This diagnostic is for the algebra-based Physics 1 course, Unit 4 (Linear Momentum). It shows you which topics to revisit. Allow about 30 minutes, then use the table at the end.
These are original Marlbridge practice questions, not past exam questions. They have not been calibrated against real exam results, so your total does not predict an exam score. Use a scientific calculator and g = 9.8 m/s². Ignore air resistance. Every question states its axis: use it for every sign.
Question 1 (multiple choice · 4.1)
Object A has 3 times the mass of object B. The two objects have momenta of the same size. What is the ratio of their kinetic energies, K_A : K_B?
- (A) 3 : 1
- (B) 1 : 9
- (C) 1 : 3
- (D) 1 : 1
Answer and explanation
Answer: (C). Substitute v = p ÷ m into K = ½mv² to get K = p² ÷ (2m). With the same p, kinetic energy is inversely proportional to mass, so the heavier object has one third of the kinetic energy.
- (A) assumes that more mass always means more kinetic energy. Here B is 3 times faster.
- (B) squares the mass ratio. In K = p² ÷ (2m) the mass is not squared.
- (D) treats momentum and kinetic energy as the same quantity. Momentum scales with v; kinetic energy with v².
If you missed this: read Worked example 2 in the Topic 4.1 study guide.
Question 2 (multiple choice · 4.1)
Take +y up. A ball is thrown straight up, rises to its highest point and falls back to the thrower’s hand. Which describes the graph of the ball’s momentum p_y against time for the whole flight?
- (A) A line falling to zero at the highest point, then rising again, like a V
- (B) A straight line with a constant negative slope, crossing zero at the highest point
- (C) A horizontal line, because momentum is conserved during the flight
- (D) A curve that falls more and more steeply, because the ball speeds up on the way down
Answer and explanation
Answer: (B). p_y = m v_y, so the momentum–time graph has the same shape as the velocity–time graph, scaled by m. The velocity falls steadily from positive to negative, so p_y does too. It is zero at the top. The slope is −mg, the only force on the ball.
- (A) graphs the size of the momentum. Signed p_y keeps falling below zero after the top.
- (C) forgets that gravity is an external force on the ball.
- (D) confuses a steady change with a speeding-up change. The ball’s acceleration is constant, so the line is straight.
If you missed this: read “Graphs of momentum” in the Topic 4.1 study guide.
Question 3 (multiple choice · 4.2)
Take +x to the right. A 0.40 kg puck slides to the right at 2.0 m/s on frictionless ice. A paddle then pushes on it. The net force rises in a straight line from 0 to 6.0 N to the left over 0.20 s, then falls back to 0 in a straight line over another 0.20 s. What is the puck’s velocity after the push?
- (A) −1.0 m/s
- (B) −3.0 m/s
- (C) −4.0 m/s
- (D) +5.0 m/s
Answer and explanation
Answer: (A). Impulse = area of the triangle = ½ × 0.40 s × (−6.0 N) = −1.2 N·s. The puck starts with p₀ = 0.40 × 2.0 = +0.80 kg·m/s. After: p = 0.80 − 1.2 = −0.40 kg·m/s, so v = −0.40 ÷ 0.40 = −1.0 m/s.
- (B) ignores the puck’s starting momentum and treats −1.2 N·s as its final momentum.
- (C) uses the peak force for the whole 0.40 s (−2.4 N·s). The force is at its peak only for an instant.
- (D) gives the impulse the wrong sign. It points left, the same way as the force.
If you missed this: read “Impulse as an area” and Worked example 1 in the Topic 4.2 study guide.
Question 4 (multiple choice · 4.2)
Looking down on a smooth, level air table, a puck slides towards a straight, frictionless wall at an angle. It bounces off at the same speed and at the same angle to the wall. What is the direction of the impulse that the wall exerts on the puck?
- (A) Zero, because the puck’s speed does not change
- (B) Opposite to the puck’s velocity before the bounce
- (C) Along the wall, in the direction the puck is travelling
- (D) Perpendicular to the wall, pointing away from it
Answer and explanation
Answer: (D). Split the velocity into parts along and perpendicular to the wall. The part along the wall is unchanged; the perpendicular part reverses. So Δp, and therefore the impulse, is perpendicular to the wall, pointing away from it.
- (A) uses speed. Momentum is a vector, and its direction has changed.
- (B) is true only for a head-on bounce.
- (C) would need friction from the wall, and this wall is frictionless.
If you missed this: read “Impulse” and “The impulse–momentum theorem” in the Topic 4.2 study guide, then “Momentum in two dimensions” in the Topic 4.1 study guide.
Question 5 (multiple choice · 4.3)
Take +x forward. A 50 kg skater holding a 5.0 kg ball glides forward at 2.0 m/s on smooth ice. She throws the ball forward. Just after the throw, the ball moves at +8.0 m/s relative to the ice. What is the skater’s velocity just after the throw?
- (A) −0.80 m/s
- (B) +1.4 m/s
- (C) +1.3 m/s
- (D) +3.0 m/s
Answer and explanation
Answer: (B). System: skater + ball. The throw is internal, and the smooth ice gives no horizontal external force. Before: (55 kg)(+2.0 m/s) = +110 kg·m/s. After: (5.0)(+8.0) + (50)v = 110, so v = 70 ÷ 50 = +1.4 m/s.
- (A) leaves out the 110 kg·m/s the system had before the throw.
- (C) divides by 55 kg. After the throw, only the skater’s 50 kg moves at v.
- (D) adds the ball’s momentum instead of subtracting it. Pushing the ball forward pushes her backward.
If you missed this: read Worked example 2 in the Topic 4.3 study guide.
Question 6 (multiple choice · 4.3)
Take +x east and +y north. On smooth ice, a 1.0 kg puck moving east hits a 2.0 kg puck at rest. Afterwards the 1.0 kg puck moves at 40° north of east and the 2.0 kg puck at 40° south of east. Which statement is correct?
- (A) The pucks move at equal speeds, because their angles are equal.
- (B) The 2.0 kg puck moves twice as fast as the 1.0 kg puck.
- (C) The 1.0 kg puck moves twice as fast as the 2.0 kg puck.
- (D) No comparison is possible without the first puck’s starting speed.
Answer and explanation
Answer: (C). The total y-momentum is zero before, so it is zero after: (1.0)v₁ sin 40° = (2.0)v₂ sin 40°. The sines cancel, leaving v₁ = 2v₂.
- (A) treats equal angles as meaning equal momenta. Equal angles with unequal masses need unequal speeds.
- (B) inverts the mass ratio. The lighter puck needs the larger speed for the same size of y-momentum.
- (D) The starting speed sets the actual speeds, but the ratio comes from the y-equation alone.
If you missed this: read “Momentum in two dimensions” in the Topic 4.3 study guide.
Question 7 (multiple choice · 4.4)
Take +x to the right. Cart A moves right at speed v towards an identical cart B at rest on a level, frictionless track. There is no spring or explosive charge between them. Which outcome is impossible?
- (A) A stops; B moves at +v.
- (B) Both move at +v/2.
- (C) A moves at +v/4; B moves at +3v/4.
- (D) A moves at −v/4; B moves at +5v/4.
Answer and explanation
Answer: (D). All four outcomes conserve momentum: each pair of velocities adds to +v. So check kinetic energy, in units of ½mv². Before: 1. (A): 0 + 1 = 1, elastic. (B): ¼ + ¼ = ½, perfectly inelastic. (C): 1/16 + 9/16 = 10/16, inelastic. (D): 1/16 + 25/16 = 26/16, more than before. With no stored energy to release, a collision cannot increase the system’s kinetic energy.
- (A) is the elastic result for equal masses.
- (B) is what happens if the carts stick together.
- (C) is an ordinary inelastic collision: some kinetic energy is lost, but the carts separate.
If you missed this: read “Two conservation questions for every collision” and “Limiting cases worth knowing” in the Topic 4.4 study guide.
Question 8 (short answer · 4.2)
Take +x towards a bumper. A 0.50 kg cart hits a spring bumper at 0.80 m/s and leaves at 0.60 m/s in the opposite direction. The contact lasts 0.14 s. Friction is negligible.
(a) Find the impulse on the cart. (b) Find the average force from the bumper on the cart. (c) The force–time graph is a symmetric triangle. Find the peak force. (d) A softer bumper gives the same speeds but a contact time of 0.28 s. How do the area under the force–time graph and the average force change?
Worked answer
(a) v₀ = +0.80 m/s and v = −0.60 m/s. J = Δp = 0.50(−0.60 − 0.80) = −0.70 N·s (away from the bumper).
(b) The bumper force is the only horizontal force, so F_avg = J ÷ Δt = −0.70 ÷ 0.14 = −5.0 N (5.0 N away from the bumper).
(c) Triangle area = ½ × 0.14 × F_peak = 0.70 N·s, so F_peak = 10 N (twice the average), away from the bumper.
(d) The speeds are the same, so Δp is the same: the area stays 0.70 N·s. Over twice the time, the average force halves to 2.5 N.
Check yourself: 1 point each for opposite signs on the two velocities, −0.70 N·s, 5.0 N with its direction, 10 N from the triangle area, and (d) with “same area, half the average force”.
If you missed this: read Worked examples 1 and 2 in the Topic 4.2 study guide.
Question 9 (short answer · 4.4)
Take +x to the right. A 0.20 kg lump of clay moving at +4.0 m/s hits a 0.60 kg cart moving at −1.0 m/s on a level, frictionless track. The clay sticks to the cart.
(a) Find the velocity of the clay and cart just after the collision. (b) Find the kinetic energy transformed into other forms. (c) A student says: “Almost all the kinetic energy disappeared, so almost all the momentum must have disappeared too.” Evaluate this claim.
Worked answer
(a) Momentum before = (0.20)(+4.0) + (0.60)(−1.0) = +0.20 kg·m/s. After, the 0.80 kg pair shares one velocity: v_f = 0.20 ÷ 0.80 = +0.25 m/s.
(b) K before = ½(0.20)(4.0)² + ½(0.60)(1.0)² = 1.6 + 0.30 = 1.9 J. K after = ½(0.80)(0.25)² = 0.025 J. Transformed: 1.9 J (1.875 J, about 99%).
(c) Incorrect. Momentum is conserved: +0.20 kg·m/s before and after. Kinetic energy is a separate quantity and is not conserved in a sticky collision. The momenta largely cancelled from the start, so very little kinetic energy can remain.
Check yourself: 1 point each for signed momenta giving +0.20 kg·m/s, +0.25 m/s, both kinetic energy totals with 1.9 J transformed, and rejecting the claim because momentum and kinetic energy are separate conservation questions.
If you missed this: read Worked examples 1 and 3 in the Topic 4.4 study guide.
Your next step
Mark each question right or wrong. Count Questions 8 and 9 as right only if you earned every point.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 4.1 Linear momentum | 1, 2 | Topic 4.1 study guide |
| 4.2 Change in momentum and impulse | 3, 4, 8 | Topic 4.2 study guide |
| 4.3 Conservation of linear momentum | 5, 6 | Topic 4.3 study guide |
| 4.4 Elastic and inelastic collisions | 7, 9 | Topic 4.4 study guide |
How to use your result
- Missed one question in a topic? Re-read the section named in that answer, then try the topic’s practice set.
- Missed two or more questions in a topic? Start again with that study guide, including its worked examples, before doing any mixed practice.
- Guessed? Treat the question as missed.
- Look for patterns. Errors in Questions 3, 5, 8 and 9 usually mean a lost sign: write every velocity with its sign before you multiply. Errors in Questions 1, 7 and 9 often mean momentum and kinetic energy have been mixed up.
- All correct? Move straight to the mixed unit review, which combines the topics in longer exam-style questions.