Scalars and vectors, weight, resultant forces and work done – GCSE Physics exam questions with mark scheme

1Grade 4–54 marks

A student pushes a heavy box 8.0 m across a floor with a constant horizontal force of 250 N.

(a) Calculate the work done by the student. [2]

(b) The bottom of the box and the floor both become warmer.

Explain why. [2]

(a) $W = 250 \times 8.0$(1)
(a) 2000 J allow 2.0 kJ; allow 2000 N m(1)
(b) work is done against friction (between box and floor)(1)
(b) (so) energy is transferred to the thermal (energy) store of the box/floor allow internal energy increases; do not accept 'heat is made'(1)
Answer: (a) 2000 J (b) work done against friction transfers energy to thermal stores
Examiner tip: In (b) name the force (friction) and the store (thermal) — each is a separate mark.

2Grade 1–34 marks

A student pulls a box along the floor. The diagram shows the horizontal forces on the box.

(a) Determine the size and direction of the resultant force on the box. [2]

(b) The mass of the box is 4.0 kg. Calculate the acceleration of the box. [2]

(a) 16 N(1)
(a) to the right (in the direction of the pull)(1)
(b) 16 = 4.0 × a (allow a = 16 ÷ 4.0; allow ecf from (a))(1)
(b) a = 4.0 m/s²(1)
Answer: (a) 16 N to the right (b) 4.0 m/s²
Examiner tip: A resultant force always needs a direction as well as a size.

3Grade 1–34 marks

An astronaut has a mass of 75 kg. The gravitational field strength on the Moon is 1.6 N/kg.

(a) Calculate the weight of the astronaut on the Moon. [2]

(b) The astronaut returns to Earth. State what happens to the astronaut's mass. Give a reason for your answer. [2]

(a) W = 75 × 1.6(1)
(a) 120 N(1)
(b) mass stays the same / is still 75 kg(1)
(b) mass is the amount of matter (in the astronaut), which does not change / mass does not depend on gravitational field strength (reject: weight stays the same)(1)
Answer: (a) 120 N (b) stays the same (75 kg) — mass does not depend on gravitational field strength
Examiner tip: Weight is in newtons and changes with g; mass is in kilograms and does not.

4Grade 1–36 marks

(a) Some physical quantities are listed below.

energy force mass speed velocity

Which two of these quantities are vector quantities? [2]

(b) Give one difference between a scalar quantity and a vector quantity. [1]

(i) Name the non-contact force that attracts the magnet towards the door. [1]

(ii) Name one contact force that acts on the magnet. [1]

(iii) The weight of the magnet is a non-contact force. Explain why weight is described as a non-contact force. [1]

(a) force(1)
(a) velocity
if more than two quantities are given, deduct one mark for each extra (wrong) quantity, to a minimum of 0(1)
(b) a vector quantity has (magnitude and) direction, a scalar quantity has magnitude only
allow a scalar has no direction
allow size for magnitude
do not accept ‘a vector has direction’ alone without reference to the scalar unless the contrast is clear(1)
(c)(i) magnetic (force)
allow magnetism / magnetic attraction
ignore electrostatic(1)
(c)(ii) friction
or normal contact force (from the door)
allow reaction (force) / support force
ignore air resistance
do not accept tension(1)
(c)(iii) the Earth and the magnet are not touching / the force acts at a distance
allow it is caused by the (Earth’s) gravitational field (acting on the magnet’s mass)(1)
Answer: (a) force and velocity (b) a vector has magnitude and direction; a scalar has magnitude only (c)(i) magnetic force (ii) friction (or normal contact force) (iii) it acts without the objects touching / it is due to a gravitational field
Examiner tip: Speed and velocity are the classic trap: speed is the scalar, velocity is the vector.

5Grade 4–56 marks

A crane lifts a crate of building materials vertically at a steady speed.

(a) Write down the equation that links distance (along the line of action of the force), force and work done. [1]

(b) The crane exerts an upward force of 1800 N on the crate. The crane does 54 kJ of work on the crate.

Calculate the height the crate is lifted. [3]

One joule of work is done when a force of one newton causes a displacement of one ________ . [1]

(d) Name the energy store of the crate that increases as it is lifted. [1]

(a) work done = force × distance (moved along the line of action of the force)
allow W = F s
allow an equation written in symbols only if the symbols are the standard ones(1)
(b) 54 000 = 1800 × s
(conversion of kJ to J and substitution)(1)
(b) s = 54 000 ÷ 1800
(rearrangement)(1)
(b) s = 30 (m)
an answer of 0.03 (m) scores 2 marks (kJ not converted)
an answer of 30 m with no working scores 3 marks(1)
(c) metre
allow m
(i.e. 1 J = 1 N m)(1)
(d) gravitational potential (energy store)
allow GPE
ignore potential alone(1)
Answer: (a) work done = force × distance (b) 30 m (c) metre (1 J = 1 N m) (d) gravitational potential energy store
Examiner tip: Convert kJ to J before substituting: 54 kJ is 54 000 J.

6Grade 8–9Higher7 marks

The diagram is a free body diagram for a small drone flying in a steady wind.

(a) Determine the size and direction of the resultant of the vertical forces on the drone. [1]

(b) Draw a vector diagram to scale to determine the magnitude and direction of the resultant force on the drone. [4]

(c) The drone is then tilted so that the thrust from the propellers is no longer vertical. The weight and the force from the wind do not change. The drone now stays at rest in the air.

Determine the magnitude of the thrust from the propellers. [2]

(a) 9 N upwards(1)
(b) vertical arrow of 9 N (or 30 N and 21 N tip-to-tail) and horizontal arrow of 5 N drawn to a stated scale, tip-to-tail and at right angles
allow ecf from (a)(1)
(b) resultant drawn from the tail of the first vector to the tip of the last vector(1)
(b) magnitude 10.3 N
allow 10.0–10.6 N(1)
(b) direction 61° above the horizontal (to the right / in the direction of the wind)
allow 59°–63° above the horizontal
allow 27°–31° to the vertical
direction must be described, not just an angle(1)
(c) the thrust must balance the resultant of the weight and the wind force / the forces form a closed triangle with 21 N and 5 N at right angles
allow a correct scale diagram or √(21² + 5²)(1)
(c) 21.6 (N)
allow 21.3–21.9 (N)(1)
Answer: (a) 9 N upwards (b) 10.3 N at 61° above the horizontal, towards the direction of the wind (c) 21.6 N
Examiner tip: In a scale drawing, state your scale, and give the direction as an angle from a named reference line (e.g. ‘61° above the horizontal’).

7Grade 6–7Higher7 marks

A traveller pulls a wheeled suitcase along a level floor at a constant speed. The handle makes an angle of 37° with the horizontal. The traveller exerts a force of 50 N along the handle.

(a) Resolve the 50 N force into a horizontal component and a vertical component.

You may use a scale drawing. [3]

(b) Determine the total resistive force acting on the suitcase in the horizontal direction. Give a reason for your answer. [2]

Calculate the work done by the force from the traveller. [2]

(a) scale drawing with the 50 N force at 37° to the horizontal and two perpendicular components drawn (forming a right-angled triangle / rectangle)
allow use of 50 cos 37° and 50 sin 37°(1)
(a) horizontal component 40 N
allow 39–41 N(1)
(a) vertical component 30 N
allow 29–31 N(1)
(b) (the suitcase moves at constant speed so) the resultant (horizontal) force is zero / the forces are balanced(1)
(b) 40 N
allow ecf from horizontal component in (a)
the value of the force must be the same as the horizontal component(1)
(c) W = 40 × 25
allow ecf from (a)(1)
(c) 1000 (J)
allow 1 kJ
allow 975–1025 J from the range in (a)
an answer of 1250 J (using 50 N) scores 0 marks(1)
Answer: (a) horizontal 40 N, vertical 30 N (b) 40 N; constant speed so the resultant force is zero (c) 1000 J
Examiner tip: Work done uses only the component of the force along the direction of movement – using the full 50 N scores nothing.

8Grade 1–35 marks

Two teams take part in a tug of war. The diagram shows the force each person exerts on the rope.

(a) Calculate the total force exerted on the rope by Team A. [1]

(b) Determine the size and direction of the resultant force on the rope. [2]

(c) Later, one person in Team B pulls harder than before. All the other forces stay the same. The rope is now stationary and stays stationary.

(i) State the resultant force on the rope now. [1]

(ii) Determine how much the force from this person has increased by. [1]

(a) 650 N(1)
(b) 10 N
allow ecf from (a): their (a) − 640 N(1)
(b) towards Team A / to the left
only award if a non-zero resultant is given(1)
(c)(i) zero
allow 0 N
allow ‘the forces are balanced’(1)
(c)(ii) 10 N
allow ecf from (b)(1)
Answer: (a) 650 N (b) 10 N towards Team A (to the left) (c)(i) 0 N (ii) 10 N
Examiner tip: A resultant force answer needs a direction as well as a size.

9Grade 6–76 marks

A lorry is travelling along a level road. The driver applies the brakes and the lorry stops.

On another day the lorry carries a much heavier load. It is travelling at the same speed and the same braking force is applied.

Explain, in terms of forces, work done and energy transfers, how the brakes stop the lorry.

Explain how the stopping distance and the temperature rise of the brakes would differ when the lorry carries the heavier load. [6]

Forces: brake pads are pushed against the (brake) discs / wheels; this produces a friction force; the friction force acts in the opposite direction to the motion; there is a resultant force backwards so the lorry decelerates / slows down(1)
Work done: the friction force does work on the lorry / work is done against friction; work done = braking force × distance travelled while braking(1)
Energy transfers: the kinetic energy (store) of the lorry decreases (to zero); energy is transferred to the thermal (internal) energy store of the brakes; the temperature of the brakes increases; some energy is dissipated to the surroundings (by heating the air / sound)(1)
Heavier load – distance: the lorry has a greater mass so it has more kinetic energy at the same speed; more work must be done to stop it; with the same braking force the braking distance must be greater (so the stopping distance is greater); allow the deceleration is smaller (a = F/m) so it takes longer to stop(1)
Heavier load – temperature: more energy is transferred to the thermal store of the brakes so the temperature rise of the brakes is greater (brakes may overheat)(1)

Level	Marks	What it looks like
Level 3	5–6	A detailed, coherent explanation that links the friction force, the work done over the braking distance and the transfer of kinetic energy to the thermal store of the brakes, and then logically applies this to the heavier load to explain both the longer stopping distance and the greater temperature rise.
Level 2	3–4	A partial explanation with relevant ideas about forces, work done or energy transfer, with some attempt to explain at least one effect of the heavier load, but some links in the reasoning are missing or unclear.
Level 1	1–2	Simple, mostly unlinked statements, e.g. that brakes use friction, that brakes get hot, or that a heavier lorry takes longer to stop.
No relevant content	0	No relevant content.

Answer: Brakes produce a friction force opposing motion, so the lorry decelerates; this force does work over the braking distance, transferring kinetic energy to the thermal store of the brakes (temperature rises) and the surroundings. With the heavier load the lorry has more kinetic energy, so more work must be done: the same force must act over a greater distance (longer stopping distance) and more energy is transferred to the brakes, so their temperature rise is greater.
Examiner tip: Top-band answers link each step: more mass → more kinetic energy → more work done needed → greater distance and a greater temperature rise.

10Grade 4–57 marks

A student measures the weight of objects of different mass. The graph shows how weight depends on mass on Earth and on a planet called Planet X.

(a) Name the measuring instrument the student uses to measure weight. [1]

(b) Use the graph to determine the gravitational field strength on Planet X. [2]

(d) A robotic rover has a weight of 3600 N on Planet X.

Calculate the weight of the rover on Earth.

gravitational field strength on Earth = 9.8 N/kg [3]

(a) newtonmeter
allow spring balance / force meter
do not accept (top-pan) balance / scales(1)
(b) correct pair of values read from the Planet X line, e.g. 8.0 N and 2.0 kg
or g = weight ÷ mass with values from the graph(1)
(b) (g =) 4.0 (N/kg)
allow 3.9–4.1(1)
(c) centre of mass
allow centre of gravity(1)
(d) 3600 = m × 4.0
or m = 3600 ÷ 4.0
allow ecf from (b)
alternative method: W = 3600 × (9.8 ÷ 4.0) scores the first two marks in (d)(1)
(d) m = 900 (kg)(1)
(d) (W = 900 × 9.8 =) 8820 (N)
allow ecf for the value of m
allow 8800 (N)
an answer of 8820 N with no working scores 3 marks(1)
Answer: (a) newtonmeter (b) 4.0 N/kg (c) centre of mass (d) 8820 N
Examiner tip: Find the mass first: the mass of the rover is the same on both planets, only the weight changes.

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