What this is for. This diagnostic helps you decide which Unit 4 (Linear Momentum) topics to revisit before you move on. It has one question on Topic 4.1, three on Topic 4.2, two on Topic 4.3 and three on Topic 4.4. Three of the nine need short written working. These are original Marlbridge practice questions, not past exam questions. The set is not calibrated against real exam results, so it does not give a predicted score. Treat each wrong answer as a pointer to a topic.
How to sit it. Allow about 30 minutes. Close your notes. Do the calculus by hand; use a calculator for arithmetic. Use g = 9.8 m/s². Forces are in N, momenta in kg·m/s and t in s. Tracks and floors are level and frictionless unless stated. Every question states its axes.
Question 1 (multiple choice · 4.1)
A 4.0 kg cart and a 1.0 kg cart have the same kinetic energy. What is the ratio of their momenta, p(4.0 kg cart) ÷ p(1.0 kg cart)?
- (A) ½
- (B) 1
- (C) 2
- (D) 4
Answer and explanation
Answer: (C). From K = p²/(2m), |p| = √(2mK). With K the same, p ∝ √m, so the ratio is √(4.0 ÷ 1.0) = 2.
- (A) reasons that the faster, lighter cart must have more momentum. Mass counts too.
- (B) assumes that equal kinetic energy means equal momentum. They depend on m and v in different ways.
- (D) assumes p ∝ m at the same speed. Here the speeds are not the same.
If you missed this: read “Momentum and kinetic energy” in the Topic 4.1 study guide.
Question 2 (multiple choice · 4.2)
Take +x to the right. A 2.0 kg cart moves at +1.0 m/s. The net force on it then changes as follows: F_x rises steadily from 0 to +30 N between t = 0 and t = 0.10 s, falls steadily back to 0 at t = 0.20 s, then is a constant −10 N from t = 0.20 s to t = 0.30 s. What is the cart’s velocity at t = 0.30 s?
- (A) +1.0 m/s
- (B) +2.0 m/s
- (C) +2.5 m/s
- (D) +3.0 m/s
Answer and explanation
Answer: (B). The impulse is the signed area under the F–t graph: a triangle, ½ × 0.20 × 30 = +3.0 N·s, plus a rectangle, −10 × 0.10 = −1.0 N·s. So J = +2.0 N·s = Δp, and Δv = 2.0 ÷ 2.0 = +1.0 m/s. The final velocity is 1.0 + 1.0 = +2.0 m/s.
- (A) is the change in velocity. You must add it to the initial velocity.
- (C) uses only the triangle and ignores the negative force at the end.
- (D) counts the area below the axis as positive. Area below the time axis is negative impulse.
If you missed this: read “Impulse” and “Two graph rules” in the Topic 4.2 study guide.
Question 3 (multiple choice · 4.2)
Take +x to the right. The momentum of a cart is p_x(t) = 3.0t² − 12t. What is the net force on the cart at t = 1.0 s?
- (A) −12 N
- (B) −9.0 N
- (C) −6.0 N
- (D) 0
Answer and explanation
Answer: (C). The net force is the slope of the momentum–time graph: F_x = dp_x/dt = 6.0t − 12. At t = 1.0 s, F_x = −6.0 N.
- (A) is the force at t = 0, not at 1.0 s.
- (B) is Δp/Δt from 0 to 1.0 s, the average net force over that interval. The question asks for the force at an instant: the slope of the tangent at 1.0 s.
- (D) is the force at t = 2.0 s, where the momentum is at its minimum.
If you missed this: read “Net force is the rate of change of momentum” in the Topic 4.2 study guide.
Question 4 (multiple choice · 4.3)
Take +x forward. A rider on a 40 kg cart (the 40 kg includes the rider) carries a 10 kg sack. The cart rolls at 2.0 m/s. The rider pushes the sack off the back so that, just after, the sack is at rest relative to the ground. What is the cart’s speed just after?
- (A) 1.6 m/s
- (B) 2.0 m/s
- (C) 2.2 m/s
- (D) 2.5 m/s
Answer and explanation
Answer: (D). System: cart, rider and sack. The push is internal and there is no horizontal external force, so P_x is constant: 50 × 2.0 = 40v + 10 × 0, giving v = 2.5 m/s.
- (A) puts the masses the wrong way round, 40 × 2.0 ÷ 50. Losing mass while keeping the momentum means the cart speeds up.
- (B) would be right if the sack slid off at the cart’s velocity. Pushing it back to rest gives its momentum to the cart.
- (C) keeps the kinetic energy constant. In an explosion the rider does work, so K changes; momentum does not.
If you missed this: read “Collisions and explosions: ‘just before’ and ‘just after’” in the Topic 4.3 study guide.
Question 5 (multiple choice · 4.4)
Take +x to the right. A 2.0 kg cart moving at +3.0 m/s hits a 1.0 kg cart at rest. Just after, the 2.0 kg cart moves at +1.5 m/s and the 1.0 kg cart at +3.0 m/s. Which statement is correct?
- (A) The collision is elastic, because momentum is conserved.
- (B) The collision is inelastic: 2.25 J of kinetic energy is transformed into other forms.
- (C) The collision is perfectly inelastic, because the carts have the same momentum after.
- (D) The collision is elastic, because the 1.0 kg cart gained the kinetic energy the 2.0 kg cart lost.
Answer and explanation
Answer: (B). Momentum: 6.0 kg·m/s before; 3.0 + 3.0 = 6.0 kg·m/s after. Kinetic energy: 9.0 J before; ½(2.0)(1.5)² + ½(1.0)(3.0)² = 2.25 + 4.5 = 6.75 J after. The total fell by 2.25 J (25%), so the collision is inelastic.
- (A) Momentum is conserved in every collision of an isolated system, so it cannot tell the types apart.
- (C) Perfectly inelastic needs the same velocity (the carts stick together). Equal momenta with different masses means different velocities.
- (D) The 2.0 kg cart lost 6.75 J but the 1.0 kg cart gained only 4.5 J. Check the totals, not one object.
If you missed this: read “Two checks for every collision” in the Topic 4.4 study guide.
Question 6 (multiple choice · 4.4)
Take +x to the right. Cart A (1.0 kg) moves at +6.0 m/s towards cart B (2.0 kg), which is at rest. Which pair of velocities just after the collision is physically possible?
- (A) v_A = 0, v_B = +3.0 m/s
- (B) v_A = −3.0 m/s, v_B = +4.5 m/s
- (C) v_A = +4.0 m/s, v_B = +1.0 m/s
- (D) v_A = +1.0 m/s, v_B = +3.0 m/s
Answer and explanation
Answer: (A). Before: P = 6.0 kg·m/s and K = 18 J. For (A), P = 0 + 6.0 = 6.0 kg·m/s and K = 9.0 J. Momentum is conserved, kinetic energy falls, and B moves ahead of A.
- (B) conserves momentum, but K = 4.5 + 20.25 = 24.75 J, more than the 18 J available. A collision cannot create kinetic energy.
- (C) conserves momentum and lowers K, but A is still behind B and faster. A would pass through B.
- (D) gives P = 1.0 + 6.0 = 7.0 kg·m/s. Momentum is not conserved.
If you missed this: read “Elastic collisions” and “Inelastic and perfectly inelastic collisions” in the Topic 4.4 study guide.
Question 7 (short answer · 4.2)
Take +x perpendicular to a wall, pointing away from it. A 0.20 kg ball hits the wall head-on at 12 m/s and bounces straight back. During the contact, from t = 0 to T = 5.0 ms, the force of the wall on the ball is F_x(t) = F₀ sin(πt/T), with F₀ = 1100 N. Ignore all other forces during contact.
(a) Find the impulse from the wall. (b) Find the ball’s velocity just after the bounce. (c) Find the average force, and compare it with the ball’s weight. (d) At what time during the contact is the ball momentarily at rest?
Answer and explanation
(a) J = ∫₀ᵀ F₀ sin(πt/T) dt = (F₀T/π)[−cos(πt/T)] from 0 to T = 2F₀T/π = 2 × 1100 × 0.0050 ÷ π = +3.5 N·s.
(b) p_i = 0.20 × (−12) = −2.4 kg·m/s. p_f = −2.4 + 3.50 = +1.10 kg·m/s, so v_f = +5.5 m/s.
(c) F_avg = J ÷ T = 3.50 ÷ 0.0050 = 700 N, about 360 times the weight, 1.96 N. So ignoring gravity during contact is safe.
(d) p_x(t) = −2.4 + (F₀T/π)(1 − cos(πt/T)) = 0, so 1 − cos(πt/T) = 2.4 ÷ 1.75 = 1.37 and cos(πt/T) = −0.37. Then πt/T = 1.95 rad, giving t = 3.1 ms. This is after the force peak at 2.5 ms.
If you missed this: work through Worked example 1 in the Topic 4.2 study guide.
Question 8 (short answer · 4.3)
Take +x horizontal and +y upward. A 60 kg student stands on a 20 kg cart at rest on a level, frictionless floor. She throws a 4.0 kg ball forward at 5.0 m/s relative to the ground, 37° above the horizontal (so v_x = 4.0 m/s and v_y = 3.0 m/s).
(a) Which component of the momentum of the student, cart and ball is conserved during the throw? Explain. (b) Find the velocity of the student and cart just after the throw. (c) Find the impulse the floor exerts on the system during the throw, beyond the part that balances the system’s weight.
Answer and explanation
(a) The x-component only. The floor is frictionless, so there is no horizontal external force. Vertically, the floor pushes harder than the total weight while she throws the ball upward, so P_y is not conserved.
(b) P_x = 0 before, so 0 = 4.0 × 4.0 + 80v, giving v = −0.20 m/s (0.20 m/s backwards).
(c) The system’s P_y goes from 0 to 4.0 × 3.0 = 12 kg·m/s; the student and cart stay on the floor. So the net vertical external impulse is 12 N·s, and the floor supplies it: 12 N·s upward beyond the impulse that cancels the weight.
If you missed this: read “Choosing the system” and the sandbag example in the Topic 4.3 study guide.
Question 9 (short answer · 4.4)
Take +x to the right. Cart A (0.50 kg) at +0.80 m/s and cart B (0.30 kg) at −0.60 m/s collide head-on. Just after, A moves at −0.10 m/s.
(a) Find B’s velocity just after. (b) Classify the collision, using kinetic energy. (c) Confirm your answer with approach and separation speeds.
Answer and explanation
(a) P = 0.50(0.80) + 0.30(−0.60) = 0.22 kg·m/s. Then 0.22 = 0.50(−0.10) + 0.30v_B, so v_B = +0.90 m/s.
(b) K before = 0.16 + 0.054 = 0.214 J. K after = 0.0025 + 0.1215 = 0.124 J. Kinetic energy fell by 0.090 J (42%), so the collision is inelastic. The carts did not stick, so it is not perfectly inelastic.
(c) Approach speed = 0.80 + 0.60 = 1.40 m/s. Separation speed = 0.90 + 0.10 = 1.00 m/s. Separation is slower, but not zero: inelastic, as in (b).
If you missed this: read “A quick test in one dimension: approach and separation speeds” in the Topic 4.4 study guide.
Your next step
Count a short answer as missed if any part went wrong.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 4.1 Linear Momentum | 1 | Topic 4.1 study guide |
| 4.2 Change in Momentum and Impulse | 2, 3, 7 | Topic 4.2 study guide |
| 4.3 Conservation of Linear Momentum | 4, 8 | Topic 4.3 study guide |
| 4.4 Elastic and Inelastic Collisions | 5, 6, 9 | Topic 4.4 study guide |
How to use your result
- Read every explanation, even for questions you got right. A right answer for a wrong reason is still a gap.
- Missed one question in a topic? Read the named section, then try that topic’s practice set.
- Missed two or more in a topic, or most of a short answer? Work through the whole study guide for that topic, then its practice set and checklist.
- Missed questions across several topics? Start with Topic 4.2. Force, impulse and momentum feed every later topic.
- Got everything right? Go straight to the mixed unit review, where each question combines two or more topics.
Your result here is a guide to what to study next. It does not predict an exam score.