero net momentum ≠ zero kinetic energy.

Question 6

A 0.5 kg grenade at rest explodes into three fragments. Fragment 1 (0.1 kg) flies north at $30 \ \text{m/s}$ . Fragment 2 (0.2 kg) flies east at $20 \ \text{m/s}$ . Fragment 3 has mass 0.2 kg.

What is the speed and direction of Fragment 3 immediately after the explosion?

Fragment 3 moves at $25 \ \text{m/s}$ directed southwest, at an angle of $\arctan\left(\frac{3}{4}\right)$ west of south, found by applying conservation of momentum so that the vector sum of all three fragment momenta equals zero but with the south and west components transposed when computing the angle.
Fragment 3 moves at approximately $180 \ \text{m/s}$ directed southwest, obtained by taking the vector magnitude of the other fragments' velocities (not momenta) — $\sqrt{30^2+20^2} \approx 36 \ \text{m/s}$ — and dividing by Fragment 3's mass of 0.2 kg, incorrectly skipping the step of multiplying each velocity by its respective mass.
Fragment 3 moves at $25 \ \text{m/s}$ directed southwest, at an angle of $\arctan\left(\frac{4}{3}\right)$ west of south, obtained by correctly setting the southward and westward momentum components of Fragment 3 equal to those of Fragments 1 and 2 respectively, then computing the resultant speed and direction. (correct answer)
Fragment 3 moves at $50 \ \text{m/s}$ directed southwest along the bisector of north and east, found by simply summing the speeds of the other two fragments ( $30 + 20 = 50 \ \text{m/s}$ ) and assuming the third fragment travels in the opposite direction at that combined speed, without performing any momentum or vector analysis.

Explanation: When a stationary object explodes, the total momentum before and after must be conserved — and since the initial momentum is zero, the vector sum of all fragment momenta must equal zero. This means Fragment 3's momentum must exactly cancel the combined momenta of Fragments 1 and 2. Start by computing the momenta of the known fragments. Fragment 1 carries

p_1 = 0.1 \times 30 = 3 \ \text{kg·m/s}

northward. Fragment 2 carries

p_2 = 0.2 \times 20 = 4 \ \text{kg·m/s}

eastward. For the total to be zero, Fragment 3 must supply

3 \ \text{kg·m/s}

southward and

4 \ \text{kg·m/s}

westward. Its speed is therefore

v_3 = \frac{\sqrt{3^2 + 4^2}}{0.2} = \frac{5}{0.2} = 25 \ \text{m/s}

, directed southwest. The angle west of south is

\arctan\!\left(\frac{4}{3}\right)

— the westward component (4) over the southward component (3) — confirming answer C. Answer A reaches the right speed but swaps the south and west components when computing the angle, giving

\arctan\!\left(\frac{3}{4}\right)

instead of

\arctan\!\left(\frac{4}{3}\right)

— a small but meaningful directional error. Answer B skips multiplying velocities by mass entirely, taking

\sqrt{30^2+20^2}/0.2 \approx 180 \ \text{m/s}

, which conflates speed with momentum. Answer D simply adds the two speeds (

30+20=50

) with no vector or mass reasoning whatsoever. The key habit: always work in momentum (mass × velocity), not raw speed, and track vector components carefully — confusing the two is the most common trap on explosion and collision problems.

Question 7

Two particles, A (mass $3m$ ) and B (mass $m$ ), undergo a head-on elastic collision. Before the collision, A moves to the right at speed $u$ and B is at rest. The standard elastic collision formulas give post-collision velocities: $v_A = \frac{3m - m}{3m + m}\,u = \frac{u}{2}, \quad v_B = \frac{2(3m)}{3m + m}\,u = \frac{3u}{2}$

A student wishes to verify the given velocities are consistent with an elastic collision by checking whether the speed of approach equals the speed of separation: $|u - 0| \stackrel{?}{=} |v_B - v_A|$ . Which of the following correctly evaluates this check and best characterizes what it alone can — and cannot — confirm?

The check yields $\left|\frac{3u}{2} - \frac{u}{2}\right| = u$ , which equals the approach speed $u$ , so $e = 1$ . This single check is both necessary and sufficient to fully confirm the collision is elastic, because $e = 1$ by definition means kinetic energy is conserved.
The check yields $\left|\frac{3u}{2} + \frac{u}{2}\right| = 2u \neq u$ , so the separation speed does not equal the approach speed. This means $e \neq 1$ , indicating the provided velocity formulas are inconsistent with an elastic collision.
The check yields $\left|\frac{3u}{2} - \frac{u}{2}\right| = u$ , confirming $e = 1$ . However, this restitution formula is only valid when both particles are initially moving; since B starts at rest, a separate modified formula must be applied to validate the result for this specific case.
The check yields $\left|\frac{3u}{2} - \frac{u}{2}\right| = u$ , confirming $e = 1$ . This is a necessary condition for an elastic collision, but used in isolation — without also verifying conservation of linear momentum — it is not sufficient to fully confirm the collision is elastic, since the restitution equation alone does not constrain both unknowns. (correct answer)

Explanation: Whenever you see a question testing elastic collision verification, keep two independent conditions in mind: conservation of momentum and conservation of kinetic energy. The coefficient of restitution condition (

e = 1

, or speed of approach equals speed of separation) is mathematically equivalent to energy conservation only when combined with momentum conservation — not on its own. For the separation check:

|v_B - v_A| = \left|\frac{3u}{2} - \frac{u}{2}\right| = u

, which equals the approach speed

|u - 0| = u

. So

e = 1

is confirmed. This is the correct arithmetic result, making D the right answer — but the key insight is what this check alone cannot tell you. The restitution equation is one equation in two unknowns (

v_A

and

v_B

). To uniquely determine both post-collision velocities and fully certify the collision, you also need conservation of momentum:

3mu + 0 = 3mv_A + mv_B

. Together, these two equations fully constrain the system. Alone, neither suffices. A is wrong because it claims

e = 1

alone is sufficient to confirm an elastic collision. It isn't — you need momentum conservation too. The two conditions are jointly sufficient, not individually. B is wrong on arithmetic: it incorrectly adds the post-collision speeds instead of taking their difference, yielding

2u

instead of

u

. The separation speed is

v_B - v_A

, not

v_B + v_A

. C is wrong because the restitution formula

e = \frac{\text{speed of separation}}{\text{speed of approach}}

applies regardless of whether one particle is initially at rest — no modified formula is needed. Study tip: On collision problems, always track both conservation laws separately. A single condition (restitution or momentum) is necessary but not sufficient on its own to fully characterize elastic behavior.