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. Use g = 9.8 m/s² where gravity appears. Treat the ground as an inertial frame and strings as ideal unless a question says otherwise. Round final answers to 2 significant figures unless told otherwise. Any scientific calculator is fine.
Question 1 (multiple choice · foundation)
A lift and its passengers are pulled up by a steel cable. The lift moves upward at a constant speed of 2.0 m/s. Air resistance and friction are negligible. How does the tension in the cable compare with the total weight of the lift and passengers?
- (A) The tension is greater than the weight, because the lift is moving up.
- (B) The tension equals the weight.
- (C) The tension is less than the weight, because the lift is not speeding up.
- (D) The tension equals the weight only if the speed is very small.
Answer and explanation
Answer: (B). Constant velocity means zero net force (first law). Only two forces act: the tension upward and the weight downward. ΣF_y = T − W = 0, so T = W.
- (A) is the idea that motion needs a force in the direction of motion. An upward net force would make the lift speed up, not move steadily.
- (C) would give a downward net force, so the lift would slow down.
- (D) is wrong because the first law does not depend on how fast the object moves, only on whether its velocity changes.
Question 2 (multiple choice · foundation)
In which situation is the net force on the object not zero?
- (A) A car travels at a steady 25 m/s along a straight, level road.
- (B) A book rests on a table.
- (C) A ball thrown straight up is at the highest point of its flight.
- (D) A raindrop falls at a constant speed near the ground.
Answer and explanation
Answer: (C). At the highest point the ball’s velocity is zero for an instant, but gravity still acts on it and nothing balances it (air resistance is negligible at that moment). The net force is mg downward, and its velocity is changing from upward to downward.
- (A) has constant velocity, so the forces on the car add to zero: the road’s forward friction on the tyres balances air resistance and rolling resistance.
- (B) is at rest and staying at rest: weight and normal force balance.
- (D) has constant velocity: air resistance has grown to balance the drop’s weight. Constant speed in a straight line means zero net force, even while falling.
Question 3 (multiple choice · core)
A puck slides in a straight line across frictionless ice at a steady speed. At t = 1.0 s a player taps it with a stick, in the direction it is already moving. The tap lasts 0.05 s. After that nothing horizontal touches the puck. Which description of the puck’s velocity–time graph from t = 0 to t = 3.0 s is correct?
- (A) Horizontal, then a short steep rise at 1.0 s, then horizontal at a higher value until 3.0 s.
- (B) Horizontal, then a short steep rise at 1.0 s, then a gradual fall back to the original value.
- (C) Horizontal, then a short steep rise at 1.0 s, then a gradual fall to zero.
- (D) Horizontal at the same value for the whole 3.0 s, because the tap is so short.
Answer and explanation
Answer: (A). Before the tap, ΣF = 0, so the velocity is constant (flat line). During the tap a forward net force acts, so the velocity increases (a steep rise over 0.05 s). After the tap, ΣF = 0 again, so the velocity stays at its new value (flat again, higher up).
- (B) assumes the puck “returns” to its old motion. Nothing acts to slow it.
- (C) assumes moving objects slow down on their own. On frictionless ice nothing acts against the motion.
- (D) ignores the tap. A short force still changes the velocity; the rise is steep, not absent.
Question 4 (multiple choice · core)
Take +x east and +y north. Three horizontal forces act on a sled that slides across frictionless ice at constant velocity. Two of them are 12 N east and 5.0 N north. What is the third force?
- (A) 13 N, at about 23° south of west
- (B) 13 N, at about 23° north of east
- (C) 17 N, at about 23° south of west
- (D) 13 N, at about 67° south of west
Answer and explanation
Answer: (A). Constant velocity means ΣF_x = 0 and ΣF_y = 0. The third force must have components −12 N (west) and −5.0 N (south). Its size is √(12² + 5.0²) = 13 N. Its direction is tan⁻¹(5.0 ÷ 12) ≈ 23° south of west.
- (B) is the sum of the two given forces. The third force must be opposite to that sum.
- (C) adds the sizes, 12 + 5.0 = 17 N. Forces at right angles add as vectors, not as numbers.
- (D) has the right size but measures the angle from the wrong axis: tan⁻¹(12 ÷ 5.0) ≈ 67° is the angle from the south direction, not from west.
Question 5 (constructed response · core)
A box of mass m is pushed across a level floor at constant velocity. The push has size F and is directed at angle θ below the horizontal (the person pushes forwards and down on the handle). The floor exerts a normal force F_N and a friction force f on the box.
(a) Draw the free-body diagram of the box. (b) Derive expressions for f and F_N in terms of F, θ, m and g. (c) Evaluate f and F_N for m = 30 kg, F = 120 N and θ = 25°. (d) The person now pulls with a force of the same size F, at the same angle θ but above the horizontal. The box stays on the floor. Without new calculation, state whether F_N is larger or smaller than in (c), and by how much.
Worked solution and suggested Marlbridge rubric
(a) Four forces from a dot: push F, forwards and down at θ below horizontal; weight mg, down; normal force F_N, up; friction f, backwards.
(b) Take +x forwards and +y up. Constant velocity, so equilibrium. x: F cos θ − f = 0, so f = F cos θ. y: F_N − mg − F sin θ = 0, so F_N = mg + F sin θ.
(c) f = 120 × cos 25° ≈ 110 N (108.8 N). F_N = 30 × 9.8 + 120 × sin 25° = 294 + 50.7 ≈ 340 N (344.7 N).
(d) The box does not move vertically, so the vertical forces still balance. The pull’s vertical component now points up, so F_N = mg − F sin θ = 294 − 50.7 ≈ 240 N. That is smaller, by 2F sin θ ≈ 100 N (101.4 N). Topic 2.7 shows that friction usually grows with the normal force, which is why pulling at an upward angle is often easier than pushing down at the same angle.
| Point | What earns it |
|---|---|
| 1 | Free-body diagram with four labelled forces in the correct directions |
| 1 | Uses ΣF_x = 0 and ΣF_y = 0, justified by constant velocity |
| 1 | f = F cos θ |
| 1 | F_N = mg + F sin θ, with the push’s vertical component adding to the weight |
| 1 | Correct values in (c) with units |
| 1 | (d) smaller, by 2F sin θ (about 100 N), because the pull’s vertical component now points up |
A common error is F_N = mg. That ignores the vertical component of the push and loses the fourth and sixth points.
Question 6 (constructed response · core)
A train runs along a straight, level track. A small ball rests on the smooth floor of one carriage. While the train moves at a constant 15 m/s, the ball stays at rest relative to the carriage. Then the train starts to speed up, and a passenger sees the ball roll towards the back of the carriage.
(a) Describe the ball’s motion, while the train speeds up, as seen by an observer standing beside the track. Justify your answer with Newton’s first law. (b) Explain why the carriage is not an inertial frame while the train speeds up. (c) Was the carriage an inertial frame while the train moved at 15 m/s? Justify your answer.
Model answer and suggested Marlbridge rubric
(a) The ball keeps moving forwards at a constant 15 m/s. The floor is smooth, so no horizontal force acts on it; its weight and the normal force balance vertically. With zero net force, its velocity stays constant. The carriage speeds up beneath it, so the ball ends up nearer the back.
(b) In the carriage’s frame, the ball starts at rest and then moves backwards, so its velocity changes. But the net force on it is zero. An object with zero net force that changes velocity breaks the first law, so the first law does not hold in this frame. A frame where the first law fails is not inertial.
(c) Yes. While the train moved at constant velocity, the ball (zero net force) stayed at rest relative to the carriage, so the first law held. A frame moving at constant velocity relative to an inertial frame is also inertial.
| Point | What earns it |
|---|---|
| 1 | Ground frame: ball continues at constant velocity (15 m/s forwards) |
| 1 | Reason: no horizontal force on the ball, so ΣF = 0 |
| 1 | Carriage frame: velocity changes with zero net force, so the first law fails |
| 1 | Concludes the accelerating carriage is not inertial |
| 1 | (c) inertial, because the ball obeys the first law (or the carriage moves at constant velocity) |
An answer that says “a force pushes the ball backwards” loses the second and third points: no object exerts that force.
Question 7 (constructed response · stretch)
Take +x and +y as two perpendicular directions in space. A space probe, far from any planet or star, drifts with its engines off at 4.0 m/s in the +x direction. From t = 0 to t = 2.0 s, a side thruster pushes it with a constant force in the +y direction. At t = 2.0 s the thruster stops, and the probe’s y-velocity is then 1.5 m/s.
(a) Sketch, on separate axes, v_x against t and v_y against t from t = 0 to t = 4.0 s. Label key values. (b) Justify the shape of each graph using the forces on the probe. (c) Find the probe’s speed and direction at t = 3.0 s. (d) A student claims: “Some of the thrust went sideways, so the probe must now move more slowly in the x-direction.” Evaluate this claim.
Model answer and suggested Marlbridge rubric
(a) v_x: a horizontal line at 4.0 m/s from 0 to 4.0 s. v_y: a straight line rising from 0 at t = 0 to 1.5 m/s at t = 2.0 s (slope 0.75 m/s²), then horizontal at 1.5 m/s from 2.0 s to 4.0 s.
(b) No force acts along x at any time, so ΣF_x = 0 and v_x stays constant. From 0 to 2.0 s, ΣF_y is a constant non-zero force, so v_y changes at a steady rate (straight sloping line). After 2.0 s, ΣF_y = 0, so v_y stays at its new value.
(c) At t = 3.0 s, v_x = 4.0 m/s and v_y = 1.5 m/s. Speed = √(4.0² + 1.5²) ≈ 4.3 m/s (4.27 m/s). Direction: tan⁻¹(1.5 ÷ 4.0) ≈ 21° from +x towards +y.
(d) The claim is incorrect. The thrust acts only along +y, so no force acts along x and ΣF_x = 0. The first law applies to each axis separately: v_x stays 4.0 m/s. The probe’s speed actually increases, from 4.0 to 4.3 m/s, because a new y-component has been added.
| Point | What earns it |
|---|---|
| 1 | v_x graph flat at 4.0 m/s throughout |
| 1 | v_y graph: linear rise to 1.5 m/s at 2.0 s, then flat |
| 1 | Justifies both flat parts by zero net force along that axis |
| 1 | Justifies the rising part by a constant non-zero force along y |
| 1 | Speed 4.3 m/s and direction about 21° from +x |
| 1 | Rejects the claim: forces along x are balanced, so v_x is unchanged |
A v_y graph that drops back to zero after 2.0 s loses the second and third points.
How did you do?
- Q1 or Q2 wrong: re-read “Translational equilibrium” and “Newton’s first law” in the study guide. Constant velocity, not rest, is the test.
- Q3 or Q7 wrong: study Figure 1 and “What the first law looks like on a graph”.
- Q4 wrong: go back to “Net force is a vector sum”.
- Q5 incomplete: compare your work with Worked examples 1 and 2: one equation per axis, and check the sign of the push’s vertical component.
- Q6 incomplete: revisit “Inertial reference frames”.
Then tick off the topic checklist.