OxfordAQA International GCSE Physics: Resultant forces and Newton's laws (9203) – 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 follow section 3.1.3, Resultant forces (statements a to h), of the 9203 OxfordAQA International GCSE Physics specification for exams from May/June 2018. It is untiered. Learn the content from the study guide and recall it with the revision notes. More practice: the forces practice set, the OxfordAQA IGCSE Physics hub, the printable checklist and the free diagnostics.

Questions

1. A book rests on a table. The table pushes upwards on the book. Identify the other force in the Newton’s Third Law pair with this push. Give its size compared with the push, its direction and the object it acts on. [2]

2. A trolley is pushed to the right by a 640 N force. Friction of 150 N and air resistance of 90 N both act to the left. Calculate the resultant force on the trolley and state its direction. [3]

3. Two forces act on a buoy at the same point on the water surface. One is 10 N. The other is 6.0 N. The angle between their directions is 60°. Find the resultant force by scale drawing. State your scale, the size of the resultant and its angle from the 10 N force. [4]

4. A cyclist increases her velocity from 2.0 m/s to 11 m/s in 6.0 s.

(a) Calculate her acceleration. [2] (b) She then brakes and comes to rest from 11 m/s in 5.5 s. Calculate her acceleration during braking. [2]

5. A boat travels round a circular course at a constant speed of 8 m/s. A learner says: “The speed is not changing, so there is no acceleration.” Explain whether the learner is correct. [3]

6. The table shows the velocity of a train moving in a straight line.

Time (s)	0	5	12	16
Velocity (m/s)	0	10	10	0

Between each pair of readings the velocity changes at a steady rate.

(a) Calculate the acceleration from 0 to 5 s. [2] (b) State the resultant force on the train from 5 s to 12 s and give a reason. [1] (c) Calculate the acceleration from 12 s to 16 s. [2] (d) Calculate the total distance travelled. [2]

7. A bus brakes sharply. Passengers who are not holding on lurch forwards. Use Newton’s First Law to explain why. [3]

8. (a) A resultant force of 2400 N acts on a car of mass 800 kg. Calculate the acceleration. [2] (b) A resultant force of 55 N gives a trolley an acceleration of 2.5 m/s². Calculate the mass of the trolley. [2]

9. A lorry of mass 5000 kg starts from rest. Its engine provides a forward force of 9500 N. Resistive forces of 4000 N act on it. Assume both forces stay constant.

(a) Find the size of the resultant force. [1] (b) Work out the acceleration. [2] (c) Calculate the time taken to reach 22 m/s. [2] (d) Calculate the distance travelled in this time. [2]

10. A car of mass 1100 kg travelling at 20 m/s brakes with a constant resultant force and stops in 8.0 s.

(a) Calculate the acceleration of the car. [2] (b) Calculate the size of the braking force. [2] (c) Calculate the braking distance. [2]

11. A learner pulls a trolley with a series of different constant forces and measures its acceleration each time. Friction has been compensated for.

Force (N)	0.80	1.6	2.4	3.2
Acceleration (m/s²)	0.50	1.00	1.50	2.00

(a) Describe how the acceleration depends on the force. [1] (b) Calculate the gradient of a graph of acceleration against force, using the point (3.2 N, 2.00 m/s²) and the origin. [2] (c) Use your answer to find the mass of the trolley. [2]

12. A footballer kicks a ball of mass 0.43 kg. The ball accelerates at 60 m/s².

(a) Calculate the resultant force on the ball, giving your answer to 2 significant figures. [2] (b) State the size and direction of the force the ball exerts on the footballer’s foot during the kick, and name the law that tells you. [2]

Answers

1. The book pushes downwards on the table [1] with the same size as the push, and it acts on the table (not the book) [1]. [2] Examiner insight: Naming the object each force acts on is what separates a Third Law pair from two balanced forces on the same object, so always state it.

2. Take right as positive: 640 - 150 - 90 [1] = 400 [1] so the resultant is 400 N to the right [1]. [3] Examiner insight: A force is a vector, so a number with no direction is an incomplete answer; write the direction as part of it.

3. Scale 1 cm to 1 N, so draw a 10.0 cm line [1]. From its end draw a 6.0 cm line at 60° to the first, head to tail [1]. Join the start to the end: the line measures about 14.0 cm, so the resultant is about 14 N [1], at about 22° from the 10 N force [1]. [4] Examiner insight: A drawing gives a range, not an exact value, so a careful answer states the scale, shows the measurement and names the force the angle is measured from.

4. (a) Δv = 11 - 2.0 = 9.0 m/s [1]; a = 9.0 / 6.0 = 1.5 m/s² [1] (b) Δv = 0 - 11 = -11 m/s [1]; a = -11 / 5.5 = -2.0 m/s², a deceleration of 2.0 m/s² [1]. [4] Examiner insight: Subtract the starting velocity from the final velocity before dividing, and keep the minus sign for slowing down.

5. The learner is not correct [1]. Velocity includes direction, and the boat’s direction is continually changing, so its velocity changes [1]. A change in velocity is acceleration, so the boat is accelerating (with a resultant force towards the centre of the circle) [1]. [3] Examiner insight: Link speed, direction and velocity explicitly; saying only “direction changes” leaves the step to acceleration unstated.

6. (a) a = Δv / t = 10 / 5 [1] = 2.0 m/s² [1] (b) Zero, because the velocity is constant (the forces are balanced) [1]. (c) a = (0 - 10) / (16 - 12) [1] = -2.5 m/s² [1] (d) Area = 1/2 x 5 x 10 + 7 x 10 + 1/2 x 4 x 10 = 25 + 70 + 20 [1] = 115 m [1]. [7] Examiner insight: In part (c) the time is the length of that stage (4 s), not the clock reading of 16 s; check which interval your gradient uses.

7. The bus slows because of a force on it from the brakes and road [1]. No resultant force acts on a passenger who is not holding on, so their velocity stays the same (First Law) [1]. They keep moving forwards at the original velocity, so they move forwards relative to the slowing bus [1]. [3] Examiner insight: Avoid saying “a force pushes them forward”; the explanation is that no force slows them, not that one speeds them forward.

8. (a) a = F / m = 2400 / 800 [1] = 3.0 m/s² [1] (b) m = F / a = 55 / 2.5 [1] = 22 kg [1]. [4] Examiner insight: Writing the rearranged equation before substituting makes your method clear to the reader and helps you spot slips.

9. (a) 9500 - 4000 = 5500 N [1] (b) a = 5500 / 5000 [1] = 1.1 m/s² [1] (c) t = Δv / a = 22 / 1.1 [1] = 20 s [1] (d) Velocity rises steadily from 0, so distance = 1/2 x 20 x 22 [1] = 220 m [1]. [7] Examiner insight: Put the resultant, not the engine force, into F = ma; using 9500 N here would give an acceleration that is too large.

10. (a) a = (0 - 20) / 8.0 [1] = -2.5 m/s² [1] (b) F = ma = 1100 x 2.5 [1] = 2750 N (opposite to the motion) [1] (c) Distance = 1/2 x 8.0 x 20 [1] = 80 m [1]. [6] Examiner insight: Show a negative acceleration in part (a) and then use its size in part (b), saying the force acts against the motion.

11. (a) Acceleration rises in proportion to force: each doubling of force doubles the acceleration [1]. (b) Gradient = 2.00 / 3.2 [1] = 0.625 m/s² per N [1] (c) a = F / m, so the gradient is 1 / m [1]; m = 1 / 0.625 = 1.6 kg [1]. [5] Examiner insight: To support “proportional” quote numbers from two rows, such as the force doubling from 0.80 N to 1.6 N while the acceleration doubles from 0.50 to 1.00 m/s².

12. (a) F = ma = 0.43 x 60 [1] = 25.8, so 26 N (2 s.f.) [1] (b) 26 N [1], in the opposite direction to the force on the ball; this is Newton’s Third Law [1]. [4] Examiner insight: Name the law in words (“Newton’s Third Law”) and state both size and direction, since “an equal force” alone does not give the direction.

Where marks are usually lost

• Giving a force without a direction.
• Putting only the driving force into F = ma.
• Taking Δv as the final velocity instead of final minus initial.
• Dropping the minus sign for a deceleration, or using it twice.
• Using the clock time rather than the interval for a stage of a graph.
• Forgetting the half when finding the area of a triangle.
• Saying Third Law forces cancel instead of stating they act on different objects.
• Saying a steady-speed circular path involves no acceleration.
• Giving an angle from a scale drawing without saying which force it is measured from.
• Leaving out units or rounding to the wrong number of significant figures.

Next steps

• Recall the content using the revision notes.
• Return to the study guide.
• Go to the OxfordAQA IGCSE Physics hub or tick off the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Source document: OxfordAQA International GCSE Physics (9203), version 5.2, exams May/June 2018 onwards. Scope here: section 3.1.3, a to h.