A Level H2 Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics H2 A-Level (Mechanics)

Answer Key and Marking Scheme

Total Marks: 60

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Section A: Principles and Definitions (Questions 1–5)

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

Answer: The total momentum of a system remains constant (1) provided no net external force acts on the system (1).

Alternative acceptable answer: In a closed/isolated system, the total momentum before an interaction equals the total momentum after the interaction (1), provided no external forces act (1).

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2. Define the term impulse and state its SI unit. [2 marks]

Answer: Impulse is the product of the average force acting on an object and the time for which it acts (1). SI unit: N s (or kg m s⁻¹) (1).

Accept: Impulse = change in momentum (1) with correct unit (1).

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3. State Newton's second law of motion in terms of momentum. [2 marks]

Answer: The rate of change of momentum of an object is directly proportional to the net force acting on it (1) and takes place in the direction of the force (1).

Accept: F = Δp/Δt or F = dp/dt with explanation (1 mark each for rate of change and direction/proportionality).

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4. Define work as used in mechanics and state the condition under which no work is done by a force acting on a moving object. [2 marks]

Answer: Work is the product of the force and the displacement in the direction of the force (1). No work is done when the force is perpendicular to the displacement (1).

Accept: W = Fs cos θ with explanation (1 mark for definition, 1 mark for perpendicular condition).

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5. State the principle of conservation of energy. [2 marks]

Answer: Energy cannot be created or destroyed (1); it can only be transferred or transformed from one form to another (1). The total energy of an isolated system remains constant (1).

Award 2 marks for any two of the above points.

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Section B: Calculations and Applications (Questions 6–15)

6. Ball dropped and rebounding.

(a) Speed just before hitting ground. [2 marks]

v² = u² + 2as = 0 + 2(9.81)(2.5) = 49.05 (1) v = √49.05 = 7.00 m s⁻¹ (1)

(b) Speed just after leaving ground. [2 marks]

v² = u² + 2as → 0 = u² + 2(-9.81)(1.8) (1) u² = 35.316 → u = 5.94 m s⁻¹ (1)

(c) Change in momentum. [3 marks]

Taking downward as positive: Initial momentum = 0.45 × 7.00 = 3.15 kg m s⁻¹ downward (1) Final momentum = 0.45 × (-5.94) = -2.67 kg m s⁻¹ (i.e., 2.67 kg m s⁻¹ upward) (1) Change in momentum = -2.67 - 3.15 = -5.82 kg m s⁻¹ Magnitude = 5.82 kg m s⁻¹, direction: upward (1)

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7. Car acceleration.

(a) Acceleration. [2 marks]

a = (v - u)/t = (25 - 0)/8.0 (1) = 3.125 m s⁻² ≈ 3.13 m s⁻² (1)

(b) Net force. [2 marks]

F_net = ma = 1200 × 3.125 (1) = 3750 N ≈ 3.75 × 10³ N (1)

(c) Driving force. [2 marks]

F_driving - F_resistive = F_net (1) F_driving = 3750 + 600 = 4350 N ≈ 4.35 × 10³ N (1)

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8. Simple harmonic motion.

(a) Angular frequency. [2 marks]

ω = 2π/T = 2π/2.5 (1) = 2.51 rad s⁻¹ (1)

(b) Maximum acceleration. [2 marks]

a_max = ω²x₀ = (2.51)² × 0.080 (1) = 0.504 m s⁻² (1)

(c) Maximum kinetic energy. [2 marks]

v_max = ωx₀ = 2.51 × 0.080 = 0.201 m s⁻¹ (1) KE_max = ½mv_max² = ½ × 0.20 × (0.201)² = 4.04 × 10⁻³ J (1)

Alternative: KE_max = ½mω²x₀² = ½ × 0.20 × (2.51)² × (0.080)² = 4.03 × 10⁻³ J

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9. Block on incline.

(a) Component of weight down plane. [2 marks]

mg sin θ = 3.0 × 9.81 × sin 30° (1) = 3.0 × 9.81 × 0.5 = 14.7 N (1)

(b) Acceleration down plane. [2 marks]

F = ma → a = F/m = 14.7/3.0 (1) = 4.90 m s⁻² (1)

(c) Speed after 4.0 m. [2 marks]

v² = u² + 2as = 0 + 2(4.90)(4.0) = 39.2 (1) v = √39.2 = 6.26 m s⁻¹ (1)

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10. Trolley collision.

(a) Velocity of trolley B after collision. [3 marks]

By conservation of momentum: m_A u_A + m_B u_B = m_A v_A + m_B v_B (1) (2.0)(3.0) + (1.0)(0) = (2.0)(1.0) + (1.0)(v_B) (1) 6.0 = 2.0 + v_B → v_B = 4.0 m s⁻¹ (1)

(b) Elastic or inelastic? [3 marks]

Initial KE = ½(2.0)(3.0)² + 0 = 9.0 J (1) Final KE = ½(2.0)(1.0)² + ½(1.0)(4.0)² = 1.0 + 8.0 = 9.0 J (1) Since initial KE = final KE, the collision is elastic (1).

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11. Stone in vertical circle.

(a) Centripetal force at top. [2 marks]

At top: mg + T = F_c (1) F_c = (0.50 × 9.81) + 2.0 = 4.905 + 2.0 = 6.91 N (1)

(b) Speed of stone. [2 marks]

F_c = mv²/r → v² = F_c × r/m = 6.91 × 1.2/0.50 = 16.58 (1) v = √16.58 = 4.07 m s⁻¹ (1)

(c) Tension at bottom. [2 marks]

At bottom: T - mg = F_c (1) T = F_c + mg = 6.91 + 4.905 = 11.8 N (1)

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12. Satellite orbit.

(a) Gravitational force. [2 marks]

F = GMm/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(500)/(7.0 × 10⁶)² (1) = (2.001 × 10¹⁷)/(4.9 × 10¹³) = 4.08 × 10³ N (1)

(b) Orbital speed. [2 marks]

F = mv²/r → v² = Fr/m = (4.08 × 10³)(7.0 × 10⁶)/500 = 5.71 × 10⁷ (1) v = √(5.71 × 10⁷) = 7.56 × 10³ m s⁻¹ (1)

Alternative: v = √(GM/r) = √[(6.67 × 10⁻¹¹)(6.0 × 10²⁴)/(7.0 × 10⁶)] = 7.56 × 10³ m s⁻¹

(c) Period of orbit. [2 marks]

T = 2πr/v = 2π(7.0 × 10⁶)/(7.56 × 10³) (1) = (4.40 × 10⁷)/(7.56 × 10³) = 5.82 × 10³ s (1)

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13. Spring and block.

(a) Elastic potential energy. [2 marks]

EPE = ½kx² = ½ × 200 × (0.15)² (1) = 2.25 J (1)

(b) Speed of block. [2 marks]

EPE → KE: ½mv² = 2.25 (1) v² = 2 × 2.25/0.50 = 9.0 → v = 3.0 m s⁻¹ (1)

(c) Distance on rough surface. [3 marks]

Friction force: f = μmg = 0.40 × 0.50 × 9.81 = 1.962 N (1) Work done by friction = loss in KE: f × d = ½mv² (1) 1.962 × d = 2.25 → d = 2.25/1.962 = 1.15 m (1)

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14. Projectile motion.

(a) Components of initial velocity. [2 marks]

u_x = 30 cos 40° = 30 × 0.7660 = 23.0 m s⁻¹ (1) u_y = 30 sin 40° = 30 × 0.6428 = 19.3 m s⁻¹ (1)

(b) Time to maximum height. [2 marks]

At max height, v_y = 0: v_y = u_y - gt → 0 = 19.3 - 9.81t (1) t = 19.3/9.81 = 1.97 s (1)

(c) Horizontal range. [2 marks]

Total time of flight = 2 × 1.97 = 3.94 s (1) Range = u_x × t_total = 23.0 × 3.94 = 90.6 m (1)

Alternative: Range = (u² sin 2θ)/g = (30² × sin 80°)/9.81 = 90.4 m

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15. Uniform rod.

(a) Diagram. [2 marks]

Diagram showing:

• Rod AB horizontal, pivoted at A (1)
• Weight 50 N acting downward at centre (1.0 m from A)
• Force F acting upward at B (2.0 m from A)
• Reaction force R at pivot A acting upward (1)

(b) Force F by moments. [2 marks]

Taking moments about A: Clockwise moment = anticlockwise moment (1) 50 × 1.0 = F × 2.0 → F = 25 N (1)

(c) Reaction at pivot. [2 marks]

Vertical equilibrium: R + F = 50 (1) R = 50 - 25 = 25 N, acting upward (1)

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Section C: Data Analysis and Experimental Skills (Questions 16–20)

16. Pendulum experiment.

(a) Quantities for straight-line graph. [2 marks]

y-axis: log T (or ln T) (1) x-axis: log L (or ln L) (1)

(b) Value of n. [3 marks]

T = kLⁿ → log T = n log L + log k Using two points: n = (log T₂ - log T₁)/(log L₂ - log L₁) (1) = (log 2.19 - log 1.27)/(log 1.20 - log 0.40) (1) = (0.3404 - 0.1038)/(0.0792 - (-0.3979)) = 0.2366/0.4771 = 0.496 ≈ 0.50 (1)

(c) Value of k with units. [2 marks]

Using L = 1.00 m, T = 2.00 s: k = T/Lⁿ = 2.00/(1.00)⁰·⁵⁰ = 2.00 (1) Units: s m⁻⁰·⁵⁰ (1)

Note: The theoretical value is T = 2π√(L/g), so n = 0.5 and k = 2π/√g ≈ 2.01 s m⁻⁰·⁵⁰.

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17. Determination of g.

(a) Why h vs t² is straight line through origin. [2 marks]

h = ½gt² (since u = 0) (1) This is of the form y = mx with y = h, x = t², and gradient = ½g (1) Therefore, the graph is a straight line passing through the origin.

(b) Value of g from gradient. [2 marks]

Gradient = ½g = 4.7 m s⁻² (1) g = 2 × 4.7 = 9.4 m s⁻² (1)

(c) Sources of systematic error and reduction. [4 marks]

Award 2 marks for each valid error with reduction method (1 mark for error, 1 mark for reduction):

1. Reaction time error in starting/stopping timer (1) — use an electronic timing system with automatic triggering (e.g., photogate) (1)
2. Parallax error in measuring height h (1) — use a metre rule with a set square or measure at eye level (1)
3. Air resistance affecting the fall (1) — use a dense, streamlined object or perform in a vacuum (1)
4. Delay in trapdoor mechanism (1) — calibrate the trapdoor or use a contact sensor (1)

Accept any two valid suggestions.

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18. Dynamics cart experiment.

(a) Rearrangement for straight-line graph. [2 marks]

a = mg/(M + m) 1/a = (M + m)/mg = M/(mg) + 1/g (1) 1/a = (M/g)(1/m) + 1/g (1)

(b) Gradient and y-intercept. [2 marks]

Gradient = M/g (1) y-intercept = 1/g (1)

(c) Calculate M and g. [3 marks]

y-intercept = 1/g = 0.10 → g = 1/0.10 = 10 m s⁻² (1) Gradient = M/g = 0.51 → M = 0.51 × 10 = 5.1 kg (1) Values: M = 5.1 kg, g = 10 m s⁻² (1)

Accept g = 10.0 m s⁻² with appropriate significant figures.

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19. Oscillating mass on spring.

(a) Graphical method to verify SHM. [3 marks]

For SHM, a = -ω²x, so acceleration is proportional to displacement and directed towards equilibrium (1). Plot a graph of acceleration a (y-axis) against displacement x (x-axis) (1). If the graph is a straight line passing through the origin with negative gradient, the motion is SHM (1).

Alternative: Measure period T for different amplitudes. If T is constant (independent of amplitude), the motion is SHM (1). Plot x against t and verify sinusoidal form (1). Check that a ∝ -x from the data (1).

(b) Additional measurements for spring constant k. [3 marks]

Need to measure the mass m attached to the spring (1). For a mass-spring system: ω² = k/m, so k = mω² (1). ω can be determined from the period: ω = 2π/T (1).

(c) Decreasing amplitude and precaution. [3 marks]

The amplitude decreases due to energy loss from the system (1), primarily due to air resistance/damping (1). Precaution: Use a more streamlined mass or enclose the system to reduce air currents (1).

Accept: Use a stiffer spring to reduce amplitude of air displacement; lubricate any guide rods; perform in a partial vacuum.

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20. Conservation of momentum experiment.

(a) Expected velocity after collision. [2 marks]

By conservation of momentum: m_A u_A + m_B u_B = (m_A + m_B)v (1) (0.30)(0.50) + 0 = (0.50)v → 0.15 = 0.50v → v = 0.30 m s⁻¹ (1)

(b) Precautions for accuracy and safety. [4 marks]

Award 1 mark for each valid precaution with explanation (up to 4 marks):

1. Ensure the air track is level (1) — to eliminate the effect of gravity along the track, ensuring no external horizontal forces (1)
2. Use light gates or motion sensors to measure velocities accurately (1) — reduces timing errors compared to manual stopwatch (1)
3. Minimise friction by ensuring adequate air flow (1) — reduces external forces that would invalidate conservation of momentum (1)
4. Ensure the gliders move along a straight line (1) — prevents oblique collisions that complicate analysis (1)
5. Safety: Keep hands clear of moving gliders; secure the track to prevent tipping (1) — prevents injury from fast-moving masses (1)

Accept any four valid points with clear accuracy or safety justification.

(c) Percentage difference and reason. [2 marks]

Percentage difference = |theoretical - experimental|/theoretical × 100% = |0.30 - 0.28|/0.30 × 100% = 6.67% (1)

Reason: Small amount of friction on the air track (1) OR air resistance acting on the gliders (1) OR imperfect alignment causing some energy loss (1).

Accept any valid reason for the discrepancy.

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END OF ANSWER KEY

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