OxfordAQA International AS and A-level Physics: Vectors and moments (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.2.1 (Scalars and vectors) and 3.2.2 (Moments). This is shared International AS and A-level content. Take g = 9.81 N kg⁻¹ where needed. Learn the ideas first from the study guide or the revision notes. Other links: Physics hub, printable checklist, free 10-minute diagnostics.

Questions

1. Identify which three of these quantities are vectors: speed, displacement, mass, acceleration, distance, weight. [3]

2. A cyclist rides 6.5 km north and then 7.2 km east, taking 0.50 h in total.

(a) Calculate the distance travelled and the magnitude of the displacement. [2] (b) Calculate the bearing of the final position from the start. [2] (c) Calculate the average speed and the average velocity magnitude. [2]

3. Two forces of 30 N and 45 N act at a point with 110° between their directions. Calculate the magnitude of the resultant and its angle from the 30 N force. [4]

4. A child pulls a trolley of mass 40 kg with a rope that exerts 250 N at 32° above the horizontal.

(a) Find the components of the pull along the horizontal and along the vertical. [2] (b) The trolley does not accelerate vertically. Calculate the normal contact force from the ground. [3]

5. Forces of 35 N and 50 N act at a point with 60° between them.

(a) Describe how you would find the resultant by scale drawing. [3] (b) Calculate the resultant and its angle from the 35 N force, to check your drawing. [3]

6. A picture of mass 2.4 kg hangs from a string that passes over a nail. Each side of the string makes 35° with the horizontal. Calculate the tension in the string. [4]

7. A crate of mass 55 kg rests on a smooth ramp at 25° to the horizontal, held by a rope parallel to the ramp. Calculate the tension in the rope and the normal contact force. [4]

8. A parachutist of mass 80 kg falls at a constant speed of 5.0 m s⁻¹, directly downwards.

(a) Calculate the upward drag force on the parachutist. [2] (b) Explain whether the parachutist is in equilibrium. [2]

9. (a) A wheel brace is turned by two equal and opposite forces of 90 N, each perpendicular to the brace, at its two ends 0.42 m apart. Calculate the moment of this couple. [2] (b) Explain why a couple turns an object without moving its centre of mass. [2] (c) A single force of 150 N acts on a spanner 0.28 m from the nut, at 50° to the spanner. Calculate its moment about the nut. [3]

10. A uniform plank of length 4.0 m and mass 16 kg rests on two supports, P and Q, 0.5 m and 3.2 m from its left end. A child of mass 40 kg stands 1.8 m from the left end.

(a) Find the upward force exerted by each of the two supports. [5] (b) The child leaves. Calculate the greatest mass that can be placed on the right-hand end before the plank starts to lift off P. [3]

11. A uniform rod of length 2.0 m and mass 1.5 kg carries a 0.80 kg mass at its left end and a 0.40 kg mass at its right end.

(a) Calculate the distance of the centre of mass of the loaded rod from the left end. [3] (b) A student says that a uniform ring has its centre of mass in the metal of the ring. Correct this statement. [2]

12. A uniform horizontal beam of length 2.4 m and mass 18 kg is hinged at a wall. A cable joins the free end of the beam to the wall, above the hinge, at 40° to the beam. A lamp of mass 12 kg hangs from the beam 1.8 m from the wall.

(a) Calculate the tension in the cable. [4] (b) Calculate the horizontal component of the force exerted by the hinge on the beam, and state its direction. [2] (c) Calculate the vertical component of the hinge force. [2]

Answers

1. Displacement [1], acceleration [1] and weight [1]. [3] Examiner insight: Naming exactly three earns the marks; listing extra quantities such as speed alongside the right ones tends to lose credit.

2. (a) Distance = 6.5 + 7.2 = 13.7 km [1]. Displacement = √(6.5² + 7.2²) = 9.7 km [1]. (b) tan θ = 7.2/6.5 gives θ = 47.9° east of north [1], so the bearing is 048° (to the nearest degree) [1]. (c) Average speed = 13.7/0.50 = 27.4 km h⁻¹ [1]. Average velocity = 9.7/0.50 = 19.4 km h⁻¹ [1]. Examiner insight: A bearing is measured clockwise from north, so here the angle from north is the bearing; an angle with no stated reference line is a common way to lose the mark.

3. Take the 30 N force along x. x-component = 30 + 45 cos 110° = 14.6 N [1]. y-component = 45 sin 110° = 42.3 N [1]. R = √(14.6² + 42.3²) = 44.7 N [1]. Angle = tan⁻¹(42.3/14.6) = 70.9° from the 30 N force [1]. [4] Examiner insight: The cosine of 110° is negative, so the x-component is smaller than 30 N; checking that the resultant is smaller than 75 N, the sum of the magnitudes, catches sign slips.

4. (a) Horizontal = 250 cos 32° = 212 N [1]; vertical = 250 sin 32° = 132 N [1]. (b) Weight = 40 × 9.81 = 392 N [1]. N + 132 = 392 [1], so N = 260 N [1]. Examiner insight: The rope lifts the trolley slightly, so a normal force equal to the full weight is a frequent error; answers show the vertical balance explicitly.

5. (a) State a scale such as 1 cm to 10 N [1]. Draw the 35 N vector, then the 50 N vector from its tip at the correct angle to it, drawn tip to tail (the angle between them tail to tail is 60°, so the angle at the join is 120°) [1]. Join the start to the end; measure its length and convert using the scale, and measure its angle with a protractor [1]. (b) x = 35 + 50 cos 60° = 60 N and y = 50 sin 60° = 43.3 N [1]. R = 74.0 N [1]. Angle = tan⁻¹(43.3/60) = 35.8° from the 35 N force [1]. [6] Examiner insight: A description gains credit for naming the scale, the tip-to-tail joining and the measured quantities; vague phrases such as “draw them accurately” add nothing.

6. Weight = 2.4 × 9.81 = 23.5 N [1]. By symmetry the two sides have equal tension T [1]. Vertical balance: 2T sin 35° = 23.5 [1]. T = 20.5 N [1]. [4] Examiner insight: Using 2T cos 35° is the typical slip when the angle is taken from the horizontal; labelling the angle on a diagram helps you pick the right function.

7. Weight = 55 × 9.81 = 540 N [1]. Along the ramp: T = 540 sin 25° [1] = 228 N [1]. Perpendicular: R = 540 cos 25° = 489 N [1]. [4] Examiner insight: Answers that resolve along and perpendicular to the slope reach the result in two lines; resolving horizontally and vertically works but invites extra unknowns.

8. (a) Constant speed in a straight line means zero resultant force [1], so drag = weight = 80 × 9.81 = 785 N [1]. (b) The resultant force is zero [1], so the parachutist moves with constant velocity and is in equilibrium, even though the parachutist is moving [1]. [4] Examiner insight: A reason referring to the resultant force or constant velocity is needed; “because the forces are balanced” alone is thin.

9. (a) Moment = 90 × 0.42 [1] = 37.8 N m [1]. (b) The forces cancel as vectors, so the net force is zero [1], so there is no acceleration of the centre of mass, but the forces have different lines of action so they produce a turning effect [1]. (c) Perpendicular distance = 0.28 sin 50° = 0.214 m [1]. Moment = 150 × 0.214 [1] = 32.2 N m [1]. [7] Examiner insight: Credit goes to the perpendicular distance, not to 150 × 0.28 = 42 N m; state which distance you use.

10. (a) Moments about P (so R_P drops out): R_Q × 2.7 [1] = 16 × 9.81 × 1.5 [1] + 40 × 9.81 × 1.3 [1]. R_Q = 276 N [1]. Total weight = 56 × 9.81 = 549 N, so R_P = 549 − 276 = 273 N [1]. (b) The plank lifts off P when R_P = 0, so take moments about Q: 16 × 9.81 × 1.2 [1] = m × 9.81 × 0.8 [1], giving m = 24 kg [1]. [8] Examiner insight: Taking moments about a support removes one unknown; mixing pivots within one equation is the usual way marks are lost.

11. (a) Take moments about the left end. Total mass = 1.5 + 0.80 + 0.40 = 2.7 kg [1]. 2.7 x = 1.5 × 1.0 + 0.80 × 0 + 0.40 × 2.0 = 2.3 [1]. x = 0.85 m (0.852 m) from the left end [1]. (b) The centre of mass of a uniform ring is at its centre [1], in the hole, not in the material [1]. [5] Examiner insight: Show each mass multiplied by its distance from the stated point; answers without the reference point named are hard to credit.

12. (a) The component of T perpendicular to the beam is T sin 40° [1]. Moments about the hinge (the hinge force is eliminated): T sin 40° × 2.4 [1] = 18 × 9.81 × 1.2 + 12 × 9.81 × 1.8 [1]. T = 275 N [1]. (b) Horizontal component = T cos 40° = 210 N [1], directed away from the wall (the cable pulls the beam towards the wall) [1]. (c) Total weight = 30 × 9.81 = 294 N. The vertical hinge force is 294 − 275 sin 40° = 118 N upwards [1] (the cable supports 177 N; the hinge provides the remainder) [1]. [8] Examiner insight: Moments about the hinge give the tension in one step; many answers resolve first and gain nothing because the hinge force is unknown.

Where marks are usually lost

• Using the rod’s full length as the arm of an angled force.
• Stating a resultant with no direction, or an angle with no reference line.
• Choosing sin where cos is needed because the angle was not marked on a diagram.
• Forgetting that a tension or support force also has a vertical or horizontal component.
• Placing the weight of a uniform beam anywhere except its midpoint.
• Using different pivots on each side of a moments equation.
• Describing equilibrium as “not moving” rather than zero resultant force.
• Describing a couple as a force; its resultant force is zero.
• Dropping units or writing N m for a force.

Next steps

• Review the revision notes and the study guide.
• Visit the Physics hub and tick off the checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, sections 3.2.1 and 3.2.2, published by OxfordAQA.