Paper 4 style practice paper: Theory (Extended) — mark scheme

a = Δv/Δt with Δv = g × t = 9.8 × 3.0 = 29 m/s (2 s.f.).

• use of a = Δv/Δt rearranged [1]
• 29 m/s [1]

The distance is the area under the speed–time graph: a triangle of base 3.0 s and height 29.4 m/s. Distance = ½ × 3.0 × 29.4 = 44 m.

• area under a speed–time graph = distance [1]
• triangle with base 3.0 and height 29.4 [1]
• 44 m [1]

Air resistance grows with speed, reducing the resultant force, so the acceleration falls towards zero at terminal velocity.

• acceleration decreases as drag grows (terminal velocity) [1]

ΔEk = ½mv² = ½ × 2.5 × 8.0² = 80 J.

• use of ½mv² [1]
• 80 J [1]

F = W/d = 80 ÷ 4.0 = 20 N.

• use of W = Fd rearranged [1]
• 20 N [1]

A resultant upward force is needed for the upward acceleration, so the thrust must exceed the weight (Newton’s second law). Weight = 2.5 × 9.8 = 24.5 N. Thrust = weight + resultant = 24.5 + 20 = 44.5 N.

• acceleration needs a resultant force / thrust > weight [1]
• weight = 24.5 N calculated [1]
• thrust = weight + resultant force [1]
• 44.5 N [1]

pV = constant: 100 × 120 = p₂ × 40, so p₂ = 12 000 ÷ 40 = 300 kPa.

• pV = constant used [1]
• substitution 100 × 120 = p₂ × 40 [1]
• 300 kPa [1]

The particles move at the same average speed (same temperature) but are now confined to a smaller volume, so they collide with the walls more frequently each second, giving a larger force per unit area.

• same average speed at constant temperature [1]
• more frequent collisions with the walls [1]
• greater force per unit area [1]

It rises further: faster particles strike the walls harder and more often, increasing the force per unit area.

• pressure rises [1]
• faster particles hit walls harder and more often [1]

I = P/V = 36 ÷ 12 = 3.0 A.

• use of I = P/V [1]
• 3.0 A [1]

E = Pt = 36 × 300 = 10 800 J.

• time converted to 300 s [1]
• 10 800 J [1]

Δθ = ΔE/(mc) = 10 800 ÷ (0.60 × 2000) = 9.0 °C.

• rearranged Δθ = ΔE/(mc) [1]
• 9.0 °C [1]

Some energy heats the container and the heater itself, or leaks to the surroundings, so the oil receives less than 10 800 J.

• energy to container/heater [1]
• or lost to surroundings [1]

sin r = sin i / n = sin 40° ÷ 1.5 = 0.643 ÷ 1.5 = 0.429, so r = 25°.

• n = sin i / sin r rearranged [1]
• sin r = 0.429 [1]
• r = 25° [1]

sin c = 1/n = 1 ÷ 1.5 = 0.667, so c = 42°.

• sin c = 1/n [1]
• 42° [1]

65° exceeds the 42° critical angle, and the light is travelling inside the denser medium, so the ray is totally internally reflected — none escapes at that face.

• 65° > 42° compared [1]
• inside the denser medium noted [1]
• total internal reflection stated [1]

Optical fibres carrying telephone and internet data (or prisms in periscopes and binoculars).

• valid application named [1]
• brief link to TIR [1]

R = V²/P = 144 ÷ 24 = 6.0 Ω (or I = P/V = 2.0 A when normal, R = 12 ÷ 2.0 = 6.0 Ω).

• I = P/V or equivalent first step [1]
• 6.0 Ω [1]

The resistor R shares the 12 V with the lamp, so the lamp receives less than its rated 12 V; its current (1.5 A) is below the rated 2.0 A, so it runs below normal brightness.

• voltage shared between R and lamp [1]
• lamp gets less than 12 V / 1.5 A < 2.0 A [1]
• so below normal brightness [1]

R_total = V/I = 12 ÷ 1.5 = 8.0 Ω.

• use of R = V/I [1]
• 8.0 Ω [1]

R = 8.0 − 6.0 = 2.0 Ω. V_R = IR = 1.5 × 2.0 = 3.0 V (leaving 9.0 V across the lamp).

• R subtracted from total [1]
• 2.0 Ω [1]
• 3.0 V across R [1]

I = P/V = 800 × 10⁶ ÷ 400 × 10³ = 2000 A.

• I = P/V used [1]
• 2000 A [1]

P = I²R = 2000² × 3.0 = 12 MW.

• use of I²R [1]
• 12 MW [1]

I = 800 × 10⁶ ÷ 25 × 10³ = 32 000 A. P = I²R = 32 000² × 3.0 = 3072 MW — far more than the station generates.

• 32 000 A [1]
• recognised as 16× the current [1]
• I²R applied [1]
• 3072 MW with comment [1]

Induction needs a changing magnetic field: the alternating primary current produces a continuously changing field in the core, which induces an e.m.f. in the secondary. A steady direct current gives a constant field and no secondary e.m.f.

• changing field needed for induction [1]
• d.c. gives a constant field, so no secondary e.m.f. [1]

A − Z = 24 − 11 = 13 neutrons.

• 13 [1]

60 h = 4 half-lives. Corrected start = 6400 − 100 = 6300 counts/min. After 4 halvings: 6300 → 3150 → 1575 → 787.5 → 394 counts/min.

• background subtracted first [1]
• four half-lives identified [1]
• 394 counts/min (accept 393–394) [1]

²⁴₁₁X → ²⁴₁₂Mg + ⁰₋₁e. Both the nucleon numbers (24 = 24 + 0) and the proton numbers (11 = 12 − 1) balance.

• daughter ²⁴₁₂Mg [1]
• beta written as ⁰₋₁e [1]
• both rows balance [1]

Beta particles carry a negative electric charge and are light, so the magnetic field exerts a force on them and curves their path; gamma radiation has no charge, so the field exerts no force on it.

• beta is charged [1]
• field exerts a force on moving charge [1]
• gamma uncharged, undeflected [1]

v = 2πr/T = 2π × 8.0 × 10⁶ ÷ 2.0 × 10⁴ = 2513 ≈ 2500 m/s.

• substitution into 2πr/T [1]
• evaluation of 2πr/T [1]
• ≈ 2500 m/s [1]

The moon’s field from the planet is much weaker at its greater distance, so it orbits more slowly: orbital speeds decrease with distance from the planet, so the moon moves slower than 2500 m/s.

• field weaker further out [1]
• slower orbital speed at greater radius [1]
• so moon slower than the satellite [1]

d = vt = 3.0 × 10⁸ × 6.0 × 10³ = 1.8 × 10¹² m.

• d = vt used [1]
• 1.8 × 10¹² m [1]

The star (with its planet) is receding from us; redshifts of distant galaxies across the sky show the Universe is expanding, supporting the Big Bang Theory.

• star receding [1]
• Universe expanding / Big Bang support [1]