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 strings, ropes and pulleys 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 loaded delivery van and a small hatchback collide head-on. The van’s mass is about four times the hatchback’s. During the collision, how does the size of the force the van exerts on the hatchback compare with the size of the force the hatchback exerts on the van?
- (A) It is about four times larger.
- (B) It is exactly the same.
- (C) It is about four times smaller.
- (D) It depends on which vehicle was moving faster.
Answer and explanation
Answer: (B). The two forces are a third-law pair: the contact force of van on hatchback and the contact force of hatchback on van. They are always equal in size and opposite in direction, whatever the masses or speeds.
- (A) confuses force with effect. The hatchback’s velocity changes about four times as much, because its mass is smaller, but the force on it is not larger.
- (C) reverses the same confusion.
- (D) is the idea that the “attacker” pushes harder. Speeds change how large the force is, but both forces of the pair change together and stay equal.
Question 2 (multiple choice · foundation)
A flower pot rests on a windowsill. Which force is the third-law partner of the gravitational force exerted by Earth on the pot?
- (A) The normal force exerted by the windowsill on the pot
- (B) The normal force exerted by the pot on the windowsill
- (C) The gravitational force exerted by the pot on Earth
- (D) The gravitational force exerted by Earth on the windowsill
Answer and explanation
Answer: (C). Swap the objects and keep the type: “gravitational force of Earth on pot” becomes “gravitational force of pot on Earth”. It points up, towards the pot, and acts on Earth.
- (A) is the classic trap. It is equal and opposite to the weight here only because the pot is at rest, but it is a normal force (different type) and it acts on the same object (the pot).
- (B) is the partner of (A), not of the weight.
- (D) involves the windowsill, not the pot, so it is not part of the pot–Earth interaction.
Question 3 (multiple choice · core)
A light spring scale is joined into the middle of an ideal rope. Two students pull on the two ends of the rope in opposite directions, each with a steady 120 N force, and nothing moves. What does the scale read?
- (A) 0 N
- (B) 60 N
- (C) 120 N
- (D) 240 N
Answer and explanation
Answer: (C). A scale reads the tension in the rope where it sits. The tension in an ideal rope is the same everywhere. Each end is pulled with 120 N, and each piece of rope pulls its neighbour with 120 N, so the scale reads 120 N.
- (A) assumes the two pulls “cancel” inside the rope. They balance, so the rope does not accelerate, but the rope is still being stretched by both.
- (B) splits the force between the two halves of the rope. Tension is not shared out like that.
- (D) adds the two pulls. Tension is not the sum of the forces at the two ends.
Question 4 (multiple choice · core)
Take +x to the right. Two carts sit at rest on a level track with a compressed light spring between them. Cart A has mass 0.50 kg and cart B has mass 1.5 kg. The spring is released and pushes the carts apart. Friction is negligible. Which statement is correct while the spring is pushing?
- (A) The forces on the carts are equal in size, and A’s acceleration is three times B’s.
- (B) The forces on the carts are equal in size, and the accelerations are equal in size.
- (C) The force on B is three times the force on A, so the accelerations are equal in size.
- (D) The force on A is three times the force on B, so A’s acceleration is three times B’s.
Answer and explanation
Answer: (A). The spring pushes on both carts at once; the carts push on each other through it. The forces on A and on B are equal in size and opposite in direction. The same force changes the velocity of a smaller mass more, in inverse proportion to the mass (Topic 2.5 states this as a = F_net ÷ m). B’s mass is 1.5 ÷ 0.50 = 3 times A’s, so A’s acceleration is 3 times B’s.
- (B) gets the forces right but forgets that equal forces on unequal masses give unequal accelerations.
- (C) assumes the heavier cart needs a bigger force “to keep up”. The forces are fixed by the third law, not by the result.
- (D) states the right acceleration ratio but makes the forces unequal. If the force on A really were three times the force on B, A’s acceleration would be nine times B’s.
Question 5 (constructed response · core)
A student (60 kg) pushes horizontally on a heavy crate (40 kg) with a force of 160 N. Neither the student nor the crate moves. Both stand on a rough floor.
(a) Draw separate free-body diagrams for the student and for the crate. Show each force as an arrow from a dot and label it “type, A on B”. (b) Identify two third-law pairs that include forces shown on your diagrams. For each, say which object the partner force acts on. (c) The student says: “The crate does not move because it pushes back on me as hard as I push on it.” Evaluate this claim.
Worked solution and suggested Marlbridge rubric
(a) Student: gravitational force of Earth on student (down, 588 N); normal force of floor on student (up); contact force of crate on student (horizontal, 160 N, pointing away from the crate); static friction of floor on student (horizontal, towards the crate). Crate: gravitational force of Earth on crate (down, 392 N); normal force of floor on crate (up); contact force of student on crate (160 N, away from the student); static friction of floor on crate (towards the student).
(b) Any two, with the partner object named:
- student on crate ↔ crate on student (both are on the diagrams: one on each);
- friction of floor on crate ↔ friction of crate on floor (partner acts on the floor);
- friction of floor on student ↔ friction of student on floor (partner acts on the floor);
- normal force of floor on crate ↔ crate on floor; Earth on student ↔ student on Earth.
(c) The claim is incorrect. The crate’s push acts on the student, so it cannot affect the crate’s motion. To explain why the crate stays still, look only at forces on the crate: the student’s 160 N push is balanced by the floor’s static friction on the crate (160 N the other way). The third-law pair explains why the student feels a push, not why the crate stays still.
| Point | What earns it |
|---|---|
| 1 | Crate diagram: four forces, correct directions, labelled with agent and object |
| 1 | Student diagram: four forces, with the crate’s push and the floor’s friction in opposite horizontal directions |
| 1 | Two correct third-law pairs, each with the partner’s object named |
| 1 | States that the crate’s push acts on the student, not on the crate |
| 1 | Uses the floor’s friction on the crate to explain why the crate stays at rest |
A diagram that puts “crate on student” on the crate’s diagram loses the first two points (the crate gains a wrong force and the student loses one) and usually the fourth.
Question 6 (constructed response · core)
Two astronauts float at rest inside a space station, far from its walls, holding opposite ends of a light rope 6.0 m long. Astronaut P has mass 60 kg (with suit) and astronaut Q has mass 90 kg. Each pulls the rope hand over hand until they meet. Treat the station as an inertial frame, and ignore air resistance and any force from the station on the astronauts.
(a) Compare the sizes of the forces the rope exerts on P and on Q at any instant. Justify your answer. (b) Explain why the centre of mass of the P + Q + rope system does not move. (c) How far does each astronaut move before they meet?
Worked solution and suggested Marlbridge rubric
(a) Equal. The rope is light (ideal), so its tension is the same at both ends, and it pulls each astronaut with that tension.
(b) The only forces are between P, Q and the rope. With all three inside the system, every force is internal and comes with an equal and opposite partner inside the system. Internal forces cannot change the motion of the centre of mass, so it stays at rest where it started.
(c) The astronauts meet at the centre of mass. Measured from P along the rope: x_cm = (60 × 0 + 90 × 6.0) ÷ (60 + 90) = 540 ÷ 150 = 3.6 m. So P moves 3.6 m and Q moves 6.0 − 3.6 = 2.4 m.
Check. The distances are in the ratio 3.6 ÷ 2.4 = 1.5 = 90 ÷ 60, the inverse of the mass ratio. The lighter astronaut moves further.
| Point | What earns it |
|---|---|
| 1 | Equal forces, with a reason (ideal rope tension, or third-law chain) |
| 1 | Identifies all forces as internal to the system |
| 1 | Concludes the centre of mass stays at rest because no external force acts |
| 1 | Locates the meeting point at the centre of mass, 3.6 m from P |
| 1 | P moves 3.6 m and Q moves 2.4 m |
Alternative method. Equal forces for equal times on masses in the ratio 2 : 3 give velocities, and so distances, in the ratio 3 : 2. Splitting 6.0 m in the ratio 3 : 2 gives 3.6 m and 2.4 m. Full credit if the inverse ratio is justified.
Question 7 (constructed response · stretch)
A long, heavy rope lies on smooth ice. A student grabs one end and pulls it so that the whole rope speeds up in a straight line. Points X (at the student’s hand), Y (the middle of the rope) and Z (the far end, which is free) are marked on the rope.
(a) Rank the tensions at X, Y and Z from greatest to least. Explain using the forces that pieces of the rope exert on each other. (b) The heavy rope is replaced by a light (ideal) rope of the same length tied to a sled. The student pulls so that the rope and sled speed up. How do the tensions at the hand end and the sled end compare now? Explain what is different.
Model answer and suggested Marlbridge rubric
(a) T_X > T_Y > T_Z, with T_Z = 0. Think of the rope as segments. The part of the rope beyond any point is speeding up, so something must pull it forward: the segment just in front of that point. The part beyond X is the whole rope, the part beyond Y is half the rope, and nothing lies beyond Z. A larger piece of rope needs a larger forward pull to speed up at the same rate, so the tension is largest at the hand and falls to zero at the free end. (Each segment pulls back on the segment in front with an equal force, by the third law; that backward pull is what the student feels.)
(b) The tensions are equal. A light rope has negligible mass, so no forward pull is needed to speed up the rope itself. Every segment passes the pull on unchanged, and the rope pulls on the sled with the same force that the student pulls on the rope. The difference from (a) is the rope’s mass: in (a) the rope’s own mass needs a net forward force, so the tension changes along it.
| Point | What earns it |
|---|---|
| 1 | Correct ranking with T_Z = 0 |
| 1 | Reasons from the amount of rope that each point must pull along |
| 1 | Uses segment-on-segment forces (third-law pairs) to describe tension |
| 1 | Equal tensions for the light rope, linked to its negligible mass |
How did you do?
- Q1 or Q4 wrong: re-read Worked example 1 and “Predicting with the third law” in the study guide. Equal forces do not mean equal effects.
- Q2 wrong: use the “swap the objects” recipe and study Figure 1.
- Q3 or Q7 wrong: go back to “Tension: the string pulls both ways”.
- Q5 incomplete: your reasoning must use only the forces on the object whose motion you explain. See Worked example 2.
- Q6 incomplete: revisit “Internal and external forces”.
Then tick off the topic checklist.