A Level H1 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 2 (Structured & Free Response) Version: 4 of 5 Total Marks: 80

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Section A: Structured Questions [50 marks]

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Question 1: Kinematics of a Falling Object [6 marks]

(a) Calculate the height of the cliff. [2 marks]

Answer: s = ut + ½at² [M1] s = 0 + ½(9.81)(3.20)² s = ½ × 9.81 × 10.24 s = 50.2 m [A1]

Award [M1] for correct substitution into equation of motion; [A1] for correct answer with unit.

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(b) Calculate the speed of the stone just before it hits the sea. [2 marks]

Answer: v = u + at [M1] v = 0 + (9.81)(3.20) v = 31.4 m s⁻¹ [A1]

Accept use of v² = u² + 2as. [M1] for correct method; [A1] for correct answer with unit.

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(c) Sketch the velocity-time graph. [2 marks]

Answer:

• Straight line through origin [B1]
• Line extends to t = 3.20 s, v = 31.4 m s⁻¹ (or approximately 31 m s⁻¹) [B1]
• Axes correctly labelled: v / m s⁻¹ (vertical), t / s (horizontal)

[B1] for correct shape (straight line, positive gradient); [B1] for correct endpoint values labelled.

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Question 2: Forces and Equilibrium [7 marks]

(a) Draw a free-body diagram. [3 marks]

Answer: Forces to be shown:

• Weight (W = 250 N) acting vertically downward at the centre of the ladder (2.50 m from either end) [B1]
• Normal reaction from wall (R_w) acting horizontally to the right at the top of the ladder [B1]
• Normal reaction from ground (R_g) acting vertically upward at the foot of the ladder [B1]
• Frictional force (F) acting horizontally to the left at the foot of the ladder

[B1] each for any three correctly drawn and labelled forces. All four forces must be present for full marks. Deduct 1 mark if weight is not shown at the centre.

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(b) Calculate the reaction force exerted by the wall. [2 marks]

Answer: Take moments about foot of ladder: Clockwise moment = Anticlockwise moment [M1] R_w × (5.00 sin 60°) = 250 × (2.50 cos 60°) R_w × 4.33 = 250 × 1.25 R_w = 312.5 / 4.33 R_w = 72.2 N [A1]

[M1] for correct moment equation; [A1] for correct answer with unit. Accept 72 N.

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(c) Calculate the frictional force. [2 marks]

Answer: For horizontal equilibrium: F = R_w [M1] F = 72.2 N [A1]

[M1] for recognising horizontal equilibrium; [A1] for correct answer (ecf from part b).

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Question 3: Momentum and Collisions [8 marks]

(a) State the principle of conservation of linear momentum. [2 marks]

Answer: In a closed/isolated system, the total linear momentum remains constant [B1] provided no external forces act on the system [B1].

Accept: "The total momentum of a system before a collision equals the total momentum after the collision, provided no external forces act."

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(b) Calculate the velocity of trolley B after the collision. [3 marks]

Answer: Total momentum before = Total momentum after [M1] m_A × u_A + m_B × u_B = m_A × v_A + m_B × v_B (2.00)(4.00) + (3.00)(0) = (2.00)(0.80) + (3.00)(v_B) [M1] 8.00 = 1.60 + 3.00 v_B v_B = 6.40 / 3.00 v_B = 2.13 m s⁻¹ to the right [A1]

[M1] for stating conservation principle; [M1] for correct substitution; [A1] for correct answer with direction.

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(c) Determine whether the collision is elastic or inelastic. [3 marks]

Answer: Total KE before = ½(2.00)(4.00)² + 0 = 16.0 J [M1] Total KE after = ½(2.00)(0.80)² + ½(3.00)(2.13)² = 0.640 + 6.81 = 7.45 J [M1] Since KE after < KE before, the collision is inelastic. [A1]

[M1] for calculating KE before; [M1] for calculating KE after; [A1] for correct conclusion with justification. Accept ecf from part (b).

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Question 4: Work, Energy and Power [7 marks]

(a) Calculate the tension in the lifting cable. [2 marks]

Answer: At constant speed, net force = 0 [M1] T = mg = 500 × 9.81 = 4905 N ≈ 4.91 × 10³ N [A1]

[M1] for recognising equilibrium condition; [A1] for correct answer with unit.

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(b) Calculate the power output of the crane motor. [2 marks]

Answer: P = Fv [M1] P = 4905 × 0.40 = 1962 W ≈ 1.96 × 10³ W [A1]

[M1] for correct formula; [A1] for correct answer with unit. Accept ecf from part (a).

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(c) Calculate the electrical power input. [2 marks]

Answer: Efficiency = (useful power output) / (power input) × 100% [M1] 75 = (1962 / P_in) × 100 P_in = 1962 / 0.75 = 2616 W ≈ 2.62 × 10³ W [A1]

[M1] for correct use of efficiency formula; [A1] for correct answer. Accept ecf from part (b).

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(d) Explain why actual power input is greater than useful power output. [1 mark]

Answer: Energy is dissipated/lost as heat due to friction in the motor and mechanical components, and as sound. [B1]

Accept any valid reason relating to energy losses in the system.

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Question 5: Current Electricity [7 marks]

(a) Calculate the current in the circuit. [2 marks]

Answer: Total resistance = R + r = 5.20 + 0.80 = 6.00 Ω [M1] I = ε / R_total = 12.0 / 6.00 = 2.00 A [A1]

[M1] for calculating total resistance; [A1] for correct current.

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(b) Calculate the terminal potential difference. [2 marks]

Answer: V = ε - Ir [M1] V = 12.0 - (2.00)(0.80) = 12.0 - 1.60 = 10.4 V [A1]

Accept V = IR = 2.00 × 5.20 = 10.4 V. [M1] for correct method; [A1] for correct answer.

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(c) Calculate the power dissipated in the external resistor. [1 mark]

Answer: P = I²R = (2.00)² × 5.20 = 20.8 W [A1]

Accept P = VI = 10.4 × 2.00 = 20.8 W or P = V²/R.

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(d) Explain why terminal p.d. is less than e.m.f. [2 marks]

Answer: The internal resistance acts as a resistor in series with the external circuit [B1]. The battery and internal resistance together form a potential divider, so some of the e.m.f. is 'dropped' across the internal resistance, reducing the voltage available at the terminals [B1].

[B1] for mentioning internal resistance as series resistor; [B1] for linking to potential divider concept or voltage drop across internal resistance.

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Question 6: D.C. Circuits [8 marks]

(a) State the function of a potential divider circuit. [1 mark]

Answer: A potential divider provides a fraction of the input voltage / divides the supply voltage into smaller voltages. [B1]

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(b) Calculate the voltmeter reading when R₂ = 200 Ω. [3 marks]

Answer: V_out = [R₂ / (R₁ + R₂)] × V_in [M1] V_out = [200 / (300 + 200)] × 9.00 [M1] V_out = (200/500) × 9.00 = 0.400 × 9.00 = 3.60 V [A1]

[M1] for correct potential divider formula; [M1] for correct substitution; [A1] for correct answer.

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(c) Calculate R₂ when voltmeter reads 6.00 V. [2 marks]

Answer: 6.00 = [R₂ / (300 + R₂)] × 9.00 [M1] 6.00(300 + R₂) = 9.00 R₂ 1800 + 6.00 R₂ = 9.00 R₂ 1800 = 3.00 R₂ R₂ = 600 Ω [A1]

[M1] for setting up equation; [A1] for correct answer.

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(d) Suggest one practical application of a potential divider circuit. [2 marks]

Answer: Any one of:

• Volume control in audio equipment [B1] with brief explanation of how varying resistance changes output voltage [B1]
• Light sensor circuit using an LDR [B1] with explanation [B1]
• Temperature sensor using a thermistor [B1] with explanation [B1]

[B1] for naming a valid application; [B1] for correct explanation linking to potential divider principle.

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Question 7: Nuclear Physics [7 marks]

(a) Write the nuclear equation for alpha decay of polonium-210. [2 marks]

Answer: ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He [B2]

[B1] for correct daughter nucleus (lead-206); [B1] for correct alpha particle. Accept ⁴₂α for alpha particle.

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(b) Calculate the number of nuclei remaining after 414 days. [3 marks]

Answer: Number of half-lives = 414 / 138 = 3 [M1] After 3 half-lives: N = N₀(½)³ [M1] N = 8.00 × 10¹² × (1/8) = 1.00 × 10¹² nuclei [A1]

[M1] for determining number of half-lives; [M1] for correct application of decay formula; [A1] for correct answer.

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(c) Explain why activity decreases exponentially with time. [2 marks]

Answer: The number of undecayed nuclei decreases by a constant fraction in equal time intervals [B1]. Since activity is proportional to the number of undecayed nuclei (A = λN), the activity also decreases exponentially with the same half-life [B1].

[B1] for linking to constant fraction decay; [B1] for linking activity to number of nuclei.

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Section B: Free Response Questions [30 marks]

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Question 8: Mechanics – Projectile Motion and Energy [15 marks]

(a) Calculate the time taken to reach the ground. [3 marks]

Answer: Vertical motion: s_y = u_y t + ½a_y t² [M1] 2.50 = 0 + ½(9.81)t² [M1] t² = 5.00 / 9.81 = 0.5097 t = 0.714 s [A1]

[M1] for identifying vertical motion equation; [M1] for correct substitution (u_y = 0); [A1] for correct answer.

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(b) Calculate the horizontal distance travelled. [2 marks]

Answer: Horizontal motion: s_x = u_x × t [M1] s_x = 24.0 × 0.714 = 17.1 m [A1]

[M1] for correct formula; [A1] for correct answer. Accept ecf from part (a).

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(c) Calculate magnitude and direction of velocity just before impact. [4 marks]

Answer: Horizontal component: v_x = 24.0 m s⁻¹ (constant) [B1] Vertical component: v_y = u_y + a_y t = 0 + 9.81 × 0.714 = 7.00 m s⁻¹ [M1] Magnitude: v = √(v_x² + v_y²) = √(24.0² + 7.00²) = √(576 + 49.0) = √625 = 25.0 m s⁻¹ [M1] Direction: θ = tan⁻¹(v_y / v_x) = tan⁻¹(7.00/24.0) = 16.3° below the horizontal [A1]

[B1] for horizontal component; [M1] for vertical component calculation; [M1] for magnitude calculation; [A1] for correct direction with reference to horizontal.

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(d) Calculate kinetic energy just before impact. [2 marks]

Answer: KE = ½mv² [M1] KE = ½(0.057)(25.0)² = ½ × 0.057 × 625 = 17.8 J [A1]

[M1] for correct formula; [A1] for correct answer. Accept ecf from part (c).

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(e) Compare horizontal range and time of flight for 30° launch. [4 marks]

Answer: Time of flight: The initial vertical velocity is now upward (u_y = 24.0 sin 30° = 12.0 m s⁻¹ upward). The ball will rise before falling, so it takes longer to reach the ground. Time of flight increases. [B2]

Horizontal range: The horizontal component of velocity is reduced (u_x = 24.0 cos 30° = 20.8 m s⁻¹). However, the increased time of flight may compensate. The range depends on the product u_x × t. A full calculation would be needed to determine whether range increases or decreases, but the longer flight time suggests the range could be greater despite the lower horizontal speed. [B2]

[B1] for recognising initial upward vertical component increases flight time; [B1] for clear explanation; [B1] for recognising reduced horizontal component; [B1] for discussing trade-off between speed and time. Award marks for qualitative reasoning without full calculation.

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Question 9: Electricity – Circuit Analysis and Power [15 marks]

(a) Calculate resistance of lamp at rated voltage. [2 marks]

Answer: P = V²/R [M1] R = V²/P = (6.0)²/12 = 36/12 = 3.0 Ω [A1]

[M1] for correct formula; [A1] for correct answer.

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(b) Draw circuit diagram. [2 marks]

Answer:

• Battery symbol with 9.0 V label [B1]
• Lamp and variable resistor in series with battery, all symbols correct [B1]

[B1] for correct battery symbol and label; [B1] for correct series arrangement with correct component symbols.

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(c) Lamp operating at rated voltage: [5 marks]

(i) Current in the circuit. [2 marks]

Answer: P = VI [M1] I = P/V = 12/6.0 = 2.0 A [A1]

[M1] for correct formula; [A1] for correct answer.

(ii) Potential difference across variable resistor. [1 mark]

Answer: V_R = 9.0 - 6.0 = 3.0 V [A1]

(iii) Resistance of variable resistor. [2 marks]

Answer: R = V/I [M1] R = 3.0/2.0 = 1.5 Ω [A1]

[M1] for correct formula; [A1] for correct answer.

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(d) Two identical lamps in series across 9.0 V battery. [3 marks]

Answer: Each lamp has resistance 3.0 Ω (at rated temperature). Total resistance = 6.0 Ω. [M1] Current = 9.0/6.0 = 1.5 A. [M1] Voltage across each lamp = 1.5 × 3.0 = 4.5 V (or 9.0/2 = 4.5 V). Since each lamp receives only 4.5 V instead of its rated 6.0 V, the power dissipated is less than 12 W. Neither lamp operates at rated brightness. [A1]

[M1] for calculating total resistance; [M1] for calculating current or voltage per lamp; [A1] for correct conclusion with justification. Note: In reality, filament lamp resistance changes with temperature, but at this level, assume constant resistance for simplicity.

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(e) Two lamps in parallel across 9.0 V battery. [3 marks]

Answer: Each lamp receives 9.0 V, which exceeds its rated 6.0 V [B1]. This would cause excessive current, overheating, and likely damage the lamps. This arrangement is not suitable [B1].

Assuming the resistance remains 3.0 Ω (though in practice it would increase with temperature): Current through each lamp = 9.0/3.0 = 3.0 A Total current = 6.0 A [B1]

[B1] for recognising overvoltage issue; [B1] for concluding unsuitability; [B1] for current calculation with stated assumption.

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Question 10: Waves and Nuclear Physics [15 marks]

(a) Photoelectric effect calculations: [8 marks]

(i) Energy of a single photon in joules. [3 marks]

Answer: E = hf = hc/λ [M1] E = (6.63 × 10⁻³⁴)(3.00 × 10⁸) / (4.50 × 10⁻⁷) [M1] E = 1.989 × 10⁻²⁵ / 4.50 × 10⁻⁷ E = 4.42 × 10⁻¹⁹ J [A1]

[M1] for correct formula; [M1] for correct substitution; [A1] for correct answer.

(ii) Show photon energy is approximately 2.76 eV. [1 mark]

Answer: E (eV) = (4.42 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹) = 2.76 eV [A1]

Must show conversion for the mark.

(iii) Maximum kinetic energy of photoelectrons. [2 marks]

Answer: K_max = hf - Φ [M1] K_max = 4.42 × 10⁻¹⁹ - (2.30 × 1.60 × 10⁻¹⁹) K_max = 4.42 × 10⁻¹⁹ - 3.68 × 10⁻¹⁹ K_max = 7.40 × 10⁻²⁰ J [A1]

[M1] for applying Einstein's photoelectric equation; [A1] for correct answer. Accept 7.4 × 10⁻²⁰ J.

(iv) Stopping potential. [2 marks]

Answer: eV_s = K_max [M1] V_s = (7.40 × 10⁻²⁰) / (1.60 × 10⁻¹⁹) = 0.463 V [A1]

[M1] for relating stopping potential to maximum KE; [A1] for correct answer.

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(b) Effect of doubling intensity: [4 marks]

(i) Effect on maximum kinetic energy. [2 marks]

Answer: No change [B1]. The maximum kinetic energy depends only on the photon energy (frequency/wavelength) and the work function, not on intensity. Doubling intensity increases the number of photons but not the energy per photon [B1].

[B1] for correct answer; [B1] for correct explanation.

(ii) Effect on photoelectric current. [2 marks]

Answer: The photoelectric current doubles [B1]. Doubling intensity doubles the number of photons incident per second, which doubles the number of photoelectrons emitted per second, doubling the current [B1].

[B1] for correct answer; [B1] for correct explanation linking photon number to electron emission rate.

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(c) Calculate half-life of radioactive source. [3 marks]

Answer: A = A₀(½)^(t/T₁/₂) [M1] 600 = 4800(½)^(24/T₁/₂) 600/4800 = (½)^(24/T₁/₂) 1/8 = (½)^(24/T₁/₂) [M1] (½)³ = (½)^(24/T₁/₂) Therefore: 3 = 24/T₁/₂ T₁/₂ = 8.0 hours [A1]

[M1] for correct decay equation; [M1] for recognising 1/8 = (½)³; [A1] for correct answer with unit.

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END OF ANSWER KEY