Mock Paper 1 — 40 MCQ

Answer: D — Weight. Weight is a force (gravitational pull), and forces are vectors — they have magnitude AND direction (downward). Mass, speed, and temperature are all scalars (magnitude only). Common confusion: speed is scalar but velocity is vector.

Answer: B — Low accuracy, high precision. Precision: readings are very close together (12.3–12.4 cm range is very small) → HIGH precision. Accuracy: mean ≈ 12.34 cm is far from the true value of 13.0 cm → LOW accuracy. This indicates a systematic error (e.g. zero error on the ruler).

Answer: A — 4.05 × 10⁻³ m. Leading zeros are not significant. Significant figures: 4, 0, 5 → 3 s.f. = 4.05. The value is 0.004050 = 4.050 × 10⁻³ → rounded to 3 s.f. = 4.05 × 10⁻³ m. Note the trailing zero in 4.050 IS significant but we drop it at 3 s.f.

Answer: B — Time 20 complete oscillations and divide by 20. Timing many oscillations reduces the effect of reaction time error. Any timing error is divided over 20 oscillations, making the period measurement much more precise. Divide total time by the number of oscillations (20), not by 40.

Answer: B — 9 m/s. Average speed = total distance / total time = (120 + 60) / (10 + 10) = 180 / 20 = 9 m/s. Do NOT average the two speeds (12 and 6) — that gives the wrong answer of 9 m/s only by coincidence here. Always use total distance ÷ total time.

Answer: C — Uniform deceleration. On a v-t graph: gradient = acceleration. A negative gradient means velocity is decreasing → deceleration. A straight line (constant gradient) means the deceleration is uniform (constant). Horizontal line = constant velocity. Curved line = changing acceleration.

Answer: C — 30 m/s. v = u + at = 0 + 10 × 3 = 30 m/s. The ball starts from rest (u = 0) and accelerates at g = 10 m/s². Common error: using s = ½gt² to find distance (which gives 45 m) and confusing it with speed.

Answer: A — 100 m. s = ½(u+v)t = ½(10+30)(5) = ½ × 40 × 5 = 100 m. Alternatively: a = (30−10)/5 = 4 m/s². s = ut + ½at² = 10(5) + ½(4)(25) = 50 + 50 = 100 m. The area under the v-t graph (trapezium) also gives ½(10+30)×5 = 100 m.

Answer: C — Distance travelled. Key v-t graph facts: gradient = acceleration; area under graph = distance (displacement) travelled. This is one of the most frequently tested kinematics facts in Paper 1.

Answer: C — 12 N. F = ma = 4 × 3 = 12 N. Always use F = ma with the RESULTANT (net) force, not any individual force. Units check: kg × m/s² = N.

Answer: B — A resultant force acts on it. Newton's 1st Law: no resultant (net) force → no change in motion. The object either stays at rest or continues at the same velocity. A balanced force (zero resultant) keeps the object in its current state of motion.

Answer: C — Weight equals air resistance. At terminal velocity, the resultant force = 0 (Newton's 1st Law). Therefore the two forces acting (weight downward, air resistance upward) must be equal. The skydiver moves at constant velocity — not accelerating, not decelerating.

Answer: B — The cart pulling the horse backward with equal force. Newton's 3rd Law: action-reaction pairs are equal in magnitude, opposite in direction, and act on DIFFERENT objects. Horse pulls cart forward → cart pulls horse backward (same magnitude). These act on different objects (horse and cart respectively).

Answer: B — 128 N. W = mg = 80 × 1.6 = 128 N. Mass (80 kg) is the same everywhere. Weight depends on gravitational field strength g. On Earth W = 80 × 10 = 800 N. On Moon W = 80 × 1.6 = 128 N — about 1/6 of Earth weight.

Answer: C — 40 N. Constant velocity → zero resultant force (Newton's 1st Law). Therefore: push force = friction force = 40 N. They are balanced. If friction were less than 40 N, the box would accelerate. If friction were more, it would decelerate.

Answer: A — 10 000 kg/m³. Volume = 5×4×3 = 60 cm³ = 60 × 10⁻⁶ m³. Mass = 600 g = 0.6 kg. Density = 0.6 / (60×10⁻⁶) = 0.6 / 0.00006 = 10 000 kg/m³. Key: convert cm³ to m³ by dividing by 10⁶, not 100.

Answer: B — 20 000 Pa. P = F/A = 300/0.015 = 20 000 Pa. Units: N/m² = Pa. Always check the area unit is in m² (not cm²).

Answer: B — 5 m. P = ρgh → h = P/(ρg) = 50 000/(1000×10) = 50 000/10 000 = 5 m. The fluid pressure formula P = ρgh gives pressure due to the liquid only (not including atmospheric pressure unless stated).

Answer: B — 100 N. Principle of moments: clockwise moment = anticlockwise moment. ACW: 200 × 0.5 = 100 Nm. CW: F × 1 = 100 Nm → F = 100 N. The beam is uniform so its own weight acts at the midpoint (pivot) — zero moment about the pivot.

Answer: C — Low centre of gravity, wide wheelbase. Stability requires: (1) low centre of gravity so the object must tilt further before CofG moves outside the base; (2) wide base so the object can tilt more before toppling. Racing cars are designed flat and wide for maximum stability at high speed.

Answer: B — 600 kg/m³. When floating, weight = upthrust. Upthrust = weight of fluid displaced = ρ_water × V_submerged × g. Weight of block = ρ_wood × V_total × g. Setting equal: ρ_wood × V = ρ_water × 0.6V → ρ_wood = 0.6 × 1000 = 600 kg/m³. The fraction submerged = density of object / density of fluid.

Answer: B — 36 J. KE = ½mv² = ½ × 2 × 6² = ½ × 2 × 36 = 36 J. The most common error is forgetting to square the velocity: ½ × 2 × 6 = 6 J is wrong. Always square v first.

Answer: C — 400 J. W = Fd = 50 × 8 = 400 J. Work is only done when the force causes movement in its direction. No movement = no work done. Units: N × m = J.

Answer: C — 70%. Efficiency = (useful output / total input) × 100 = (560/800) × 100 = 70%. Efficiency is always less than 100%. The wasted energy (240 J) is converted to heat (and sound) — not lost, but not useful.

Answer: B — 100 W. Work done = Fd = 200 × 5 = 1000 J. Power = W/t = 1000/10 = 100 W. Alternatively: P = Fv, but v is not constant here so use P = W/t.

Answer: C — 20 m/s. GPE lost = KE gained: mgh = ½mv² → gh = ½v² → v² = 2gh = 2×10×20 = 400 → v = 20 m/s. The mass cancels — speed at the bottom is independent of mass. Or use v² = u² + 2as = 0 + 2(10)(20) = 400 → v = 20 m/s.

Answer: C — Radiation only. Radiation (infrared electromagnetic waves) does not need a medium — it can travel through a vacuum. Conduction requires particle contact (solid). Convection requires fluid (liquid/gas) particle movement. This is why the Sun's heat reaches Earth through the vacuum of space.

Answer: A — 756 000 J. Q = mcΔT = 3 × 4200 × (85−25) = 3 × 4200 × 60 = 756 000 J. ΔT = 85−25 = 60°C. Common error: using 85 instead of ΔT = 60, giving 1 071 000 J.

Answer: C — Remains constant. During melting, the energy supplied goes into breaking intermolecular bonds (increasing potential energy), NOT into increasing kinetic energy. Since temperature measures average KE, temperature stays constant during the state change. This is latent heat — "hidden" heat that causes no temperature change.

Answer: B — Average kinetic energy of particles. Temperature is a measure of the average kinetic energy of particles. Increasing temperature → particles move faster → higher average KE. The number, mass, and size of particles do not change. Higher KE means more frequent and harder collisions with container walls → higher pressure (if volume is fixed).

Answer: B — 340 m/s. v = fλ = 200 × 1.7 = 340 m/s. This is approximately the speed of sound in air — consistent with a sound wave of this frequency and wavelength. Always use v = fλ for wave speed calculations.

Answer: C — Frequency. When light moves between media: speed changes (slows in glass), wavelength changes (shortens — v=fλ with same f), direction changes (bends toward normal). Frequency NEVER changes between media — it is set by the source and is a property of the wave, not the medium.

Answer: D — Gamma rays. EM spectrum from longest to shortest wavelength (lowest to highest frequency): Radio → Microwave → Infrared → Visible → Ultraviolet → X-rays → Gamma rays. Gamma rays have the shortest wavelength and highest frequency and energy.

Answer: B — Sound requires particles to vibrate and transmit energy. Sound is a longitudinal mechanical wave — it is transmitted by the compression and rarefaction (expansion) of particles. A vacuum has no particles, so there is nothing to vibrate. Sound is also longitudinal (not transverse like EM waves).

Answer: A — 0.5 A. Series: R_total = 12 + 6 = 18 Ω. I = V/R = 9/18 = 0.5 A. In series circuits the same current flows through every component. The voltage is shared: V₁ = 0.5×12 = 6 V; V₂ = 0.5×6 = 3 V. Check: 6+3 = 9 V ✓

Answer: A — 5 A; 46 Ω. P = IV → I = P/V = 1150/230 = 5 A. R = V/I = 230/5 = 46 Ω. Or R = V²/P = 230²/1150 = 52900/1150 = 46 Ω. Double-check: P = I²R = 25×46 = 1150 W ✓

Answer: C — 2 Ω. 1/R = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 → R = 2 Ω. Key rule: parallel resistance is ALWAYS less than the smallest individual resistor (here smallest is 6 Ω, so answer must be less than 6 Ω). For n identical resistors in parallel: R_total = R/n = 6/3 = 2 Ω.

Answer: A — 60. Vs/Vp = Ns/Np → Ns = Np × Vs/Vp = 1200 × 12/240 = 1200 × 0.05 = 60. This is a step-DOWN transformer (voltage decreases from 240V to 12V) so secondary turns must be fewer than primary turns (60 < 1200) ✓

Answer: C — 100 g. 24 days ÷ 8 days per half-life = 3 half-lives. Mass = 800 × (½)³ = 800 × 1/8 = 100 g. After each half-life: 800 → 400 → 200 → 100 g. Always count the number of half-lives first, then apply (½)ⁿ.

Answer: C — Gamma (γ). Penetration: α stopped by paper/5 cm air; β stopped by 3 mm aluminium; γ reduced (never fully stopped) by thick lead or concrete. Ionising power is in the opposite order: α most ionising, γ least ionising. This inverse relationship between penetration and ionisation is a key exam fact.