A Level H1 Physics Mechanics Quiz

A-Level Physics H1 Quiz - Mechanics — Answer Key and Marking Scheme

Total Marks: 50

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Section A: Short Answer and Structured Response (20 marks)

1. State the principle of conservation of linear momentum. [2 marks]

Answer:

• [B1] The total momentum of a closed/isolated system remains constant
• [B1] provided no external forces act on the system / in the absence of external forces

Accept: "In a closed system, the total momentum before an interaction equals the total momentum after the interaction." Accept: "The total momentum of a system is conserved if the net external force is zero."

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2. Write down, in terms of mass m and velocity v: [2 marks]

(a) momentum p [1 mark]

• [B1] p = mv

(b) kinetic energy K [1 mark]

• [B1] K = ½mv²

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3. Car of mass 1200 kg at 15 m s⁻¹ [2 marks]

(a) Momentum [1 mark]

• [M1] p = mv = 1200 × 15
• [A1] p = 1.8 × 10⁴ N s (or 18 000 kg m s⁻¹)

(b) Kinetic energy [1 mark]

• [M1] K = ½mv² = ½ × 1200 × (15)²
• [A1] K = 1.35 × 10⁵ J (or 135 000 J)

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4. Ball falling with air resistance [3 marks]

(a) Graph sketch [1 mark]

• [B1] Curve starting at origin, increasing with decreasing gradient, approaching a horizontal asymptote (terminal velocity)

(b) Explanation [2 marks]

• [B1] Initially, net force = weight, so acceleration = g; speed increases rapidly.
• [B1] As speed increases, air resistance increases, reducing net downward force. Acceleration decreases. Eventually, air resistance equals weight, net force = 0, and speed becomes constant (terminal velocity).

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5. Trolley collision [5 marks]

(a) Velocity after collision [2 marks]

• [M1] Conservation of momentum: m₁u₁ + m₂u₂ = (m₁ + m₂)v
  • (2.0 × 3.0) + (1.0 × 0) = (2.0 + 1.0)v
  • 6.0 = 3.0v
• [A1] v = 2.0 m s⁻¹

(b) Elastic or inelastic? [3 marks]

• [M1] Calculate KE before: KE_before = ½ × 2.0 × (3.0)² + 0 = 9.0 J
• [M1] Calculate KE after: KE_after = ½ × 3.0 × (2.0)² = 6.0 J
• [A1] KE is not conserved (9.0 J ≠ 6.0 J), therefore the collision is inelastic.
• Award [B1] for correct conclusion with justification even if calculation has minor arithmetic error.

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Section B: Calculation and Application (20 marks)

6. Uniform plank with person [5 marks]

(a) Force diagram [2 marks]

• [B1] Four forces shown and labelled:
  • Weight of plank (200 N) acting downward at centre (2.0 m from either end)
  • Weight of person (600 N) acting downward at 1.0 m from left end
  • Reaction force at left support (R_L) acting upward at left end
  • Reaction force at right support (R_R) acting upward at right end
• [B1] All forces correctly positioned and directions indicated

(b) Reaction at right support [2 marks]

• [M1] Taking moments about left support:
  • Clockwise moments = (200 × 2.0) + (600 × 1.0) = 400 + 600 = 1000 N m
  • Anticlockwise moment = R_R × 4.0
  • For equilibrium: R_R × 4.0 = 1000
• [A1] R_R = 250 N

(c) Reaction at left support [1 mark]

• [M1] Vertical equilibrium: R_L + R_R = 200 + 600 = 800
• [A1] R_L = 800 − 250 = 550 N

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7. Define the newton in base SI units [1 mark]

• [B1] 1 N = 1 kg m s⁻²
• Accept: "The force required to accelerate a mass of 1 kg at 1 m s⁻²."

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8. Momentum 24 N s, KE 72 J [3 marks]

(a) Show mass is 4.0 kg [2 marks]

• [M1] From p = mv: v = p/m
  • Substitute into K = ½mv²: K = ½m(p/m)² = p²/(2m)
  • Therefore: m = p²/(2K)
• [M1] m = (24)²/(2 × 72) = 576/144 = 4.0 kg
• [A1] Shown

(b) Velocity [1 mark]

• [M1] v = p/m = 24/4.0
• [A1] v = 6.0 m s⁻¹

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9. Force 50 N on 10 kg mass [5 marks]

(a) Acceleration [1 mark]

• [M1] F = ma → a = F/m = 50/10
• [A1] a = 5.0 m s⁻²

(b) Velocity after 4.0 s [1 mark]

• [M1] v = u + at = 0 + 5.0 × 4.0
• [A1] v = 20 m s⁻¹

(c) Distance in 4.0 s [1 mark]

• [M1] s = ut + ½at² = 0 + ½ × 5.0 × (4.0)²
• [A1] s = 40 m

(d) Work done in 4.0 s [2 marks]

• [M1] W = F × s = 50 × 40
• [A1] W = 2000 J (or 2.0 × 10³ J)
• Alternative: W = ΔKE = ½ × 10 × (20)² − 0 = 2000 J [M1][A1]

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10. Stone projected horizontally from cliff [6 marks]

(a) Time to reach ground [2 marks]

• [M1] Vertical motion: s = ut + ½at²
  • 45 = 0 + ½ × 9.81 × t²
  • t² = 45 × 2 / 9.81 = 90/9.81 = 9.17
• [A1] t = 3.03 s (accept 3.0 s if g = 10 used: t = 3.0 s)

(b) Horizontal distance [1 mark]

• [M1] s_h = v_h × t = 20 × 3.03
• [A1] s_h = 60.6 m (or 60 m if t = 3.0 s)

(c) Vertical component of velocity [1 mark]

• [M1] v_v = u_v + at = 0 + 9.81 × 3.03
• [A1] v_v = 29.7 m s⁻¹ (or 30 m s⁻¹ if g = 10)

(d) Speed just before hitting ground [2 marks]

• [M1] v = √(v_h² + v_v²) = √(20² + 29.7²) = √(400 + 882) = √1282
• [A1] v = 35.8 m s⁻¹ (accept 36 m s⁻¹)

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Section C: Data Analysis and Extended Response (10 marks)

11. Car acceleration [4 marks]

(a) Acceleration [1 mark]

• [M1] a = (v − u)/t = (25 − 0)/10
• [A1] a = 2.5 m s⁻²

(b) Force [1 mark]

• [M1] F = ma = 800 × 2.5
• [A1] F = 2000 N

(c) Power at 25 m s⁻¹ [2 marks]

• [M1] P = Fv = 2000 × 25
• [A1] P = 5.0 × 10⁴ W (or 50 kW)
• Alternative: P = ΔKE/t = (½ × 800 × 25²)/10 = 250 000/10 = 25 000 W average power [M1], but instantaneous power at 25 m s⁻¹ is Fv = 50 000 W [A1].

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12. Trolley experiment [6 marks]

(a) Graph [3 marks]

• [B1] Axes correctly labelled with units (Acceleration / m s⁻² on y-axis; Hanging mass / kg on x-axis)
• [B1] All 5 points plotted correctly (± half small square)
• [B1] Best-fit straight line drawn through origin and points

(b) Gradient [1 mark]

• [M1] Gradient = Δy/Δx = (2.45 − 0)/(0.50 − 0) = 2.45/0.50
• [A1] Gradient = 4.9 m s⁻² kg⁻¹ (accept 4.9 to 5.0)

(c) Determine g [2 marks]

• [M1] For the system: mg = (M + m)a, where M = mass of trolley, m = hanging mass
  • Rearranging: a = [g/(M + m)] × m
  • Gradient = g/(M + m) = g/2.0
• [M1] g = gradient × 2.0 = 4.9 × 2.0
• [A1] g = 9.8 m s⁻² (accept 9.8 to 10.0)

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13. Ball thrown vertically [6 marks]

(a) Maximum height above ground [2 marks]

• [M1] Using v² = u² + 2as:
  • 0 = (12)² + 2(−9.81)s
  • s = 144/(2 × 9.81) = 144/19.62 = 7.34 m
  • This is height above point of projection
• [A1] Maximum height above ground = 7.34 + 2.0 = 9.34 m (accept 9.3 m or 9.4 m)
• Alternative using energy: ½mv² = mgh → h = v²/(2g) = 144/19.62 = 7.34 m [M1]; total height = 9.34 m [A1]

(b) Speed at 5.0 m above ground on way down [2 marks]

• [M1] Height fallen from maximum = 9.34 − 5.0 = 4.34 m
  • Using v² = u² + 2as: v² = 0 + 2 × 9.81 × 4.34 = 85.15
• [A1] v = 9.23 m s⁻¹ (accept 9.2 m s⁻¹)
• Alternative using energy: Loss in PE = Gain in KE → mg(9.34 − 5.0) = ½mv² → v = √(2 × 9.81 × 4.34) = 9.23 m s⁻¹ [M1][A1]

(c) Explanation using energy [2 marks]

• [B1] In the absence of air resistance, mechanical energy is conserved. The total mechanical energy (KE + PE) at the point of projection equals the total mechanical energy when the ball returns to the same point.
• [B1] Since the height is the same, the gravitational potential energy is the same. Therefore, by conservation of energy, the kinetic energy must also be the same, meaning the speed is the same (12 m s⁻¹).
• Accept: "Energy is conserved. At the same height, PE is the same, so KE must be the same, giving the same speed."

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14. State Newton's second law of motion. [1 mark]

• [B1] The rate of change of momentum of an object is directly proportional to the net force acting on it, and takes place in the direction of the net force.
• Accept: F = ma, or "The net force acting on an object is equal to the product of its mass and acceleration."

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15. Cyclist braking [3 marks]

(a) Deceleration [2 marks]

• [M1] Using v² = u² + 2as: 0 = (8.0)² + 2a(20)
  • 0 = 64 + 40a
  • a = −64/40 = −1.6 m s⁻²
• [A1] Deceleration = 1.6 m s⁻²

(b) Average braking force [1 mark]

• [M1] F = ma = 70 × 1.6
• [A1] F = 112 N (or 110 N if 1.6 used)
• Direction opposite to motion.

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Section D: Further Mechanics and Problem Solving (10 marks)

16. Box pulled on rough surface [7 marks]

(a) Free-body diagram [2 marks]

• [B1] Four forces shown:
  • Weight (mg = 5.0 × 9.81 = 49.05 N) acting downwards
  • Normal reaction (N) acting upwards
  • Applied force (30 N) at 30° above horizontal
  • Friction (f) acting opposite to direction of motion (horizontally)
• [B1] All forces correctly labelled and directions indicated.

(b) Normal reaction force [2 marks]

• [M1] Vertical equilibrium: N + 30 sin 30° = mg
  • N + 15 = 5.0 × 9.81 = 49.05
• [A1] N = 49.05 − 15 = 34.05 N ≈ 34 N

(c) Acceleration [3 marks]

• [M1] Friction force: f = μN = 0.40 × 34.05 = 13.62 N
• [M1] Net horizontal force: F_net = 30 cos 30° − f = 25.98 − 13.62 = 12.36 N
• [A1] a = F_net / m = 12.36 / 5.0 = 2.47 m s⁻² ≈ 2.5 m s⁻²

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17. Spring and ball [5 marks]

(a) Elastic potential energy [1 mark]

• [M1] EPE = ½kx² = ½ × 200 × (0.10)²
• [A1] EPE = 1.0 J

(b) Speed leaving spring [2 marks]

• [M1] Conservation of energy: EPE = KE → 1.0 = ½ × 0.050 × v²
• [A1] v = √(2 × 1.0 / 0.050) = √40 = 6.32 m s⁻¹ ≈ 6.3 m s⁻¹

(c) Maximum height on ramp [2 marks]

• [M1] KE at bottom = PE at top: ½mv² = mgh → h = v²/(2g)
  • h = (6.32)² / (2 × 9.81) = 40 / 19.62
• [A1] h = 2.04 m ≈ 2.0 m

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18. Satellite orbit [6 marks]

(a) Gravitational force [2 marks]

• [M1] F = GMm/r² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 500) / (7.0 × 10⁶)²
  • = (6.67 × 6.0 × 500 × 10¹³) / (49 × 10¹²) = (20010 × 10¹³) / (49 × 10¹²)
• [A1] F = 4084 N ≈ 4.1 × 10³ N

(b) Orbital speed [2 marks]

• [M1] Gravitational force provides centripetal force: GMm/r² = mv²/r → v = √(GM/r)
  • v = √[(6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (7.0 × 10⁶)] = √(4.002 × 10¹⁴ / 7.0 × 10⁶) = √(5.717 × 10⁷)
• [A1] v = 7.56 × 10³ m s⁻¹ ≈ 7.6 × 10³ m s⁻¹

(c) Period of orbit [2 marks]

• [M1] T = 2πr / v = 2π × 7.0 × 10⁶ / (7.56 × 10³)
  • = (4.398 × 10⁷) / (7.56 × 10³)
• [A1] T = 5817 s ≈ 5.8 × 10³ s (or about 1.6 hours)

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19. Pendulum and elastic collision [5 marks]

(a) Speed at point B [2 marks]

• [M1] Conservation of energy: mgh = ½mv² → v = √(2gh)
  • v = √(2 × 9.81 × 0.10) = √1.962
• [A1] v = 1.40 m s⁻¹ ≈ 1.4 m s⁻¹

(b) Velocity after elastic collision [3 marks]

• [M1] For elastic collision, both momentum and KE are conserved.
  • m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ and ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂²
  • With u₂ = 0: v₁ = (m₁ − m₂)u₁ / (m₁ + m₂)
• [M1] v₁ = (0.20 − 0.30) × 1.40 / (0.20 + 0.30) = (−0.10 × 1.40) / 0.50 = −0.28 m s⁻¹
• [A1] Velocity of bob = 0.28 m s⁻¹ in the opposite direction (i.e., rebounds).

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20. Car on slope [5 marks]

(a) Force diagram [1 mark]

• [B1] Diagram showing:
  • Weight component down slope: mg sin 10°
  • Resistive force (500 N) down slope
  • Driving force (F) up slope
  • Normal reaction perpendicular to slope (not required for parallel forces, but acceptable if shown)

(b) Driving force [3 marks]

• [M1] Component of weight down slope = mg sin 10° = 1500 × 9.81 × sin 10° = 1500 × 9.81 × 0.1736 = 2555 N
• [M1] Constant speed → net force = 0 → F = mg sin 10° + resistive force
  • F = 2555 + 500
• [A1] F = 3055 N ≈ 3.06 × 10³ N

(c) Power output [1 mark]

• [M1] P = Fv = 3055 × 20
• [A1] P = 6.11 × 10⁴ W ≈ 61 kW

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END OF ANSWER KEY