These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This is the algebra-based course, so no calculus is needed. Momentum is in kg·m/s. Round final answers to 2 significant figures unless told otherwise, and keep unrounded values in your calculator until the end. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
Take +x to the right. A 0.45 kg ball rolls to the left at 12 m/s. What is the ball’s momentum?
- (A) +5.4 kg·m/s
- (B) −5.4 kg·m/s
- (C) −2.7 kg·m/s
- (D) 32 kg·m/s
Answer and explanation
Answer: (B). p_x = m v_x = 0.45 kg × (−12 m/s) = −5.4 kg·m/s. The minus sign means the momentum points left, the same way as the velocity.
- (A) has the right size but ignores the direction of motion. With +x to the right, leftward motion needs a negative sign.
- (C) includes a factor of ½, borrowed from the kinetic energy formula. Momentum has no ½.
- (D) is the kinetic energy, ½ × 0.45 × 12² = 32 J. It is a scalar in joules, not a momentum.
Question 2 (multiple choice · core)
A trolley’s mass is halved and its speed is tripled. By what factor does its momentum change?
- (A) 1.5
- (B) 3
- (C) 6
- (D) 4.5
Answer and explanation
Answer: (A). Momentum is the product mv, so the factors multiply: ½ × 3 = 1.5.
- (B) ignores the change in mass.
- (C) doubles the mass instead of halving it: 2 × 3 = 6.
- (D) is the factor for kinetic energy, ½ × 3² = 4.5. Kinetic energy depends on v², momentum only on v.
Question 3 (multiple choice · core)
Take +x to the right. Cart A (2.0 kg) moves right at 3.0 m/s. Cart B (3.0 kg) moves left at 2.0 m/s. What is the total momentum of the two-cart system?
- (A) +12 kg·m/s
- (B) +2.5 kg·m/s
- (C) 0
- (D) +6.0 kg·m/s
Answer and explanation
Answer: (C). p_A = 2.0 × (+3.0) = +6.0 kg·m/s and p_B = 3.0 × (−2.0) = −6.0 kg·m/s. The sum is zero. Both carts are moving, but their momenta are equal in size and opposite in direction.
- (A) adds the sizes, 6.0 + 6.0, ignoring direction.
- (B) multiplies the plain mean of the two velocities, ½(3.0 − 2.0) = 0.50 m/s, by the total mass 5.0 kg. That ignores the different masses of the carts.
- (D) counts only cart A.
Question 4 (calculation · core)
Take +x east and +y north. A 0.25 kg puck slides across ice with velocity components v_x = +2.4 m/s and v_y = −1.0 m/s.
(a) Find the x- and y-components of the puck’s momentum. (b) Find the size and direction of the puck’s momentum. (c) The puck later moves at the same speed, but due north. State whether its momentum has changed, with a reason.
Worked solution
(a) p_x = 0.25 × (+2.4) = +0.60 kg·m/s; p_y = 0.25 × (−1.0) = −0.25 kg·m/s.
(b) Size: √(0.60² + 0.25²) = 0.65 kg·m/s. Direction: tan⁻¹(0.25 ÷ 0.60) = 23° south of east (below +x), the same direction as the velocity.
(c) Yes, it has changed. The size is still 0.65 kg·m/s because the speed (2.6 m/s) and mass are unchanged, but momentum is a vector, and its direction is now north instead of 23° south of east.
Suggested mark points (4): 1 for both components with correct signs; 1 for the size 0.65 kg·m/s; 1 for the direction; 1 for (c) with the reason “direction changed, so the vector changed”.
Question 5 (graph · core)
Take +x to the right. A 0.50 kg cart moves right at a constant 2.0 m/s from t = 0 to t = 1.0 s. It then meets a spring bumper. From 1.0 s to 2.5 s its velocity changes steadily from +2.0 m/s to −1.0 m/s, after which it leaves the bumper.
(a) Sketch a graph of the cart’s momentum p_x against time from 0 to 2.5 s. Label the values at 0, 1.0 s and 2.5 s. (b) At what time is the cart’s momentum zero? What is the cart doing at that instant? (c) On one set of axes, sketch p_x against v_x for this cart and for a second cart of mass 1.0 kg. Say how the two lines differ. (d) Compare the size of the cart’s momentum at 0.5 s with its size at 2.5 s.
Worked solution
(a) A horizontal line at p_x = +1.0 kg·m/s (0.50 × 2.0) from 0 to 1.0 s. Then a straight line falling from +1.0 kg·m/s at 1.0 s to −0.50 kg·m/s (0.50 × −1.0) at 2.5 s, crossing the time axis on the way. The shape matches the velocity–time graph, scaled by 0.50 kg.
(b) The velocity falls by 3.0 m/s in 1.5 s, which is 2.0 m/s each second. It reaches zero 1.0 s after the bumper contact starts, at t = 2.0 s. At that instant the cart is momentarily at rest, about to move back to the left.
(c) Both are straight lines through the origin, continuing into negative values. The slope of each line is the mass, so the 1.0 kg line is twice as steep as the 0.50 kg line.
(d) At 0.5 s, |p_x| = 1.0 kg·m/s. At 2.5 s, |p_x| = 0.50 kg·m/s. The size at 0.5 s is twice the size at 2.5 s, and the directions are opposite.
Suggested mark points (5): 1 for the flat section at +1.0 kg·m/s; 1 for the straight sloping section ending at −0.50 kg·m/s and crossing zero; 1 for t = 2.0 s with “momentarily at rest”; 1 for two lines through the origin with the heavier one steeper; 1 for the comparison in (d).
Question 6 (constructed response · core)
(a) For each interaction below, decide whether it is best modelled as a collision, an explosion, or neither. Justify each choice by comparing internal and external forces.
(i) Two gliders on a level air track meet and push on each other through magnetic bumpers for 0.08 s, then move apart. (ii) A skateboarder standing on a stationary skateboard jumps off the back, and the board rolls forward. (iii) A loaded sled slides across rough snow and comes to rest over 6 seconds.
(b) For interaction (i), explain why each glider can be treated as a single object, even though the bumpers change shape during contact.
Model answer and suggested Marlbridge rubric
(i) Collision. The air track removes almost all friction, and the weights are balanced by the air cushion, so the net external force is close to zero. The magnetic forces between the gliders are internal to the two-glider system and are much larger.
(ii) Explosion. The skateboarder and board start together at rest. The push between the skateboarder’s feet and the board is internal to the skateboarder–board system and sends the two parts apart.
(iii) Neither. The force that slows the sled is friction from the snow. That force is external to the sled and it is the main force acting, over a long time. Nothing inside the system is much stronger than it.
(b) In the collision model we only compare the state just before contact with the state just after. How the bumpers deform during the 0.08 s does not enter that comparison, so each glider can be described by one mass and one velocity: the object model.
| Point | What earns it |
|---|---|
| 1 | (i) collision, because internal forces are much larger than the net external force |
| 1 | (ii) explosion, because an internal force pushes the parts apart |
| 1 | (iii) neither, because an external force (friction) dominates |
| 1 | (b) only initial and final states are analysed, so internal structure does not matter |
An answer that calls (iii) a collision “between the sled and the snow” earns no credit for (iii): the snow is outside the sled system and the interaction is neither short nor dominated by internal forces.
Question 7 (constructed response · stretch)
A student claims: “If two objects have the same kinetic energy, they must have the same momentum.”
Object P has mass 0.20 kg. Object Q has mass 0.80 kg. Each has kinetic energy 1.6 J.
(a) Calculate the speed and the size of the momentum of each object. (b) Starting from p = mv and K = ½mv², derive an expression for the size of the momentum p in terms of m and K only. (c) Use your expression to evaluate the student’s claim, and predict the ratio of the momenta for any two objects with the same kinetic energy whose masses differ by a factor of 4.
Model answer and suggested Marlbridge rubric
(a) From K = ½mv², v = √(2K ÷ m). P: v = √(2 × 1.6 ÷ 0.20) = 4.0 m/s; p = 0.20 × 4.0 = 0.80 kg·m/s. Q: v = √(2 × 1.6 ÷ 0.80) = 2.0 m/s; p = 0.80 × 2.0 = 1.6 kg·m/s.
(b) From p = mv, v = p ÷ m. Substitute into K = ½mv²: K = ½m(p ÷ m)² = p² ÷ (2m). Rearranging, p = √(2mK).
(c) The claim is false. With the same K, momentum depends on √m, so the heavier object has more momentum. Q has 4 times the mass of P, so its momentum is √4 = 2 times larger, which matches (a): 1.6 ÷ 0.80 = 2. This holds for any pair with equal K and a mass ratio of 4.
| Point | What earns it |
|---|---|
| 1 | Both speeds correct (4.0 m/s and 2.0 m/s) |
| 1 | Both momenta correct (0.80 and 1.6 kg·m/s) |
| 1 | Derives p = √(2mK) with a clear substitution step |
| 1 | States the claim is false and that momentum scales with √m at fixed K |
| 1 | Predicts a factor of 2 and links it to the numbers in (a) |
Alternative method. Writing the ratio directly, p_Q ÷ p_P = √(m_Q ÷ m_P) = √4 = 2, earns the last two points if the reason (same K) is stated.
How did you do?
- Q1 or Q3 wrong: re-read “Momentum is a vector” and Worked example 1 in the study guide. Keep signs all the way through.
- Q2 or Q7 wrong: go back to “Factors of change” and Worked example 2. Momentum scales with v, kinetic energy with v².
- Q4 wrong: revisit “Momentum in two dimensions”.
- Q5 incomplete: see “Graphs of momentum” and Figure 2.
- Q6 incomplete: your reasoning needs the comparison of internal and external forces. See “Collisions and explosions as models”.
Then tick off the topic checklist.