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 set is for the calculus-based Physics C: Mechanics course. Use g = 9.8 m/s². Draw free-body diagrams to the course conventions: a dot for the center of mass, one straight labelled arrow per force starting on the dot, same-direction forces side by side, and no components. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
A hockey puck has left the stick and slides to the right across smooth, level ice. Friction and air resistance are negligible. Which forces act on the puck while it slides?
- (A) The gravitational force and the normal force only.
- (B) The gravitational force, the normal force and a forward push from the stick.
- (C) The gravitational force, the normal force and a forward “force of motion”.
- (D) The gravitational force only, because the puck is moving.
Answer and explanation
Answer: (A). Earth pulls the puck down (long-range force) and the ice pushes it up (contact force). Nothing touches it in the horizontal direction, so there is no horizontal force. It keeps sliding without one.
- (B) The stick is no longer touching the puck. A contact force needs contact.
- (C) A “force of motion” has nothing that exerts it. Moving does not need a force (Topic 2.4).
- (D) The ice is still touching the puck and pushing up on it. Motion does not remove the normal force.
Question 2 (multiple choice · core)
A child sits in a toy cart on a smooth, level floor. She pulls hard on a rope tied to the front of the cart. Take the system to be the child, the rope and the cart together. Which statement is correct?
- (A) The rope pulls the cart forward, so the system moves forward.
- (B) The rope’s forces are internal to the system, so they give no net force on it and cannot set the system moving.
- (C) Her pull on the rope is larger than the rope’s pull on her, so the system moves forward.
- (D) The rope’s pull on the cart belongs on the system’s free-body diagram as a forward arrow.
Answer and explanation
Answer: (B). The rope, the child and the cart are all inside the boundary. The rope pulls the cart forward and pulls the child’s hands backward, and these internal forces cancel. A system cannot exert a net force on itself.
- (A) looks at the cart alone. For the child–rope–cart system, the backward pull on the child cancels it.
- (C) Forces between two objects are always equal in size (Topic 2.3); the claim is false.
- (D) A system’s free-body diagram shows only forces from the environment. The rope is part of the system.
Question 3 (multiple choice · core)
A student draws a free-body diagram for a crate pulled by a rope at an angle above the horizontal while two friends push it from behind, both horizontally. Which version follows the course conventions?
- (A) A dot with separate arrows for gravity, the normal force, friction and the rope, plus dashed arrows for the rope’s horizontal and vertical components.
- (B) A dot with one arrow labelled “pushes” whose length equals the two friends’ forces added together.
- (C) A dot with each force as its own straight arrow from the dot, with the two pushes drawn as separate arrows side by side.
- (D) A sketch of the crate with each arrow drawn at the point where that force acts on the crate.
Answer and explanation
Answer: (C). Each force is one straight arrow starting on the dot. The two pushes point the same way, so they are drawn side by side, not merged and not overlapping.
- (A) Components are not drawn on the free-body diagram; work them out separately.
- (B) Merging two forces hides that there are two interactions. Draw each force.
- (D) The course draws forces from a dot that represents the center of mass, not from a sketch of the object.
Question 4 (calculation · core)
A 4.0 kg box is on a smooth slope inclined at 30° to the horizontal. A student pushes it with a horizontal force P = 50 N directed towards the slope (so P has a component up the slope and a component into the slope). Use tilted axes: +x up the slope and +y perpendicular to the slope, away from it.
(a) List the forces on the box, naming what exerts each. (b) The box does not accelerate perpendicular to the slope, so the y-components balance. Find the normal force. (c) Find the sum of the force components along the slope, and state its direction.
Worked solution
(a) F_g (Earth on box), straight down; F_N (slope on box), perpendicular to the slope; P (student on box), horizontal. The slope is smooth, so there is no friction.
(b) Components along y: gravity −mg cos 30° = −(39.2)(0.866) = −33.9 N; the push −P sin 30° = −25.0 N (it presses the box into the slope); normal +F_N. So F_N − 33.9 − 25.0 = 0, giving F_N = 59 N (58.9 N).
(c) Along x: +P cos 30° − mg sin 30° = 43.3 − 19.6 = +24 N (23.7 N), up the slope.
Suggested mark points (3): 1 for all three forces with their sources and no extra forces; 1 for both y-components with correct signs, giving 59 N; 1 for the x-sum with direction.
Common errors: F_N = mg cos 30° = 34 N (forgets that the push presses the box into the slope), or F_N = 34 − 25 = 8.9 N (gives the push’s y-component the wrong sign).
Question 5 (diagram · core)
A 3.0 kg flower pot hangs from a single cord. A steady horizontal wind pushes it with a force of 6.0 N to the right, and the pot hangs at rest with the cord at an angle θ to the vertical.
(a) Draw the free-body diagram of the pot. (b) Take +x to the right and +y up. Using the fact that the forces balance, find the tension in the cord and the angle θ. (c) Explain why the tension is greater than the pot’s weight.
Worked solution and suggested Marlbridge rubric
(a) A dot with three separate arrows: F_g (Earth on pot) 29.4 N straight down; F_w (air on pot) 6.0 N to the right; F_T (cord on pot) up and to the left, along the cord. No components drawn.
(b) x: F_w − F_T sin θ = 0, so F_T sin θ = 6.0 N. y: F_T cos θ − mg = 0, so F_T cos θ = 29.4 N. Dividing: tan θ = 6.0 ÷ 29.4, so θ = 12° (11.5°). Squaring and adding: F_T = √(6.0² + 29.4²) = 30 N (30.0 N).
(c) The cord must supply a vertical component equal to the weight and a horizontal component to balance the wind. Its total size is therefore larger than either component alone.
| Point | What earns it |
|---|---|
| 1 | Three forces, each a separate straight arrow from the dot, correctly directed and labelled |
| 1 | No components drawn on the diagram, and no extra forces |
| 1 | Both component equations set up with correct signs |
| 1 | θ = 12° and F_T = 30 N |
| 1 | (c) Tension has to provide both a vertical and a horizontal component |
Question 6 (constructed response · stretch)
Two crates are on a smooth slope inclined at 20°. Crate P (8.0 kg) is higher up the slope. Crate Q (5.0 kg) is below it, tied to P by a rope parallel to the slope. A cable parallel to the slope pulls P, and both crates move up the slope at a steady speed. Use +x up the slope and +y perpendicular to it, away from the slope.
(a) Draw free-body diagrams for P, for Q, and for the system P + Q. (b) Which forces from your diagrams in (a) do not appear on the system diagram, and why? (c) Find the normal force on Q and the tension in the rope between the crates. (d) Find the tension in the cable.
Worked solution and suggested Marlbridge rubric
(a) P: F_g (Earth on P) down; F_N (slope on P) perpendicular to the slope; F_cable (cable on P) up the slope; F_rope (rope on P) down the slope. Q: F_g (Earth on Q) down; F_N (slope on Q) perpendicular to the slope; F_rope (rope on Q) up the slope. P + Q: F_g (Earth on P + Q) down; F_N (slope on P) and F_N (slope on Q), both perpendicular to the slope, so drawn side by side; F_cable up the slope.
(b) The two rope forces (rope on P and rope on Q). The rope connects two parts of the chosen system, so its forces are internal and cancel. Only forces from outside the boundary appear.
(c) For Q, y: F_N − m_Q g cos 20° = 0, so F_N = 5.0 × 9.8 × cos 20° = 46 N. x (steady speed, so balanced): F_rope − m_Q g sin 20° = 0, so F_rope = 5.0 × 9.8 × sin 20° = 17 N (16.8 N).
(d) For the system, x: F_cable − (m_P + m_Q) g sin 20° = 0, so F_cable = 13 × 9.8 × sin 20° = 44 N (43.6 N).
| Point | What earns it |
|---|---|
| 1 | Correct diagrams for P and Q, with rope forces in opposite directions on the two crates |
| 1 | System diagram with no rope forces, and the two normal forces side by side |
| 1 | (b) Identifies the rope forces as internal and explains that they cancel |
| 1 | (c) Normal force on Q, 46 N, from the y-components |
| 1 | (c) Rope tension, 17 N, from the x-components of Q |
| 1 | (d) Cable tension, 44 N, using the system (or P with the rope force) |
Alternative method for (d). Using P alone: F_cable − F_rope − m_P g sin 20° = 0 gives 16.8 + 26.8 = 43.6 N. This earns the point.
Question 7 (explanation · stretch)
A front-wheel-drive car speeds up from rest along a level road. Air resistance is negligible at this low speed.
A student writes: “The engine is what pushes the car forward. The road just holds the car up, because a road can’t push anything sideways.”
(a) Draw the free-body diagram of the car, treating it as a single system. (b) Explain what is wrong with each of the student’s two claims, naming the object that exerts the forward force. (c) Explain, at the level of atoms, how the road can exert a force on the tyres at all.
Model answer and suggested Marlbridge rubric
(a) A dot with: F_g (Earth on car) down; F_N (road on front tyres) up and F_N (road on rear tyres) up, drawn as two separate arrows side by side; F_f (road on front tyres) forward. The two upward arrows together are as long as F_g.
(b) The engine is part of the car, so its forces on the wheels and axles are internal; a system cannot exert a net force on itself. The engine turns the front wheels, the tyres push backward on the road, and the road pushes the tyres forward by static friction. So the road does more than hold the car up: it supplies the forward force as well as the upward normal forces.
(c) Contact forces are electric forces between atoms. Where the tyre presses on the road, the electrons of the surface atoms repel each other (giving the normal force), and atoms in the tyre and road surface attract and catch on one another, resisting sliding (giving friction). On a large scale this appears as forces from the road on the tyres.
| Point | What earns it |
|---|---|
| 1 | Diagram with gravity, the normal force(s) and a forward friction force, no engine force |
| 1 | Two same-direction normal forces side by side (or one combined normal force with a note that it acts at both axles) |
| 1 | Explains that the engine is internal, so it cannot give a net force on the car |
| 1 | Names the road (static friction on the drive tyres) as the source of the forward force |
| 1 | Links contact forces to electric forces between atoms |
How did you do?
- Q1 or Q3 wrong: re-read “What a free-body diagram shows” and the method for finding every force in the study guide.
- Q4 or Q5 wrong: revisit “Choosing axes” and Worked example 1. Make a component table before you write equations.
- Q2, Q6 or Q7 incomplete: go back to “Nothing can push itself” and Worked example 3. Name the system, then ask which forces cross its boundary.
Then tick off the topic checklist.