9th Class Physics — MCQs & Practice Questions

78 multiple-choice questions and 63 exam-style questions with mark schemes, organised by chapter, with answers you can check as you go. Free, no sign-up.

New to a topic? Read the 9th Class Physics notes first, then come back to practise.

01

Physical Quantities and Measurements

Multiple choice · 6

Q1A student measures the same length five times and gets 4.71, 4.72, 4.71, 4.72, 4.71 cm. The true length is 5.20 cm. The measurements are:

  1. AAccurate and precise
  2. BPrecise but not accurate
  3. CAccurate but not precise
  4. DNeither accurate nor precise
Show answer

Correct answer: B — Precise but not accurate

The readings agree with each other to within 0.01 cm, so they are precise. They all sit about 0.49 cm below the true value, so they are not accurate. A consistent offset like this is the signature of a systematic error — a zero error on the instrument, most likely.

Q2Which of these is NOT an SI base unit?

  1. Akilogram
  2. Bnewton
  3. Ckelvin
  4. Dmole
Show answer

Correct answer: B — newton

The newton is derived: 1 N = 1 kg m s⁻², built from three base units. The kilogram, kelvin and mole are all base units. The tempting mistake is assuming that any famous unit must be fundamental.

Q3The dimensions of pressure are:

  1. AM L T⁻²
  2. BM L⁻¹ T⁻²
  3. CM L² T⁻²
  4. DM L⁻² T⁻¹
Show answer

Correct answer: B — M L⁻¹ T⁻²

Pressure is force ÷ area = (M L T⁻²) ÷ L² = M L⁻¹ T⁻². Option A is force itself and option C is energy, which is why both look plausible if you stop one step early.

Q42.5 m is multiplied by 3.14159 m. To the correct number of significant figures the answer is:

  1. A7.853975 m²
  2. B7.85 m²
  3. C7.9 m²
  4. D8 m²
Show answer

Correct answer: C — 7.9 m²

The least precise input, 2.5, carries two significant figures, so the answer does too: 7.9 m². Writing 7.85 keeps three and quietly claims the 2.5 was really 2.50.

Q5Repeating a measurement many times and averaging will reduce:

  1. ASystematic error only
  2. BRandom error only
  3. CBoth equally
  4. DNeither
Show answer

Correct answer: B — Random error only

Random errors scatter either side of the true value, so they partly cancel in an average. A systematic error pushes every single reading the same way, so it survives averaging untouched — you have to find its cause instead.

Q6A length is recorded as 0.00470 m. How many significant figures does it have?

  1. ATwo
  2. BThree
  3. CFive
  4. DSix
Show answer

Correct answer: B — Three

Three: 4, 7 and the trailing 0. The leading zeros only place the decimal point and never count, but the final zero is after the decimal point and after a non-zero digit, so it is a genuine claim about precision.

Exam-style questions · 5

Q1[2 marks]
Differentiate between base and derived units, giving one example of each.
Answer

A base unit is one of the seven independent SI units, such as the metre. A derived unit is built from base units by multiplication or division, such as the newton, 1 N = 1 kg m s⁻².

Q2[2 marks]
What is meant by the least count of a measuring instrument?
Answer

The smallest measurement the instrument can read — the value of one division on its scale. A metre rule has a least count of 1 mm; a vernier calliper, 0.1 mm; a screw gauge, 0.01 mm.

Q3[2 marks]
Why is the mean of several readings more reliable than a single reading?
Answer

Random errors scatter either side of the true value, so averaging makes them partly cancel. A single reading may happen to be one of the extreme ones.

Q4[3 marks]
The length of a rod is measured as 12.5 cm with an uncertainty of 0.1 cm. Express this in metres and state the percentage uncertainty.
Mark scheme
  1. Convert: 12.5 cm = 0.125 m, 0.1 cm = 0.001 mdivide by 100[1]
  2. Percentage uncertainty = (Δl / l) × 100 = (0.1 / 12.5) × 100same units top and bottom, so the conversion cancels[1]
  3. = 0.8%[1]

0.125 ± 0.001 m, or 0.8%

Q5[4 marks]
A student measures the diameter of a wire five times with a screw gauge and records: 0.42, 0.43, 0.42, 0.51, 0.43 mm.
  1. Identify the anomalous reading and state what should be done with it.
  2. Calculate the mean diameter using the remaining readings.
Mark scheme
  1. 0.51 mm is anomalous — it lies well away from the others[1]
  2. It should be excluded from the mean, but reported rather than deletedthe reporting half is the mark most often missed[1]
  3. Mean = (0.42 + 0.43 + 0.42 + 0.43) / 4four readings, not five[1]
  4. = 0.425 ≈ 0.43 mm[1]

(a) 0.51 mm, excluded but reported (b) 0.43 mm

02

Kinematics

Multiple choice · 12

Q1An athlete runs exactly one lap of a 400 m circular track in 50 s. Their average velocity is:

  1. A8 m s⁻¹
  2. B0 m s⁻¹
  3. C400 m s⁻¹
  4. D4 m s⁻¹
Show answer

Correct answer: B — 0 m s⁻¹

Average velocity is displacement ÷ time, and after a complete lap the displacement is zero — start and finish are the same point. The 8 m s⁻¹ answer is the average speed, which uses distance instead.

Q2A car moves at a constant 20 m s⁻¹ around a circular bend. Which statement is true?

  1. AIts velocity is constant
  2. BIts acceleration is zero
  3. CIt is accelerating because its direction is changing
  4. DIt cannot accelerate while its speed is constant
Show answer

Correct answer: C — It is accelerating because its direction is changing

Velocity is a vector, so changing direction changes the velocity even at constant speed — and a changing velocity is what acceleration means. The acceleration points toward the centre of the bend.

Q3An object has negative velocity and negative acceleration. It is:

  1. AMoving forwards and slowing down
  2. BMoving backwards and speeding up
  3. CMoving backwards and slowing down
  4. DStationary
Show answer

Correct answer: B — Moving backwards and speeding up

Matching signs mean the acceleration acts in the same direction as the motion, so the object speeds up. The common error is reading "negative acceleration" as "deceleration", which is only true when the velocity is positive.

Q4On a velocity–time graph, the area between the line and the time axis represents:

  1. AAcceleration
  2. BDisplacement
  3. CSpeed
  4. DForce
Show answer

Correct answer: B — Displacement

Velocity × time = displacement, and the area is that product accumulated. The gradient of the same graph gives acceleration — the two are the pair most often swapped.

Q5A horizontal line on a velocity–time graph means the object is:

  1. AStationary
  2. BMoving at constant velocity
  3. CAccelerating uniformly
  4. DChanging direction
Show answer

Correct answer: B — Moving at constant velocity

Constant velocity, so zero acceleration. A stationary object would be a horizontal line sitting on the time axis itself, at v = 0 — a special case, not the general meaning.

Q6A stone is dropped from rest and falls for 3.0 s. Taking g = 10 m s⁻², how far does it fall?

  1. A15 m
  2. B30 m
  3. C45 m
  4. D90 m
Show answer

Correct answer: C — 45 m

s = ut + ½at² with u = 0 gives s = ½ × 10 × 3.0² = 45 m. Answer B is the mistake of calculating v = at = 30 and calling it a distance; answer A comes from forgetting to square the time.

Q7Which equation of motion would you choose if the question gives u, a and s, and asks for v?

  1. Av = u + at
  2. Bs = ut + ½at²
  3. Cv² = u² + 2as
  4. Ds = ½(u + v)t
Show answer

Correct answer: C — v² = u² + 2as

The unknown you neither have nor want is t, so use the equation that omits t. Options A and B both contain t and would need it to be found first — twice the work and twice the chance of an arithmetic slip.

Q8Two balls leave a table edge at the same moment — one dropped, one thrown horizontally at 5 m s⁻¹. Ignoring air resistance:

  1. AThe dropped ball lands first
  2. BThe thrown ball lands first
  3. CThey land at the same time
  4. DIt depends on their masses
Show answer

Correct answer: C — They land at the same time

Vertical and horizontal motion are independent. Both start with zero vertical velocity and fall the same height under the same g, so both take the same time. The thrown ball merely covers ground while it falls.

Q9At the highest point of a projectile's path, its velocity is:

  1. AZero
  2. BEqual to the horizontal component of the launch velocity
  3. CEqual to the launch velocity
  4. DDirected vertically upward
Show answer

Correct answer: B — Equal to the horizontal component of the launch velocity

Only the vertical component reaches zero at the top. Nothing acts horizontally, so that component is unchanged throughout the flight and is the whole of the velocity at the peak.

Q10A projectile is launched on level ground. Which launch angle gives the greatest range?

  1. A30°
  2. B45°
  3. C60°
  4. D90°
Show answer

Correct answer: B — 45°

Range = u² sin 2θ / g, largest when sin 2θ = 1, so 2θ = 90° and θ = 45°. A 90° launch goes straight up and lands back at the launch point with zero range — the answer that catches anyone reasoning "higher must be further".

Q11The SUVAT equations may only be used when:

  1. AThe object is falling freely
  2. BThe acceleration is constant
  3. CThe velocity is constant
  4. DAir resistance is present
Show answer

Correct answer: B — The acceleration is constant

Constant acceleration is the one condition. Free fall is a common case of it, not the requirement. Where acceleration varies, you need the gradient and area of a graph, or calculus.

Q12Air resistance acts on a projectile. Compared with the ideal path, the actual path has:

  1. AA longer range and a symmetric shape
  2. BA shorter range and a steeper descent than ascent
  3. CThe same range but a lower peak
  4. DA longer time of flight
Show answer

Correct answer: B — A shorter range and a steeper descent than ascent

Drag opposes the motion throughout, cutting both range and maximum height, and the descent becomes steeper than the ascent — so the path is no longer a symmetric parabola.

Exam-style questions · 8

Q1[4 marks]
A cyclist travelling at 4.0 m s⁻¹ accelerates uniformly to 10.0 m s⁻¹ over a distance of 42 m.
  1. Calculate the acceleration of the cyclist.
  2. Calculate the time taken.
Mark scheme
  1. Selects v² = u² + 2asthe equation without t, since t is not given in part (a)[1]
  2. 10.0² = 4.0² + 2 × a × 42 → 100 = 16 + 84a → a = 1.0 m s⁻²unit required for the mark[1]
  3. Selects v = u + at (or s = ½(u+v)t)[1]
  4. 10.0 = 4.0 + 1.0t → t = 6.0 s[1]

(a) 1.0 m s⁻² (b) 6.0 s

Q2[5 marks]
A stone is dropped from rest at the top of a cliff and hits the sea 3.2 s later. Take g = 9.81 m s⁻² and ignore air resistance.
  1. Calculate the height of the cliff.
  2. Calculate the speed at which the stone hits the water.
  3. State one effect of air resistance on your answer to (b).
Mark scheme
  1. Uses s = ut + ½at² with u = 0"dropped from rest" is what tells you u = 0[1]
  2. s = ½ × 9.81 × 3.2² = 50.2 ≈ 50 m[1]
  3. Uses v = u + ator v² = u² + 2as[1]
  4. v = 9.81 × 3.2 = 31.4 ≈ 31 m s⁻¹[1]
  5. The actual speed would be lower / the stone would reach terminal velocitya statement about direction of change is enough[1]

(a) 50 m (b) 31 m s⁻¹ (c) the speed would be less

Q3[6 marks]
The velocity–time graph of a train shows: a uniform rise from 0 to 20 m s⁻¹ over the first 40 s, a constant 20 m s⁻¹ for the next 60 s, then a uniform fall to rest over the final 30 s.
  1. Calculate the acceleration during the first 40 s.
  2. Calculate the total distance travelled.
  3. Calculate the average speed for the whole journey.
Mark scheme
  1. Acceleration = gradient = (20 − 0) / 40gradient of a velocity–time graph is acceleration[1]
  2. = 0.50 m s⁻²[1]
  3. Recognises distance = area under the graphthis is the mark most often missed[1]
  4. Triangle ½ × 40 × 20 = 400; rectangle 60 × 20 = 1200; triangle ½ × 30 × 20 = 300all three areas needed[1]
  5. Total = 1900 m[1]
  6. Average speed = 1900 / 130 = 14.6 ≈ 15 m s⁻¹total distance ÷ total time, not the mean of the velocities[1]

(a) 0.50 m s⁻² (b) 1900 m (c) 15 m s⁻¹

Q4[5 marks]
A ball is thrown horizontally at 15 m s⁻¹ from the top of a building 45 m high. Take g = 10 m s⁻² and ignore air resistance.
  1. Calculate the time the ball takes to reach the ground.
  2. Calculate the horizontal distance travelled.
  3. Explain why the time in (a) does not depend on the horizontal speed.
Mark scheme
  1. Uses vertical motion with u_y = 0: 45 = ½ × 10 × t²"thrown horizontally" means the initial vertical velocity is zero[1]
  2. t² = 9.0, t = 3.0 s[1]
  3. Uses s_x = u_x t with constant horizontal velocity[1]
  4. s_x = 15 × 3.0 = 45 m[1]
  5. Horizontal and vertical motion are independent / gravity acts only vertically, so the vertical motion is unaffected by the horizontal velocity[1]

(a) 3.0 s (b) 45 m (c) the two components are independent

Q5[3 marks]
Define displacement, and state one situation in which the magnitude of an object's displacement is smaller than the distance it has travelled.
Mark scheme
  1. Displacement is the straight-line distance from the starting point to the finishing point[1]
  2. …together with its direction / it is a vector quantitythe direction is required for the second mark[1]
  3. Any curved or non-straight path, e.g. a runner going round a bend, a car following a winding roada full circular lap, where displacement is zero, also earns this[1]
Q6[2 marks]
Differentiate between distance and displacement.
Answer

Distance is the total length of the path travelled, a scalar. Displacement is the straight line from start to finish together with its direction, a vector.

Q7[2 marks]
Can a body have zero velocity and non-zero acceleration? Explain.
Answer

Yes. A ball thrown vertically upward is momentarily at rest at the top of its flight, but gravity still acts, so its acceleration is g downward.

Q8[4 marks]
A car travelling at 25 m s⁻¹ brakes uniformly and stops in 5.0 s. Calculate the deceleration and the distance travelled while braking.
Mark scheme
  1. Uses a = (v − u)/t = (0 − 25)/5.0[1]
  2. a = −5.0 m s⁻², a deceleration of 5.0 m s⁻²the negative sign or the word deceleration, not both required[1]
  3. Uses s = ½(u + v)t or v² = u² + 2as[1]
  4. s = ½(25 + 0) × 5.0 = 62.5 m[1]

deceleration 5.0 m s⁻², distance 62.5 m

03

Dynamics

Multiple choice · 14

Q1A bullet is fired horizontally at the same instant an identical bullet is dropped from the same height. Which lands first?

  1. AThe fired bullet
  2. BThe dropped bullet
  3. CThey land at the same time
  4. DIt depends on the bullet mass
Show answer

Correct answer: C — They land at the same time

Vertical and horizontal motion are independent. Both bullets start with zero vertical velocity and fall under the same gravity, so both take identical time to reach the ground. The fired bullet simply travels much further horizontally while doing it.

Q2A projectile is launched at 30°. Which other angle gives the same horizontal range at the same speed?

  1. A45°
  2. B60°
  3. C75°
  4. DNone — 30° is unique
Show answer

Correct answer: B — 60°

Range depends on sin(2θ), and sin(60°) = sin(120°), so 30° and 60° pair up. The 60° shot goes higher and stays in the air longer but moves more slowly across; the two effects cancel exactly. Maximum range is at 45°.

Q3You push a wall and it pushes back equally. Why does nothing accelerate?

  1. AThe forces cancel out on the same object
  2. BThe two forces act on different objects, and the wall is anchored to the Earth
  3. CNewton's third law does not apply to walls
  4. DFriction removes both forces
Show answer

Correct answer: B — The two forces act on different objects, and the wall is anchored to the Earth

Your push acts on the wall; the wall's push acts on you. They never appear on the same free-body diagram, so they cannot cancel. Nothing accelerates because the wall is bolted to the ground and friction on your feet balances the force on you.

Q4A 2 kg ball is dropped from 5 m. Ignoring air resistance, its speed just before landing is about:

  1. A5 m/s
  2. B10 m/s
  3. C14 m/s
  4. D20 m/s
Show answer

Correct answer: B — 10 m/s

Energy conservation: mgh = ½mv², and the mass cancels from both sides. So v = √(2gh) = √(2 × 9.81 × 5) = √98.1 ≈ 9.9 m/s, which rounds to 10 m/s. Note that a 5 kg ball dropped from the same height arrives at exactly the same speed.

Q5A pendulum swings with friction. What happens to the total energy?

  1. AIt is destroyed
  2. BIt converts to heat and sound, so the mechanical total falls
  3. CIt stays exactly constant
  4. DIt increases as the pendulum slows
Show answer

Correct answer: B — It converts to heat and sound, so the mechanical total falls

Energy is never destroyed. Friction converts the ordered kinetic energy into disordered thermal energy in the air and pivot. The mechanical total (KE + PE) falls, but the total including heat is unchanged.

Q6Two cars collide and lock together. Which quantity is definitely conserved?

  1. AKinetic energy only
  2. BMomentum only
  3. CBoth momentum and kinetic energy
  4. DNeither
Show answer

Correct answer: B — Momentum only

Momentum is conserved in every collision, since no external horizontal force acts during the impact. Kinetic energy is not: this is a perfectly inelastic collision, and a large fraction of it goes into crumpling metal, heat and noise.

Q7The unit of momentum is:

  1. AN s⁻¹
  2. Bkg m s⁻¹
  3. CJ
  4. DN m
Show answer

Correct answer: B — kg m s⁻¹

p = mv gives kg × m s⁻¹. It is also equal to the newton second, since impulse and momentum share a unit.

Q8Momentum is conserved in:

  1. AElastic collisions only
  2. BInelastic collisions only
  3. CAll collisions with no external resultant force
  4. DNo collisions
Show answer

Correct answer: C — All collisions with no external resultant force

Momentum is always conserved provided no external resultant force acts. Kinetic energy is the quantity that separates elastic from inelastic.

Q9Two objects stick together after colliding. The collision is:

  1. AElastic
  2. BInelastic
  3. CImpossible
  4. DFrictionless
Show answer

Correct answer: B — Inelastic

Sticking together always means kinetic energy was lost to heat, sound and deformation. Momentum is still conserved.

Q10A 2 kg object moves at 3 m s⁻¹. Its momentum is:

  1. A1.5 kg m s⁻¹
  2. B6 kg m s⁻¹
  3. C9 kg m s⁻¹
  4. D0.67 kg m s⁻¹
Show answer

Correct answer: B — 6 kg m s⁻¹

p = mv = 2 × 3 = 6 kg m s⁻¹. Dividing gives 0.67 and confuses momentum with something that has no physical meaning here.

Q11Airbags reduce injury mainly because they:

  1. AReduce the change in momentum
  2. BIncrease the time of the collision, reducing the force
  3. CIncrease the mass of the passenger
  4. DAbsorb the momentum
Show answer

Correct answer: B — Increase the time of the collision, reducing the force

The change in momentum is fixed by the crash. Extending Δt reduces F, since F = Δp/Δt. Nothing "absorbs" momentum — it is transferred, not destroyed.

Q12A stationary object explodes into two fragments. Their total momentum afterwards is:

  1. AZero
  2. BEqual to the mass of the object
  3. CDoubled
  4. DImpossible to determine
Show answer

Correct answer: A — Zero

It was zero before, and momentum is conserved, so it must be zero after. The two fragments carry equal and opposite momenta.

Q13A ball of mass 0.2 kg hits a wall at 5 m s⁻¹ and rebounds at 5 m s⁻¹. The magnitude of its change in momentum is:

  1. A0
  2. B1 kg m s⁻¹
  3. C2 kg m s⁻¹
  4. D0.5 kg m s⁻¹
Show answer

Correct answer: C — 2 kg m s⁻¹

The velocity changes from +5 to −5, a change of 10 m s⁻¹, so Δp = 0.2 × 10 = 2 kg m s⁻¹. Answer A treats the speeds as equal and therefore unchanged, which ignores direction.

Q14Force is best defined as the rate of change of:

  1. AVelocity
  2. BMomentum
  3. CEnergy
  4. DDisplacement
Show answer

Correct answer: B — Momentum

F = Δp/Δt is the general form of Newton's second law, and reduces to F = ma when the mass is constant.

Exam-style questions · 11

Q1[2 marks]
State Newton's third law and give the two conditions an action–reaction pair must satisfy.
Answer

For every action there is an equal and opposite reaction. The two forces are of the same type and act on different bodies.

Q2[2 marks]
Why does a passenger lurch forward when a bus brakes suddenly?
Answer

By Newton's first law the passenger continues moving at the same velocity because no resultant force acts on them; the bus decelerates beneath them, so they move forward relative to it.

Q3[2 marks]
Define momentum and state its SI unit.
Answer

The product of mass and velocity, p = mv. A vector quantity. SI unit: kg m s⁻¹.

Q4[4 marks]
A trolley of mass 2.0 kg moving at 3.0 m s⁻¹ collides with a stationary trolley of mass 4.0 kg. They stick together. Calculate their common velocity after the collision.
Mark scheme
  1. States conservation of momentum: total before = total after[1]
  2. Before: p = 2.0 × 3.0 + 4.0 × 0 = 6.0 kg m s⁻¹[1]
  3. After: combined mass = 6.0 kg, so 6.0 = 6.0 × vthey stick together, so they share one velocity[1]
  4. v = 1.0 m s⁻¹ in the original directiondirection expected for full marks[1]

1.0 m s⁻¹ in the direction of the original motion

Q5[4 marks]
A force of 15 N acts on a 3.0 kg block resting on a surface. Friction opposing the motion is 6.0 N. Calculate the acceleration of the block.
Mark scheme
  1. Resultant force = 15 − 6.0 = 9.0 Nfriction opposes, so it subtracts[1]
  2. Uses F = ma[1]
  3. a = F/m = 9.0 / 3.0[1]
  4. a = 3.0 m s⁻²unit required[1]

3.0 m s⁻²

Q6[6 marks]
A 1200 kg car travelling at 15 m s⁻¹ collides with a stationary 800 kg car. The two lock together.
  1. Calculate the total momentum before the collision.
  2. Calculate their common velocity immediately after.
  3. State whether kinetic energy is conserved, and name the type of collision.
Mark scheme
  1. Uses p = mv[1]
  2. p = 1200 × 15 = 18 000 kg m s⁻¹the stationary car contributes nothing[1]
  3. Applies conservation: total after = 18 000 kg m s⁻¹[1]
  4. Combined mass 2000 kg, so v = 18 000 / 2000 = 9.0 m s⁻¹[1]
  5. Kinetic energy is not conserved[1]
  6. Inelastic collisionobjects sticking together is always inelastic[1]

(a) 1.8 × 10⁴ kg m s⁻¹ (b) 9.0 m s⁻¹ (c) not conserved; inelastic

Q7[5 marks]
A rifle of mass 4.0 kg fires a bullet of mass 0.020 kg at 400 m s⁻¹.
  1. Calculate the recoil velocity of the rifle.
  2. Explain, using momentum, why a heavier rifle recoils more slowly.
Mark scheme
  1. Total momentum before is zeronothing is moving[1]
  2. 0 = 0.020 × 400 + 4.0 × v[1]
  3. v = −8.0 / 4.0 = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwardsdirection required[1]
  4. The rifle's momentum must equal the bullet's in size and be opposite in direction[1]
  5. Since p = mv is fixed, a larger m gives a smaller v[1]

2.0 m s⁻¹ backwards

Q8[4 marks]
A 0.15 kg ball hits a wall at 12 m s⁻¹ and rebounds at 8.0 m s⁻¹. The contact lasts 0.050 s. Calculate the average force on the ball.
Mark scheme
  1. Takes the initial direction as positive, so the rebound velocity is −8.0 m s⁻¹the sign is the whole question[1]
  2. Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹a change of 20 m s⁻¹, not 4[1]
  3. Uses F = Δp/Δt[1]
  4. F = −3.0 / 0.050 = −60 N, i.e. 60 N away from the wall[1]

60 N, directed away from the wall

Q9[2 marks]
State the principle of conservation of momentum, including the condition under which it applies.
Answer

The total momentum of a system before an interaction equals the total momentum after it, provided no resultant external force acts on the system.

Q10[2 marks]
Explain why a cricketer moves their hands backwards while catching a fast ball.
Answer

It increases the time over which the ball is brought to rest. Since F = Δp/Δt and the change in momentum is fixed, a longer time means a smaller force on the hands.

Q11[9 marks]
A 0.045 kg golf ball is struck by a club. The ball leaves the tee at 60 m s⁻¹ and the contact lasts 0.50 ms.
  1. Calculate the change in momentum of the ball. [2]
  2. Calculate the average force exerted by the club. [3]
  3. The club has a mass of 0.30 kg and was moving at 70 m s⁻¹ before impact. Calculate its speed immediately afterwards. [4]
Mark scheme
  1. Uses Δp = m(v − u) with u = 0the ball starts at rest on the tee[1]
  2. Δp = 0.045 × 60 = 2.7 kg m s⁻¹[1]
  3. Converts the time: 0.50 ms = 5.0 × 10⁻⁴ s[1]
  4. Uses F = Δp/Δt[1]
  5. F = 2.7 / 5.0 × 10⁻⁴ = 5400 N[1]
  6. Applies conservation of momentum to club and ball together[1]
  7. Before: 0.30 × 70 = 21 kg m s⁻¹the ball contributes nothing[1]
  8. After: 21 = 0.30 × v + 2.7[1]
  9. v = 18.3 / 0.30 = 61 m s⁻¹[1]

(a) 2.7 kg m s⁻¹ (b) 5400 N (c) 61 m s⁻¹

04

Turning Effects of Force

Multiple choice · 8

Q1The moment of a force is calculated using:

  1. AForce × distance along the line of action
  2. BForce × perpendicular distance from the pivot
  3. CForce ÷ distance from the pivot
  4. DForce × time
Show answer

Correct answer: B — Force × perpendicular distance from the pivot

Only the perpendicular distance produces turning. A force pushing straight toward the pivot has zero perpendicular distance and therefore no turning effect at all, however large it is.

Q2The unit of moment is:

  1. AJ
  2. BN m
  3. CN/m
  4. DW
Show answer

Correct answer: B — N m

Newton metre. It shares its base units with the joule, but a moment and an energy are different quantities, so writing J for a moment loses the mark.

Q3A 2 N weight sits 0.6 m from a pivot. Where must a 3 N weight sit on the other side to balance it?

  1. A0.4 m
  2. B0.6 m
  3. C0.9 m
  4. D1.2 m
Show answer

Correct answer: A — 0.4 m

Anticlockwise moment = 2 × 0.6 = 1.2 N m, so 3 × d = 1.2 and d = 0.4 m. The heavier weight sits closer — the common error is placing it further out.

Q4Two equal, opposite, parallel forces with different lines of action form:

  1. AAn equilibrium
  2. BA couple
  3. CA resultant force
  4. DA moment of zero
Show answer

Correct answer: B — A couple

That is the definition of a couple. It produces pure rotation, with no resultant force, which is why a steering wheel turns without the column sliding sideways.

Q5For a uniform metre rule, the centre of gravity is at:

  1. AThe 0 cm mark
  2. BThe 50 cm mark
  3. CThe 100 cm mark
  4. DWherever it is pivoted
Show answer

Correct answer: B — The 50 cm mark

Uniform means the mass is evenly distributed, so the centre of gravity is at the geometric centre. The word "uniform" in a question is always telling you this.

Q6An object topples when:

  1. AIts centre of gravity is high
  2. BIts base is narrow
  3. CThe vertical line through its centre of gravity falls outside its base
  4. DIt is displaced at all
Show answer

Correct answer: C — The vertical line through its centre of gravity falls outside its base

A high centre of gravity and a narrow base both make toppling easier, but neither causes it on its own. The condition is the line through the centre of gravity leaving the base.

Q7When taking moments, choosing the pivot on the line of action of an unknown force is useful because:

  1. AIt makes the force larger
  2. BThat force then has zero moment and drops out
  3. CIt changes the equilibrium
  4. DIt converts N m to joules
Show answer

Correct answer: B — That force then has zero moment and drops out

Its perpendicular distance from that point is zero, so its moment is zero. One unknown vanishes and the equation solves in a single line.

Q8A racing car is built low to the ground mainly because:

  1. AIt reduces air resistance only
  2. BIt lowers the centre of gravity, improving stability
  3. CIt increases the weight
  4. DIt increases the moment of the engine
Show answer

Correct answer: B — It lowers the centre of gravity, improving stability

A lower centre of gravity means a larger tilt is needed before the vertical line through it leaves the wheelbase, so the car resists rolling in corners. Reduced drag is a genuine second benefit, but stability is the reason given in mark schemes.

Exam-style questions · 7

Q1[4 marks]
A uniform beam of weight 40 N and length 2.0 m rests on a pivot 0.50 m from its left end. A load of weight W hangs from the extreme left end, and the beam is in equilibrium.
  1. State where the weight of the beam acts.
  2. Calculate the value of W.
Mark scheme
  1. At the centre of gravity, which for a uniform beam is its midpoint, 1.0 m from the left end"uniform" is the word that tells you this[1]
  2. Takes moments about the pivot; the beam's weight acts 1.0 − 0.50 = 0.50 m to the right of it[1]
  3. Clockwise = anticlockwise: 40 × 0.50 = W × 0.50the load acts 0.50 m to the left of the pivot[1]
  4. W = 40 N[1]

(a) at the midpoint, 1.0 m from the left end (b) W = 40 N

Q2[3 marks]
Explain, in terms of centre of gravity and base, why a double-decker bus is more likely to topple when passengers stand on the upper deck than when they sit downstairs.
Mark scheme
  1. Passengers upstairs raise the centre of gravity of the bus[1]
  2. A higher centre of gravity means a smaller tilt is needed before the vertical line through it falls outside the base / wheelbasethis is the key reasoning mark[1]
  3. So the bus topples at a smaller angle / is less stable[1]
Q3[3 marks]
A force of 12 N is applied to a spanner at 30° to the handle, at a distance of 0.25 m from the nut. Calculate the moment of the force about the nut.
Mark scheme
  1. Recognises that only the perpendicular component turns the nutor equivalently uses the perpendicular distance[1]
  2. Perpendicular component = 12 sin 30° = 6.0 N[1]
  3. Moment = 6.0 × 0.25 = 1.5 N munit required[1]

1.5 N m

Q4[2 marks]
Define the moment of a force and state its SI unit.
Answer

The turning effect of a force about a pivot, equal to force × perpendicular distance from the pivot to the line of action. SI unit: the newton metre, N m.

Q5[2 marks]
Why can a moment not be measured in joules, when N m is the unit of both?
Answer

They are different quantities. Work is force acting along a displacement; a moment is force acting across a distance from a pivot. Sharing base units does not make them the same thing.

Q6[2 marks]
State two conditions that must be satisfied for a body to be in complete equilibrium.
Answer

ΣF = 0 — no resultant force, so it does not accelerate. Σ moments = 0 — no resultant turning effect, so it does not rotate.

Q7[4 marks]
A uniform metre rule is pivoted at the 50 cm mark. A weight of 3.0 N hangs at the 20 cm mark. Calculate the weight that must hang at the 70 cm mark to balance it.
Mark scheme
  1. Left distance = 50 − 20 = 30 cm = 0.30 mdistance from the pivot, not from the end of the rule[1]
  2. Anticlockwise moment = 3.0 × 0.30 = 0.90 N m[1]
  3. Right distance = 70 − 50 = 20 cm = 0.20 m; for balance W₂ × 0.20 = 0.90principle of moments[1]
  4. W₂ = 4.5 Nunit required[1]

4.5 N

05

Work, Energy and Power

Multiple choice · 6

Q1A waiter carries a tray horizontally at constant speed across a room. The work done by the waiter on the tray is:

  1. ALarge and positive
  2. BZero
  3. CNegative
  4. DEqual to the weight of the tray
Show answer

Correct answer: B — Zero

The supporting force is vertical and the movement is horizontal, so θ = 90° and cos 90° = 0. Effort is not the same as work — your arm gets tired maintaining the force, but no energy is transferred to the tray.

Q2A car doubles its speed. Its kinetic energy:

  1. ADoubles
  2. BHalves
  3. CQuadruples
  4. DStays the same
Show answer

Correct answer: C — Quadruples

KE = ½mv², so v → 2v gives (2v)² = 4v². This is why braking distance grows roughly fourfold when speed doubles: the brakes must dissipate four times the energy.

Q3A 2 kg ball is dropped from 5 m. Taking g = 10 m s⁻² and ignoring air resistance, its speed on landing is:

  1. A5 m s⁻¹
  2. B10 m s⁻¹
  3. C20 m s⁻¹
  4. D100 m s⁻¹
Show answer

Correct answer: B — 10 m s⁻¹

mgh = ½mv², and the mass cancels: v = √(2gh) = √(2 × 10 × 5) = 10 m s⁻¹. The 100 answer is v² left un-square-rooted, and the mass being given at all is a deliberate distraction.

Q4A motor lifts a 50 kg load 4 m in 10 s. Taking g = 10 m s⁻², its useful power output is:

  1. A20 W
  2. B200 W
  3. C2000 W
  4. D500 W
Show answer

Correct answer: B — 200 W

Work = mgh = 50 × 10 × 4 = 2000 J, and power = 2000 ÷ 10 = 200 W. The 2000 W answer is the work in joules read off as watts — always check whether you have divided by the time.

Q5A pendulum swings back and forth with friction present. Over time:

  1. ATotal energy is destroyed
  2. BKE and PE both fall, and the difference becomes heat
  3. CPE is converted entirely into KE with no loss
  4. DThe period grows steadily longer
Show answer

Correct answer: B — KE and PE both fall, and the difference becomes heat

Energy is never destroyed — it leaves the pendulum as heat and sound, so the mechanical total falls. The period, notably, stays essentially the same as the swing dies away, which is the isochronism that made pendulum clocks work.

Q6A machine takes in 500 J and delivers 350 J of useful output. Its efficiency is:

  1. A70%
  2. B143%
  3. C150%
  4. D35%
Show answer

Correct answer: A — 70%

350 ÷ 500 × 100 = 70%. The remaining 150 J has not vanished; it has been dissipated as heat, sound and vibration. Any efficiency above 100% means you have divided the wrong way round.

Exam-style questions · 5

Q1[2 marks]
Define work done and state the condition under which it is zero even though a force acts.
Answer

Work = force × distance moved in the direction of the force, W = Fs cos θ. It is zero when the force is perpendicular to the motion, since cos 90° = 0.

Q2[2 marks]
State the law of conservation of energy.
Answer

Energy cannot be created or destroyed, only transferred from one store to another. The total energy of a closed system remains constant.

Q3[2 marks]
A ball bounces to a lower height each time. Has energy been destroyed? Explain.
Answer

No. Energy is transferred to heat and sound in the ball and the floor at each bounce, so less remains as gravitational potential energy. The total is unchanged.

Q4[4 marks]
A pump raises 300 kg of water through a height of 12 m in 40 s. Calculate the useful power output. Take g = 10 N kg⁻¹.
Mark scheme
  1. Uses E = mghthe useful energy is the gravitational potential energy gained[1]
  2. E = 300 × 10 × 12 = 36 000 J[1]
  3. Uses P = E/t[1]
  4. P = 36 000 / 40 = 900 Wunit required[1]

900 W

Q5[4 marks]
A 1200 kg car accelerates from rest to 20 m s⁻¹. Calculate its gain in kinetic energy, and the average power developed if this takes 8.0 s.
Mark scheme
  1. Uses KE = ½mv²[1]
  2. KE = ½ × 1200 × 20² = 240 000 Jsquare the velocity before multiplying[1]
  3. Gain in KE = 240 000 J since it started from rest[1]
  4. P = 240 000 / 8.0 = 30 000 W = 30 kW[1]

2.4 × 10⁵ J and 30 kW

06

Mechanical Properties of Matter

Multiple choice · 8

Q1The density of a substance is defined as:

  1. AMass × volume
  2. BMass per unit volume
  3. CVolume per unit mass
  4. DWeight per unit area
Show answer

Correct answer: B — Mass per unit volume

ρ = m/V. Volume per unit mass is its reciprocal, and weight per unit area is pressure — both are offered because both are genuinely easy to reach for under time pressure.

Q2A block is cut exactly in half. Its density:

  1. AHalves
  2. BDoubles
  3. CStays the same
  4. DDepends on which way it is cut
Show answer

Correct answer: C — Stays the same

Both the mass and the volume halve, so their ratio is unchanged. Density is a property of the material, not of the size of the sample.

Q3The SI unit of pressure is:

  1. AN
  2. BPa
  3. CJ
  4. DN m
Show answer

Correct answer: B — Pa

The pascal, where 1 Pa = 1 N m⁻². The newton is force, the joule is energy, and N m is a moment.

Q4The pressure at the bottom of a column of liquid depends on:

  1. AThe width of the container
  2. BThe total volume of liquid
  3. CThe depth, density and g
  4. DThe shape of the container
Show answer

Correct answer: C — The depth, density and g

p = ρgh contains no term for width, volume or shape. A narrow tube 2 m tall gives exactly the same base pressure as a wide tank 2 m deep.

Q5A force of 20 N acts over an area of 0.5 m². The pressure is:

  1. A10 Pa
  2. B40 Pa
  3. C0.025 Pa
  4. D20 Pa
Show answer

Correct answer: B — 40 Pa

p = F/A = 20 ÷ 0.5 = 40 Pa. Dividing the other way round gives 0.025 and is the most common slip — check whether the answer should be larger or smaller than the force.

Q6A spring of natural length 10 cm stretches to 15 cm under a 4 N load. The spring constant is:

  1. A0.27 N m⁻¹
  2. B26.7 N m⁻¹
  3. C80 N m⁻¹
  4. D40 N m⁻¹
Show answer

Correct answer: C — 80 N m⁻¹

The extension is 5 cm = 0.05 m, so k = 4 ÷ 0.05 = 80 N m⁻¹. Using the total length of 0.15 m gives 26.7 and is precisely the error the question is set to catch.

Q7Beyond the elastic limit, a spring:

  1. AReturns to its original length
  2. BDoes not return to its original length
  3. CObeys Hooke's law exactly
  4. DHas zero spring constant
Show answer

Correct answer: B — Does not return to its original length

It is permanently deformed. The limit of proportionality is a separate and earlier point, where the force–extension graph stops being straight.

Q8Why do snowshoes stop a walker sinking into soft snow?

  1. AThey reduce the walker's weight
  2. BThey spread the weight over a larger area, lowering the pressure
  3. CThey increase the density of the snow
  4. DThey reduce the force of gravity
Show answer

Correct answer: B — They spread the weight over a larger area, lowering the pressure

The weight is unchanged — only the area over which it acts increases, so p = F/A falls. Nothing a shoe does can change a person's weight.

Exam-style questions · 7

Q1[5 marks]
A rectangular block of wood has dimensions 20 cm × 10 cm × 5.0 cm and a mass of 0.80 kg.
  1. Calculate the density of the wood in kg m⁻³.
  2. The block is placed on a table on its largest face. Calculate the pressure it exerts. Take g = 10 N kg⁻¹.
Mark scheme
  1. Volume = 0.20 × 0.10 × 0.050 = 1.0 × 10⁻³ m³converting every length to metres first[1]
  2. Density = 0.80 / 1.0 × 10⁻³ = 800 kg m⁻³[1]
  3. Weight = mg = 0.80 × 10 = 8.0 Npressure needs force, and the force here is the weight[1]
  4. Largest face area = 0.20 × 0.10 = 0.020 m²largest face gives the lowest pressure[1]
  5. Pressure = 8.0 / 0.020 = 400 Pa[1]

(a) 800 kg m⁻³ (b) 400 Pa

Q2[4 marks]
A diver is 12 m below the surface of a lake. The density of the water is 1000 kg m⁻³ and g = 10 N kg⁻¹.
  1. Calculate the pressure on the diver due to the water alone.
  2. The lake narrows sharply near the bottom. State and explain the effect of this on the pressure at 12 m depth.
Mark scheme
  1. Uses p = ρgh[1]
  2. p = 1000 × 10 × 12 = 1.2 × 10⁵ Paunit required[1]
  3. No effect / the pressure is unchanged[1]
  4. Pressure in a liquid depends only on depth, density and g — not on the shape or width of the containerthe reasoning mark; the statement alone scores 1 of 2[1]

(a) 1.2 × 10⁵ Pa (b) no change — pressure depends only on depth

Q3[4 marks]
A spring of natural length 8.0 cm extends to 12.0 cm when a load of 5.0 N is hung from it.
  1. Calculate the spring constant.
  2. Calculate the length of the spring when a load of 8.0 N is applied, assuming the limit of proportionality is not exceeded.
Mark scheme
  1. Extension = 12.0 − 8.0 = 4.0 cm = 0.040 mextension, not total length — the mark most often lost on this topic[1]
  2. k = F/x = 5.0 / 0.040 = 125 N m⁻¹[1]
  3. New extension = 8.0 / 125 = 0.064 m = 6.4 cm[1]
  4. New length = 8.0 + 6.4 = 14.4 cmadding the natural length back on[1]

(a) 125 N m⁻¹ (b) 14.4 cm

Q4[2 marks]
Define density and state its SI unit.
Answer

Mass per unit volume, ρ = m/V. SI unit: kg m⁻³.

Q5[2 marks]
Why does a camel have broad feet?
Answer

Broad feet spread the camel's weight over a larger area, so the pressure on the sand is smaller and it does not sink.

Q6[2 marks]
State Hooke's law.
Answer

The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.

Q7[4 marks]
A tank contains oil of density 800 kg m⁻³ to a depth of 1.5 m. Calculate the pressure at the base due to the oil, and the force this exerts on a base of area 2.0 m². Take g = 10 N kg⁻¹.
Mark scheme
  1. Uses p = ρgh[1]
  2. p = 800 × 10 × 1.5 = 12 000 Paunit required[1]
  3. Rearranges p = F/A to F = pA[1]
  4. F = 12 000 × 2.0 = 24 000 N[1]

p = 1.2 × 10⁴ Pa, F = 2.4 × 10⁴ N

07

Thermal Properties of Matter

Multiple choice · 8

Q1Temperature is a measure of:

  1. AThe total energy of all the particles
  2. BThe average kinetic energy of the particles
  3. CThe mass of the substance
  4. DThe rate of heat flow
Show answer

Correct answer: B — The average kinetic energy of the particles

Temperature is the average per particle. The total energy of all the particles is thermal energy, which also depends on how many there are — which is why a spark and a bath differ so completely.

Q2−273 °C in kelvin is:

  1. A0 K
  2. B273 K
  3. C−273 K
  4. D546 K
Show answer

Correct answer: A — 0 K

T(K) = θ + 273, so −273 + 273 = 0 K. This is absolute zero, the point at which particle motion is at its minimum.

Q3Which expands most for a given temperature rise?

  1. ASolids
  2. BLiquids
  3. CGases
  4. DAll expand equally
Show answer

Correct answer: C — Gases

Gases expand most, then liquids, then solids. The more freely the particles already move, the more space the additional motion requires.

Q4A bimetallic strip bends when heated because:

  1. AOne metal melts
  2. BThe two metals expand by different amounts
  3. CThe strip absorbs latent heat
  4. DThe metals contract
Show answer

Correct answer: B — The two metals expand by different amounts

Different expansion rates force the strip to curve, toward the metal that expands less. This is the basis of a simple thermostat.

Q5In E = mcΔθ, Δθ represents:

  1. AThe final temperature
  2. BThe starting temperature
  3. CThe change in temperature
  4. DThe temperature in kelvin
Show answer

Correct answer: C — The change in temperature

The change — subtract the initial from the final before substituting. Using the final temperature alone is the single most common error in this topic.

Q6During melting, the temperature of a substance:

  1. ARises steadily
  2. BFalls
  3. CStays constant
  4. DRises then falls
Show answer

Correct answer: C — Stays constant

The energy supplied breaks bonds between particles rather than raising their kinetic energy, so the thermometer does not move until all the solid has melted.

Q7Water is used as a coolant in car engines mainly because it:

  1. AIs cheap
  2. BHas a high specific heat capacity
  3. CHas a low boiling point
  4. DExpands when heated
Show answer

Correct answer: B — Has a high specific heat capacity

About 4200 J kg⁻¹ °C⁻¹ means it absorbs a great deal of energy for a small temperature rise. Being cheap and available is a real practical advantage, but not the physics being examined.

Q8Ponds freeze from the surface downward because:

  1. AIce is denser than water
  2. BWater expands between 4 °C and 0 °C, so ice floats
  3. CCold air only touches the top
  4. DThe bottom is insulated by mud
Show answer

Correct answer: B — Water expands between 4 °C and 0 °C, so ice floats

Water is anomalous below 4 °C: it expands as it cools further, so ice is less dense and floats. Cold air does only touch the top, but that alone would not stop the ice sinking once formed.

Exam-style questions · 7

Q1[5 marks]
An electric heater of power 2.0 kW is used to heat 1.5 kg of water. The specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹.
  1. Calculate the energy needed to raise the temperature of the water from 18 °C to 88 °C.
  2. Calculate the minimum time this would take.
  3. In practice the heater takes longer than your answer to (b). Give one reason.
Mark scheme
  1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
  2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
  3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
  4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
  5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

Q2[5 marks]
A student heats a block of ice at −10 °C steadily until it becomes steam. Sketch and describe the shape of the temperature–time graph.
Mark scheme
  1. Temperature rises from −10 °C to 0 °C[1]
  2. Horizontal section at 0 °C while the ice melts[1]
  3. Temperature rises from 0 °C to 100 °C[1]
  4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
  5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]
Q3[4 marks]
Explain, in terms of particles, why a bimetallic strip bends when heated, and state one use for it.
Mark scheme
  1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
  2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
  3. The strip bends toward the metal that expands less[1]
  4. Used in a thermostat / fire alarm / oven switchany valid use[1]
Q4[2 marks]
Differentiate between heat and temperature.
Answer

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

Q5[2 marks]
Why does a bimetallic strip bend on heating?
Answer

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

Q6[2 marks]
Why does the temperature remain constant while ice is melting?
Answer

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

Q7[5 marks]
Calculate the energy needed to convert 0.20 kg of ice at 0 °C completely into water at 20 °C. Take the specific latent heat of fusion of ice as 3.34 × 10⁵ J kg⁻¹ and the specific heat capacity of water as 4200 J kg⁻¹ °C⁻¹.
Mark scheme
  1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
  2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
  3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
  4. E₂ = 1.68 × 10⁴ J[1]
  5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

08

Magnetism

Multiple choice · 8

Q1A bar magnet is cut exactly in half. The result is:

  1. AOne north magnet and one south magnet
  2. BTwo magnets, each with both poles
  3. CTwo unmagnetised bars
  4. DOne magnet and one non-magnet
Show answer

Correct answer: B — Two magnets, each with both poles

Poles always occur in pairs. Every cut produces two complete magnets, each with a north and a south — you can never isolate a single pole.

Q2Which of these is NOT a magnetic material?

  1. AIron
  2. BCobalt
  3. CAluminium
  4. DNickel
Show answer

Correct answer: C — Aluminium

Aluminium is not magnetic, despite being a metal. That assumption — metal therefore magnetic — is exactly what the question tests.

Q3The only reliable test that a bar is a magnet is:

  1. AIt attracts iron
  2. BIt repels one end of another magnet
  3. CIt is made of steel
  4. DIt attracts a compass needle
Show answer

Correct answer: B — It repels one end of another magnet

Attraction happens with any magnetic material, magnetised or not. Only repulsion requires the object to be a magnet itself.

Q4Outside a bar magnet, field lines run:

  1. ASouth to north
  2. BNorth to south
  3. CIn both directions at once
  4. DIn straight lines only
Show answer

Correct answer: B — North to south

North to south outside, completing the loop south to north inside the magnet. The direction is defined by the force on a north pole.

Q5A region where field lines are close together indicates:

  1. AA weak field
  2. BA strong field
  3. CNo field
  4. DA neutral point
Show answer

Correct answer: B — A strong field

Line density represents field strength, which is why the lines crowd at the poles. A neutral point is where lines from two magnets cancel and the field is zero.

Q6Iron is preferred to steel for the core of an electromagnet because it:

  1. AIs stronger
  2. BMagnetises and demagnetises easily
  3. CRetains its magnetism permanently
  4. DIs cheaper
Show answer

Correct answer: B — Magnetises and demagnetises easily

An electromagnet must lose its magnetism the instant the current stops, so a soft material is essential. Steel would retain it and the device would not switch off.

Q7In an unmagnetised piece of iron, the domains are:

  1. AAll aligned
  2. BRandomly oriented so their effects cancel
  3. CAbsent
  4. DMade of steel
Show answer

Correct answer: B — Randomly oriented so their effects cancel

The domains exist and are individually magnetic, but point in random directions, so the sample shows no overall magnetism. Magnetising aligns them.

Q8The Earth's geographic North Pole is:

  1. AA magnetic north pole
  2. BA magnetic south pole
  3. CNot magnetic at all
  4. DBoth poles at once
Show answer

Correct answer: B — A magnetic south pole

The north-seeking pole of a compass points to it, and unlike poles attract, so it must be a magnetic south pole. The naming is historical and catches almost everyone once.

Exam-style questions · 7

Q1[4 marks]
A student is given two identical-looking steel bars. One is a permanent magnet and one is unmagnetised.
  1. Describe a test, using only the two bars, that identifies which is the magnet.
  2. Explain why attraction alone would not be enough.
Mark scheme
  1. Bring an end of one bar near the middle of the other, or bring the two ends together and reverse oneany workable procedure[1]
  2. If repulsion is observed at any point, that bar is the magnet[1]
  3. A magnet attracts any magnetic material, magnetised or not[1]
  4. So attraction does not distinguish the two; only repulsion does[1]
Q2[4 marks]
Explain, in terms of domains, what happens when a steel bar is magnetised by stroking, and why heating it strongly destroys the magnetism.
Mark scheme
  1. The bar contains domains, small regions that are already magnetic[1]
  2. Initially the domains point in random directions and their effects cancel[1]
  3. Stroking aligns the domains so they point the same way, and their effects add[1]
  4. Heating makes the domains vibrate and return to random directions, so the magnetism is lost[1]
Q3[2 marks]
State why the Earth's geographic North Pole must be a magnetic south pole.
Mark scheme
  1. The north-seeking pole of a compass points toward geographic north[1]
  2. Unlike poles attract, so the pole attracting a north pole must itself be a south pole[1]
Q4[2 marks]
Why is repulsion, and not attraction, the reliable test for a magnet?
Answer

A magnet attracts any magnetic material, whether or not it is magnetised, so attraction proves nothing. Only another magnet can be repelled, so repulsion is conclusive.

Q5[2 marks]
Explain, in terms of domains, why an iron bar is not always magnetic.
Answer

The bar contains domains that are individually magnetic. When they point in random directions their effects cancel and the bar shows no magnetism; when they are aligned, the effects add and the bar is a magnet.

Q6[2 marks]
Why is soft iron, rather than steel, used for the core of an electromagnet?
Answer

Soft iron magnetises and demagnetises easily, so the electromagnet loses its magnetism the moment the current stops. Steel would retain it and the device could not be switched off.

Q7[3 marks]
A plotting compass is placed at three points around a bar magnet: due north of the north pole, midway along the side, and due south of the south pole. State the direction the north-seeking pole of the compass points in each case, and explain your reasoning.
Mark scheme
  1. Beyond the north pole: away from the magnetfield lines leave the north pole[1]
  2. At the side: roughly parallel to the magnet, pointing from N toward Sthe compass lies along the field line[1]
  3. Beyond the south pole: toward the magnet, because field lines re-enter at the south pole[1]

Away from N, along the side N→S, and back toward S — the compass always lies along the field line.

09

Nature of Science

Multiple choice · 8

Q1For a hypothesis to be scientific it must be:

  1. ABelieved by most scientists
  2. BTestable, and capable of being shown wrong
  3. CImpossible to disprove
  4. DPublished in a journal
Show answer

Correct answer: B — Testable, and capable of being shown wrong

Testability is the defining feature. A claim that no observation could ever contradict sits outside science, however strongly it is held.

Q2In an investigation of how current affects the brightness of a lamp, the independent variable is:

  1. AThe brightness
  2. BThe current
  3. CThe resistance of the wires
  4. DThe room temperature
Show answer

Correct answer: B — The current

The current is what is deliberately changed. Brightness is measured and therefore dependent; the others would be kept constant as controls.

Q3A fair test requires that:

  1. AAll variables change together
  2. BOnly the independent variable is changed
  3. CThe result is known in advance
  4. DThe experiment is done once
Show answer

Correct answer: B — Only the independent variable is changed

If two variables change at once, the result cannot be attributed to either. That is why control variables matter — not tidiness, but interpretability.

Q4Repeating a measurement and taking a mean reduces:

  1. ASystematic error
  2. BRandom error
  3. CBoth equally
  4. DNeither
Show answer

Correct answer: B — Random error

Random errors scatter either side of the true value and partly cancel. A systematic error shifts every reading the same way and is untouched by averaging.

Q5An anomalous result should be:

  1. ADeleted without comment
  2. BIncluded in the mean regardless
  3. CIdentified, investigated, and excluded from the mean but reported
  4. DUsed as the final answer
Show answer

Correct answer: C — Identified, investigated, and excluded from the mean but reported

Excluding it from the mean is right; concealing that you excluded it is not. Reporting what you discarded is part of an honest method.

Q6Ice cream sales and drowning deaths both rise in summer. This shows that:

  1. AIce cream causes drowning
  2. BDrowning causes ice cream sales
  3. CThe two are correlated but a third factor explains both
  4. DThe data must be wrong
Show answer

Correct answer: C — The two are correlated but a third factor explains both

Hot weather drives both independently. A correlation with no mechanism and no controlled test is a prompt to look for a confounding variable, not a cause.

Q7A scientific model is best described as:

  1. AA perfect description of reality
  2. BA simplification useful within a range of conditions
  3. CA guess with no evidence
  4. DA law that can never change
Show answer

Correct answer: B — A simplification useful within a range of conditions

Every model simplifies deliberately, and every one has limits. Knowing where a model stops working is a large part of understanding it.

Q8Timing 20 swings of a pendulum and dividing by 20, rather than timing one, mainly reduces the effect of:

  1. AAir resistance
  2. BReaction time when starting and stopping the watch
  3. CThe mass of the bob
  4. DThe length of the string
Show answer

Correct answer: B — Reaction time when starting and stopping the watch

Reaction time is a roughly fixed error of a few tenths of a second. Spread across 20 swings its percentage effect falls by a factor of twenty, without needing better equipment.

Exam-style questions · 6

Q1[5 marks]
A student investigates how the length of a simple pendulum affects the time for one complete swing.
  1. Identify the independent, dependent and one control variable.
  2. The student times one swing with a stopwatch and records 1.4 s. Suggest two improvements to the method.
Mark scheme
  1. Independent: length of the pendulum[1]
  2. Dependent: time for one swing (the period)[1]
  3. Control: mass of the bob / angle of release / same stopwatch and operatorany one[1]
  4. Time 10 or 20 swings and divide, to reduce the effect of reaction timethe standard improvement for this experiment[1]
  5. Repeat each measurement and take a mean / use a fiducial marker at the centre of the swingany second valid improvement[1]
Q2[4 marks]
A newspaper reports that towns with more mobile phone masts have higher rates of a certain illness, and concludes that the masts cause the illness. Discuss whether this conclusion is justified.
Mark scheme
  1. The data shows a correlation between the two quantities[1]
  2. Correlation does not by itself establish causethe central point[1]
  3. A third factor could explain both, e.g. both are higher in larger, more densely populated townsany sensible confounding variable[1]
  4. A controlled study and a plausible mechanism would be needed before cause could be claimed[1]
Q3[3 marks]
Explain what is meant by an anomalous result, and state what a student should do with one.
Mark scheme
  1. A result that does not fit the pattern of the others / lies well away from the trend[1]
  2. It should be identified and, if possible, investigated or repeated[1]
  3. It should be excluded from the mean, but still reported rather than deleted[1]
Q4[2 marks]
What is meant by a hypothesis?
Answer

A proposed explanation for an observation, which must be testable — that is, capable of being shown wrong by an experiment.

Q5[2 marks]
Distinguish between the independent and dependent variables in an experiment.
Answer

The independent variable is the one deliberately changed by the experimenter. The dependent variable is the one measured, which responds to that change.

Q6[2 marks]
Why must control variables be kept constant?
Answer

So that any change in the dependent variable can be attributed to the independent variable alone. If two things change at once, the result cannot be interpreted.

These questions come from the 9th Class Physics lessons — each topic has its own notes, worked examples and an interactive diagram.