OCR GCSE Physics A: Interacting forces, vectors and free body diagrams (J249) – 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 statements P2.2a to P2.2h and P2.2p of the OCR GCSE (9-1) Gateway Science Physics A specification (J249, Version 5.0, July 2026, for first assessment in 2018). None of these statements is Higher Tier only. Read the study guide or the revision notes first if you need to. See also the OCR GCSE Physics hub, the checklist and OCR GCSE Physics: Forces practice. Check your level with the free 10-minute diagnostics.

Questions

1. The forces below are: normal contact force, magnetism, friction, gravity, electrostatics. Sort them into contact forces and non-contact forces. [2]

2. A fridge magnet is held to a fridge door by a magnetic force of 2.5 N.

(a) State the size and direction of the force that the door exerts on the magnet, relative to the force of the magnet on the door. [2] (b) Name the type of force. [1]

3. A boy leans on a wall with a horizontal force of 30 N.

(a) State the force the wall exerts on the boy. [2] (b) Explain why these two forces do not cancel each other. [2]

4. A lamp of weight 18 N hangs at rest from a ceiling by a cable. Describe the free body diagram for the lamp. [4]

5. A trolley is pushed to the right with 48 N. Friction of 12 N and air resistance of 9 N act to the left. Calculate the resultant force, with its direction. [3]

6. A drone is pushed by wind with a horizontal force of 24 N east. Its rotors give a force of 7.0 N north. Ignore all other forces.

(a) Describe a scale drawing to find the resultant, using 1 cm to represent 2 N. [2] (b) Calculate the size of the resultant force. [2] (c) Calculate the angle of the resultant north of east. [1]

7. A child pulls a sledge with a rope. The force in the rope is 85 N. Its horizontal and vertical sides in a scale drawing are in the ratio 15 : 8 : 17 (horizontal : vertical : rope). Use a scale of 1 cm to 5 N to find the horizontal and vertical components. [4]

8. A ring is pulled with 12 N east and 16 N north. A third force acts on it so that it is in equilibrium. Determine the size and direction of the third force, and describe the vector diagram. [5]

9. A lorry travels at a constant 20 m/s on a level road with a driving force of 5200 N.

(a) Explain what this tells you about the resistive forces. [2] (b) The driver brakes. Describe the resultant force and the change in velocity. [2]

10. A parachutist of weight 650 N opens the parachute while falling fast. At that moment drag is 1100 N.

(a) Calculate the resultant force, with its direction. [2] (b) Explain how the parachutist reaches a new constant velocity. [4]

11. A student says: “A book’s weight and the table’s normal contact force are a Newton’s third law pair, because they are equal and opposite.” Evaluate this claim. [4]

12. A child sits on a sledge on level snow. The rope force is 80 N, at an angle such that the horizontal and vertical sides are in the ratio 4 : 3 : 5 (horizontal : vertical : rope). The combined weight is 300 N. Friction is 40 N.

(a) Find the horizontal and vertical components of the rope force. [2] (b) Describe the free body diagram for the sledge. [2] (c) Calculate the horizontal resultant force. [2] (d) The vertical forces balance. Calculate the normal contact force. [1] (e) Name the interaction pair partner of the normal contact force. [1]

Answers

1. Contact: normal contact force and friction [1]. Non-contact: magnetism, gravity and electrostatics [1]. [2] Examiner insight: Both groups need to be right; one misplaced force usually costs a mark, so check each one against whether touching is needed.

2. (a) The force is 2.5 N [1] in the opposite direction, towards the door [1]. (b) Magnetic [1]. [3] Examiner insight: “Equal and opposite” needs both words addressed: give the size and the direction, not just “the same”.

3. (a) 30 N [1], horizontal, pushing the boy away from the wall [1]. (b) Each force acts on a different object [1]: the 30 N is on the wall and the 30 N is on the boy, so neither cancels the other [1]. [4] Examiner insight: Strong answers say which object each force acts on. “They are equal and opposite so they cancel” is the usual wrong answer.

4. Draw the lamp only [1]. Weight of 18 N down [1]. Tension of 18 N up from the cable [1]. The arrows are the same length because the resultant is zero (the lamp is at rest) [1]. [4] Examiner insight: Label each arrow with the object that exerts it and keep equal forces equal in length; unlabelled arrows are hard to credit.

5. Total resistance = 12 + 9 = 21 N [1]. Resultant = 48 − 21 = 27 N [1] to the right [1]. [3] Examiner insight: Direction is part of a resultant. Writing “27” with no unit or direction gives away marks.

6. (a) Draw 12.0 cm east and then 3.5 cm north from its end [1]. Join start to end and measure the line, and the angle with a protractor [1]. (b) R = √(24² + 7.0²) [1] = 25 N [1]. (c) tan⁻¹(7.0 / 24) = 16° north of east [1]. [5] Examiner insight: Convert with the scale before drawing, and show that the arrows go head to tail.

7. Rope arrow = 85 ÷ 5 = 17.0 cm [1]. Construct a right-angled triangle with the rope as the longest side [1]. Horizontal side 15.0 cm = 75 N [1]. Vertical side 8.0 cm = 40 N [1]. Check: 75² + 40² = 7225 = 85². [4] Examiner insight: Show the scale conversion for both components; answers in centimetres instead of newtons lose the final marks.

8. Resultant of the 12 N and 16 N = √(12² + 16²) = 20 N [1]. The third force is also 20 N [1] in the opposite direction [1]. Angle: tan⁻¹(16 / 12) = 53°, so it points 53° south of west [1]. Drawn head to tail the three arrows form a closed triangle [1]. [5] Examiner insight: Equilibrium means the arrows return to their start. A third force that is only the right size, with the wrong direction, does not meet the condition.

9. (a) Constant velocity means the resultant force is zero (Newton’s first law) [1], so the resistive forces total 5200 N [1]. (b) The braking force makes a resultant that acts backwards [1], so the velocity decreases and the lorry slows [1]. [4] Examiner insight: Link the motion to the resultant. “The forces are balanced” needs the reason, constant velocity.

10. (a) 1100 − 650 [1] = 450 N upward [1]. (b) Drag is greater than weight, so the resultant is upward and the parachutist slows [1]. As speed falls, drag falls [1]. This continues until drag equals 650 N, the weight [1]. The resultant is then zero, so velocity is constant, and smaller than before [1]. [6] Examiner insight: A full explanation covers the cause (drag changes with speed) and the end condition (resultant zero). Stating only “terminal velocity” is not an explanation.

11. The claim is not correct [1]. The weight and the normal contact force both act on the book, one object [1]. They are different types of force: gravitational and contact [1]. A pair is the book’s gravitational pull on the Earth with the Earth’s pull on the book, or the book’s push on the table with the table’s push on the book [1]. [4] Examiner insight: “Equal and opposite” is not sufficient to identify a pair. Check the object and the type for both forces.

12. (a) Horizontal = 80 × 4/5 = 64 N [1]; vertical = 80 × 3/5 = 48 N [1]. (b) Sledge only, with arrows for: weight 300 N down, normal contact force up, rope force at an angle (or its components) and friction 40 N backward [1]. Label each force [1]. (c) 64 − 40 [1] = 24 N forward [1]. (d) 300 − 48 = 252 N [1]. (e) The sledge’s push on the snow (a contact force of 252 N, downward) [1]. [8] Examiner insight: Treat horizontal and vertical separately, and say which direction each resultant is in. Do not draw the force of the sledge on the snow in the sledge’s diagram.

Where marks are usually lost

• Not saying which object each force acts on.
• Calling weight and normal contact force a third-law pair.
• Giving a resultant without a direction or a unit.
• Adding perpendicular forces arithmetically.
• Not converting with the scale in a vector diagram.
• Drawing the force of the object on the surroundings in a free body diagram.
• Saying “no forces” when the forces are balanced.
• Treating constant velocity as needing a constant resultant force.
• Stopping the terminal velocity explanation before the final balance.

Next steps

• Revision notes
• Study guide
• OCR GCSE Physics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0, July 2026, for first assessment in 2018, published by OCR: statements P2.2a to P2.2h and P2.2p.