Newton's laws of motion (RP7) – GCSE Physics exam questions with mark scheme

1Grade 6–76 marks

A student wants to investigate how the acceleration of a trolley depends on the resultant force acting on it.

She has a trolley, a bench pulley, string, a mass hanger with slotted masses (each of weight 1.0 N), a metre rule, a piece of card that can be fixed to the trolley, and a light gate connected to a data logger.

Describe a method she could use. Include how she would make sure the investigation is valid. [6]

attach trolley to the hanger by string passing over the pulley at the edge of the bench(1)
release trolley from rest from the same starting point each time(1)
use the light gate/data logger to measure the acceleration (allow: measure the speed at the light gate and the distance from the start with the metre rule and use v² = 2as; or measure speed at two points and the time between them)(1)
use the metre rule to keep the distance from the start position to the light gate the same each time(1)
fix a double (segmented) interrupt card to the trolley so a single light gate/data logger can record two speeds and the time between them to give the acceleration(1)
the weight of the hanging masses provides the accelerating (resultant) force(1)
change the force by moving masses from the trolley to the hanger one at a time(1)
keep the total mass of the system constant (masses not on the hanger stay on the trolley)(1)
repeat each force several times and calculate a mean acceleration(1)
compensate for friction, e.g. tilt the surface slightly (allow: place the trolley on a runway raised at one end) so the trolley just moves at constant speed when given a small push(1)
plot a graph of acceleration against force(1)
place a box/soft surface under the hanging masses to catch them (safety)(1)

Level	Marks	What it looks like
Level 3	5–6	A clear, logically ordered method that would let someone else carry out the investigation and obtain valid results; how force is varied and acceleration measured are both explained, with the mass kept constant.
Level 2	3–4	Most key steps are described; there may be gaps or an unclear order, and control of mass or measurement of acceleration may be incomplete.
Level 1	1–2	Some relevant steps are given but the method is fragmented and would not produce valid results.
No relevant content	0	Nothing worthy of credit.

Answer: Accelerating force = weight on the hanger; move 1.0 N slotted masses from trolley to hanger to change the force while the total mass stays constant; measure acceleration with the light gate; repeat and take means; plot a against F.
Examiner tip: Validity means keeping the TOTAL accelerating mass constant — the masses taken off the hanger must go back on the trolley.

2Grade 1–3Foundation3 marks

The diagram shows the forces acting on a hot-air balloon. The balloon is rising at a constant velocity.

(a) What is the resultant force on the balloon? Give a reason for your answer. [1]

(b) Calculate the weight of the balloon. [2]

(a) zero / 0 N and because the balloon moves at constant velocity (allow: constant speed in a straight line; forces are balanced)(1)
(b) $6400 - 350$ (allow: weight + 350 = 6400)(1)
(b) 6050 N (allow 6100 N; do not accept 6750 N)(1)
Answer: (a) 0 N (b) 6050 N
Examiner tip: Constant velocity means the forces are balanced, so the upward forces equal the downward forces; air resistance acts downwards here because the balloon is moving up.

3Grade 1–3Foundation3 marks

A delivery drone has a mass of 2.5 kg. It accelerates horizontally at 1.6 m/s².

(a) Write down the equation that links acceleration ($a$), mass ($m$) and resultant force ($F$). [1]

(b) Calculate the resultant force on the drone. [2]

(a) resultant force = mass × acceleration (allow $F = m\,a$ or any correct rearrangement; allow symbols)(1)
(b) $F = 2.5 \times 1.6$(1)
(b) 4.0 N (allow 4 N)(1)
Answer: (a) $F = ma$ (b) 4.0 N
Examiner tip: Learn F = ma by heart – it is not on the equations sheet, and the recall mark is lost if you write it the wrong way round (e.g. m = Fa).

4Grade 4–55 marks

The diagram shows the horizontal forces acting on an electric scooter and its rider. The total mass of the scooter and rider is 80 kg.

(a) Calculate the resultant force on the scooter and rider. [1]

(b) Calculate the acceleration of the scooter and rider. Give the unit. [3]

How does the acceleration compare with your answer to part (b)? [1]

(a) 200 N (forwards)(1)
(b) $200 = 80 \times a$ (allow ecf from (a); allow $a = \dfrac{200}{80}$)(1)
(b) $a = 2.5$(1)
(b) m/s² (allow m s$^{-2}$; unit mark is independent)(1)
(c) the acceleration is smaller (because the mass is bigger) (ignore reference to the size of the force)(1)
Answer: (a) 200 N (b) 2.5 m/s² (c) smaller
Examiner tip: Use the RESULTANT force in F = ma – using the 260 N driving force on its own gives 3.25 m/s² and loses the calculation marks.

5Grade 6–7Higher5 marks

(a) What is meant by the inertia of an object? [1]

(b) A space probe is far from any planet or star. Its thruster produces a constant resultant force of 0.45 N on the probe.

The probe starts from rest. After 600 s its speed is 0.090 m/s.

Calculate the inertial mass of the probe. [4]

(a) the tendency of an object to stay at rest or to continue moving with a constant velocity (unless a resultant force acts on it) (allow: resistance to a change in velocity / motion)(1)
(b) $a = \dfrac{0.090}{600}$(1)
(b) $a = 1.5 \times 10^{-4}$ m/s² (allow 0.00015 m/s²)(1)
(b) inertial mass $= \dfrac{0.45}{1.5 \times 10^{-4}}$ (allow ecf from their acceleration)(1)
(b) 3000 kg (allow $3.0 \times 10^{3}$ kg; an answer of 3000 kg with no working scores 4 marks)(1)
Answer: (a) the tendency to stay at rest or keep moving at constant velocity (b) 3000 kg
Examiner tip: Inertial mass = force ÷ acceleration: first find the acceleration from change in velocity ÷ time, and take care with the powers of ten.

6Grade 4–56 marks

A student investigated how the acceleration of a trolley depends on the force acting on it.

A string attached to the trolley passed over a pulley to a hanger holding slotted masses. The weight of the hanging masses provided the accelerating force. Slotted masses that were not on the hanger were kept on the trolley.

The table shows her results.

Weight of hanging masses in N	Acceleration in m/s²			Mean acceleration in m/s²
Weight of hanging masses in N	Trial 1	Trial 2	Trial 3	Mean acceleration in m/s²
0.20	0.24	0.26	0.25	0.25
0.40	0.49	0.52	0.49	0.50
0.60	0.74	0.91	0.76	X
0.80	0.98	1.01	1.01	1.00
1.00	1.27	1.24	1.24	1.25

(a) Identify the anomalous result in the table. [1]

(b) Calculate the value of X. [2]

(d) What conclusion can the student make from her results? Use data from the table in your answer. [1]

(a) 0.91 (m/s²) (trial 2 for 0.60 N)(1)
(b) $\dfrac{0.74 + 0.76}{2}$ (anomaly ignored)(1)
(b) X = 0.75 (m/s²) (an answer of 0.80 scores 0 marks)(1)
(c) so that the total mass (of the trolley and hanging masses) stays the same / is constant(1)
(c) so that the only independent variable changed is the force / mass is a control variable / it is a fair test (allow: so any change in acceleration is caused only by the change in force)(1)
(d) acceleration is directly proportional to (resultant) force, supported by data, e.g. doubling the force from 0.20 N to 0.40 N doubles the mean acceleration from 0.25 to 0.50 m/s² (‘as force increases acceleration increases’ alone scores 0)(1)
Answer: (a) 0.91 m/s² (b) 0.75 m/s² (c) keeps the total accelerated mass constant, so only the force changes (d) acceleration ∝ force
Examiner tip: The mass that is accelerated is the trolley PLUS the hanging masses, so masses must be moved from trolley to hanger (not added from the bench) to keep it constant.

7Grade 8–9Higher6 marks

A student carried out the trolley investigation (RP7) without compensating for friction. The total mass of the trolley and hanging masses was kept constant.

The graph shows her results.

(a) Explain why the line of best fit does not pass through the origin. [2]

(b) Use the graph to determine the frictional force on the trolley. [1]

(a) there is a frictional force acting on the trolley (opposing motion)(1)
(a) the weight of the hanging masses must be greater than the friction before there is a resultant force / before the trolley accelerates (allow: resultant force = weight of hanging masses − friction)(1)
(b) 0.10 N (from the intercept on the force axis)(1)
(c) a resultant force correctly found from a point on the line, e.g. at 1.00 N: resultant force = 1.00 − 0.10 = 0.90 N (allow: gradient calculated, e.g. $\dfrac{2.25}{0.90} = 2.5$; allow ecf from (b))(1)
(c) mass = resultant force ÷ acceleration, e.g. $\dfrac{0.90}{2.25}$ (allow: mass $= \dfrac{1}{\text{gradient}}$)(1)
(c) 0.40 kg (allow 0.38–0.42 kg; 0.44 kg from 1.00 ÷ 2.25 ignoring friction scores 1 mark only, for the method)(1)
Answer: (a) friction opposes motion, so the hanging weight must exceed it before there is a resultant force (b) 0.10 N (c) 0.40 kg
Examiner tip: Do not use the hanging weight as the resultant force here – subtract the friction (the x-intercept) first, or use mass = 1 ÷ gradient.

8Grade 6–76 marks

A canoeist sits in a canoe that is at rest on a lake. She starts paddling, pushing the water backwards with her paddle. She keeps paddling at the same rate.

The canoe speeds up at first, but after a while it moves at a steady speed.

Explain this motion of the canoe. Use Newton's laws of motion in your answer. [6]

the paddle pushes the water backwards(1)
Newton's third law: the water pushes the paddle (and so the canoe) forwards with an equal (and opposite) force(1)
at the start the forward force is greater than the resistive force / drag (from the water and air)(1)
so there is a resultant force forwards(1)
Newton's second law: a resultant force causes the canoe to accelerate (F = ma)(1)
as the speed increases, the drag / water resistance increases(1)
so the resultant force decreases and the acceleration decreases(1)
eventually the drag equals the forward force(1)
the resultant force is zero(1)
Newton's first law: with zero resultant force the canoe continues at a constant velocity / steady speed(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, coherent explanation using all three laws correctly: the third-law forward force from the water, a forward resultant force giving acceleration, and drag increasing until the forces balance so the canoe moves at constant speed. The steps follow in a logical order.
Level 2	3–4	An explanation that links some forces to the motion, e.g. why the canoe accelerates OR why it reaches a steady speed, with at least one of Newton's laws used correctly. Some steps may be missing or out of order.
Level 1	1–2	Simple relevant statements, e.g. the water pushes the canoe forwards or friction/drag slows it, with few or no links between force and motion.
No relevant content	0	Nothing worthy of credit.

Answer: Third law gives the forward push from the water; second law explains the initial acceleration (resultant force forwards); drag rises with speed until it equals the forward force; first law: zero resultant force gives steady speed.
Examiner tip: Name each law where you use it – top-band answers link the third law (forward push), the second law (acceleration) and the first law (steady speed) in a logical sequence.

9Grade 4–55 marks

A lamp hangs at rest from a cable fixed to a ceiling. The diagram shows the two forces acting on the lamp.

(a) State Newton's third law. [1]

(b) The weight of the lamp is 18 N. What is the size of force A? [1]

Explain why the student is wrong. [2]

(d) Describe the force that forms a Newton's third law pair with force B. [1]

(a) whenever two objects interact, the forces they exert on each other are equal and opposite (allow: if A exerts a force on B, B exerts an equal and opposite force on A; do not accept ‘every action has an equal and opposite reaction’ alone)(1)
(b) 18 N(1)
(c) force A and force B both act on the same object (the lamp)(1)
(c) the two forces in a third-law pair act on different objects
allow: a third-law pair are the same type of force, but A is a tension and B is gravitational
allow: A and B are equal only because the lamp is in equilibrium / they are balanced forces (first law)(1)
(d) the (gravitational) force of the lamp pulling the Earth upwards, 18 N (both direction ‘upwards / towards the lamp’ and object ‘Earth’ needed)(1)
Answer: (a) interacting objects exert equal and opposite forces on each other (b) 18 N (c) both act on the lamp; a third-law pair acts on two different objects (d) lamp pulls Earth upwards with 18 N
Examiner tip: Equal and opposite forces on the SAME object are balanced forces (equilibrium), not a third-law pair – a third-law pair always acts on two different objects.

10Grade 6–74 marks

A family car is waiting at traffic lights. When the lights change, the car accelerates from rest and reaches a speed of 13 m/s after 5.0 s.

(a) Estimate the mass of the car. [1]

(b) Use your estimate to calculate the mean resultant force on the car while it is accelerating. Give your answer using the symbol ≈. [3]

(a) any value from 800 kg to 2000 kg (accept 1000–1500 kg as typical)(1)
(b) $a = \dfrac{13}{5.0} = 2.6$ m/s²(1)
(b) $F = (\text{their mass}) \times 2.6$ (allow ecf from (a))(1)
(b) correct answer for their mass, given with ≈ and sensible precision, e.g. $F \approx 3000$ N for 1200 kg (expected range ≈ 2000 N to ≈ 5000 N) (allow the unrounded value, e.g. ≈ 3120 N; allow ~ in place of ≈, as AQA uses ~ for an approximate value)(1)
Answer: (a) ≈ 1200 kg (b) ≈ 3000 N (2600–3900 N for masses 1000–1500 kg)
Examiner tip: An estimate must still be realistic – a car is about a tonne (1000 kg), not 100 kg or 10 000 kg – and the answer should not be given to more significant figures than the estimate deserves.

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