Unit 4 practice test: Linear Momentum
12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value; data sets share one table, graph or scene diagram.
Covers: Linear Momentum; Change in Momentum and Impulse; Conservation of Linear Momentum; Elastic and Inelastic Collisions.
Experimental setup
An explosion on a track
Two carts, A (0.60 kg) and B (0.40 kg), sit at rest on a level, low-friction track with a compressed spring between them, held by a thread. The thread is burned. The spring pushes the carts apart, and cart A moves left at 0.80 m/s. Take right as positive.
1. What is the velocity of the two-cart system's center of mass after the carts separate?
- +1.2 m/s, the faster cart's velocity
- −0.80 m/s, the heavier cart's velocity
- +0.40 m/s, the average of the two velocities
- Zero
Show the answer
v_cm = Σp/Σm = 0/1.00 kg = 0. Internal forces cannot move the center of mass of a system.
- +1.2 m/s, the faster cart's velocity: The center of mass moves with the sum of the momenta over the total mass, which is zero.
- −0.80 m/s, the heavier cart's velocity: The center of mass is not attached to one cart; the two momenta cancel.
- +0.40 m/s, the average of the two velocities: An unweighted average ignores the masses. Weighted by mass, the velocities cancel exactly.
- Correct: Zero: Right: the system's momentum is still zero, so the center of mass stays at rest.
2. Which cart gets more kinetic energy from the spring?
- B, the lighter cart
- A, the heavier cart
- Neither: they get equal kinetic energies
- Neither: the total kinetic energy stays zero
Show the answer
K_A = ½(0.60)(0.80)² = 0.19 J; K_B = ½(0.40)(1.2)² = 0.29 J. The lighter cart moves faster with the same size of momentum.
- Correct: B, the lighter cart: Right: equal momenta, and K = p²/(2m) is larger for the smaller mass: 0.29 J against 0.19 J.
- A, the heavier cart: With equal momenta, the heavier cart is slower and gets less kinetic energy.
- Neither: they get equal kinetic energies: They get equal sizes of momentum, not equal kinetic energies.
- Neither: the total kinetic energy stays zero: The spring's stored energy becomes kinetic energy. Momentum stays zero; kinetic energy does not.
3. Why can the spring's push be treated with the explosion model?
- The spring's forces are external, so they do not change the momentum
- Its forces are internal and briefly far larger than outside forces
- The spring does no work on the carts, so the energy is unchanged
- The carts have equal masses, so their speeds are equal and opposite
Show the answer
In the collision and explosion models, the forces between the objects are internal and briefly much larger than external ones, so the system's momentum is the same before and after.
- The spring's forces are external, so they do not change the momentum: External forces are what change a system's momentum. The spring is inside the system.
- Correct: Its forces are internal and briefly far larger than outside forces: Right: so the system's momentum is the same just before and just after.
- The spring does no work on the carts, so the energy is unchanged: The spring does positive work: that is where the carts' kinetic energy comes from.
- The carts have equal masses, so their speeds are equal and opposite: They do not (0.60 kg and 0.40 kg), and the model does not need it.
Data table
Collision trials with two carts
In each trial a 0.40 kg cart A rolls into a 0.60 kg cart B at rest on a level, low-friction track. Motion sensors record the velocities just before and just after. Positive is to the right.
| Trial | v_A before (m/s) | v_A after (m/s) | v_B after (m/s) |
|---|---|---|---|
| 1 (magnets) | +1.00 | −0.20 | +0.80 |
| 2 (bare) | +1.00 | +0.10 | +0.60 |
| 3 (Velcro) | +1.00 | +0.40 | +0.40 |
4. In which trials is the total momentum of the two carts the same before and after?
- In each of the three trials
- In trial 1, the elastic one, and not the others
- In trial 3, where the carts stick, and not the others
- In no trial, because the carts' speeds change
Show the answer
Before: (0.40)(1.00) = 0.40 kg·m/s. After: trial 1: −0.08 + 0.48 = 0.40; trial 2: 0.04 + 0.36 = 0.40; trial 3: (1.00)(0.40) = 0.40.
- Correct: In each of the three trials: Right: every trial gives 0.40 kg·m/s after, as before.
- In trial 1, the elastic one, and not the others: Momentum is conserved in every collision with no net external force, elastic or not.
- In trial 3, where the carts stick, and not the others: Sticking does not make momentum special; check the products in trials 1 and 2 too.
- In no trial, because the carts' speeds change: The individual speeds change, but the total momentum does not.
5. Which trial is elastic?
- Trial 2
- Trial 1
- Trial 3
- Each of the three, because momentum is conserved in each
Show the answer
K before = ½(0.40)(1.00)² = 0.20 J. After: trial 1, 0.20 J (elastic); trial 2, 0.11 J (inelastic); trial 3, 0.080 J (perfectly inelastic).
- Trial 2: K after = 0.002 + 0.108 = 0.11 J, less than 0.20 J: inelastic.
- Correct: Trial 1: Right: K after = ½(0.40)(0.20)² + ½(0.60)(0.80)² = 0.20 J, equal to K before.
- Trial 3: The carts stick, which is perfectly inelastic: K drops to 0.080 J.
- Each of the three, because momentum is conserved in each: Conserving momentum does not make a collision elastic. Elastic means total K is unchanged too.
6. In trial 1, cart A bounces back. Did cart A's own kinetic energy stay the same?
- Yes: in an elastic collision each cart keeps its kinetic energy
- Yes: A moves back with its original speed
- No: A gave most of its K to B; the total stayed 0.20 J
- No: the total kinetic energy went down
Show the answer
In an elastic collision the total kinetic energy is the same before and after, but energy moves between the objects.
- Yes: in an elastic collision each cart keeps its kinetic energy: Elastic means the total K is unchanged; individual kinetic energies can change.
- Yes: A moves back with its original speed: A moves back at 0.20 m/s, much slower than 1.00 m/s.
- Correct: No: A gave most of its K to B; the total stayed 0.20 J: Right: K_A falls from 0.20 J to 0.008 J; B gains 0.192 J.
- No: the total kinetic energy went down: Trial 1 is elastic: the total stayed 0.20 J.
7. Trial 3 is repeated with cart B loaded to 1.60 kg. Compared with the original trial 3, the speed of the stuck carts after the collision is
- the same, because A has the same momentum
- one quarter as large
- twice as large
- one half as large
Show the answer
v_f = m_A v_A/(m_A + m_B): 0.40/1.00 = 0.40 m/s before, 0.40/2.00 = 0.20 m/s now.
- the same, because A has the same momentum: The same momentum is now shared by twice the total mass.
- one quarter as large: B's mass is a little under 3 times larger, but the total mass (which matters) doubles.
- twice as large: Adding mass to the stuck pair slows it down.
- Correct: one half as large: Right: (0.40)(1.00) = (2.00)v gives 0.20 m/s, half of 0.40 m/s.
8. A tennis racket exerts an average force of 250 N on a ball for 0.0060 s. What impulse does it deliver, in N·s?
Type a number in N·s.
Show the answer
J = F_avg Δt = (250 N)(0.0060 s) = 1.5 N·s.
- Answer: 1.5 N·s
9. A ball falls toward Earth. For which system is momentum constant (ignoring the rest of the universe)?
- The ball alone
- Earth alone
- Ball + Earth
- Neither the ball alone nor ball + Earth
Show the answer
Ball alone: Earth's pull is external, so p changes. Ball + Earth: the pull is internal, so the total momentum is constant. Earth's tiny upward speed balances the ball's.
- The ball alone: Earth's pull is external to the ball and increases its downward momentum.
- Earth alone: The ball pulls Earth up, an external force on Earth.
- Correct: Ball + Earth: Right: the gravitational forces between them are internal and cancel.
- Neither the ball alone nor ball + Earth: With both inside, the only forces are internal; the total is constant.
10. A 0.15 kg ball moving right at 20 m/s and a 1200 kg car moving left at 0.0025 m/s are compared. Which is true?
- The car has far more momentum, because of its mass
- The ball has far more momentum, because it is faster
- Same size of momentum, opposite directions
- Their momenta are equal, including direction
Show the answer
p = mv: ball +3.0 kg·m/s, car −3.0 kg·m/s. Same size, opposite directions.
- The car has far more momentum, because of its mass: The car's mass is large but its speed is tiny; the products match.
- The ball has far more momentum, because it is faster: The ball is faster but much lighter; the products match.
- Correct: Same size of momentum, opposite directions: Right: (0.15)(20) = 3.0 kg·m/s and (1200)(0.0025) = 3.0 kg·m/s.
- Their momenta are equal, including direction: They move in opposite directions, so the momentum vectors point opposite ways.
11. A 0.058 kg tennis ball arrives at 30. m/s and leaves the racket at 40. m/s in the opposite direction. What is the size of the impulse on the ball, in N·s?
Type a number in N·s.
Show the answer
Take the leaving direction as positive: Δp = (0.058)(+40.) − (0.058)(−30.) = 4.1 kg·m/s, so |J| = 4.1 N·s.
- Answer: 4.1 N·s
12. An astronaut at rest throws a wrench away. If she throws a wrench of twice the mass at the same speed, her recoil speed becomes
- 4 times as large
- √2 times as large
- the same
- 2 times as large
Show the answer
m_a v_a = m_w v_w. Doubling m_w at the same v_w doubles v_a.
- 4 times as large: Momentum is linear in mass; nothing is squared.
- √2 times as large: That would come from equal kinetic energies, which is not the condition here.
- the same: The wrench carries twice the momentum, so she must too.
- Correct: 2 times as large: Right: her momentum must balance the wrench's, which doubles.
Keep going
Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.