A Level H1 Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics H1 A-Level

ANSWER KEY AND MARKING SCHEME

Paper: Practice Paper 5 (Version 5 of 5) Total Marks: 80

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Section A: Structured Questions [50 marks]

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Question 1: Kinematics [5 marks]

(a) Calculate the acceleration of the cyclist. [2 marks]

Using s = ut + ½at² u = 0, s = 64 m, t = 8.0 s 64 = 0 + ½ × a × (8.0)² [M1] 64 = 32a a = 2.0 m s⁻² [A1]

(b) Calculate the velocity of the cyclist at t = 8.0 s. [1 mark]

v = u + at = 0 + 2.0 × 8.0 = 16.0 m s⁻¹ [A1]

(c) Calculate the total distance travelled from the start. [2 marks]

Distance during constant velocity phase: s₂ = v × t = 16.0 × 12.0 = 192 m [M1] Total distance = 64 + 192 = 256 m [A1]

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Question 2: Projectile Motion [4 marks]

(a) Calculate the horizontal and vertical components of the initial velocity. [2 marks]

v_x = 15.0 cos 40° = 15.0 × 0.766 = 11.5 m s⁻¹ [A1] v_y = 15.0 sin 40° = 15.0 × 0.643 = 9.64 m s⁻¹ [A1]

(b) Determine the maximum height reached by the ball. [2 marks]

At maximum height, v_y = 0 Using v² = u² + 2as: 0 = (9.64)² + 2(-9.81)h [M1] h = (9.64)² / (2 × 9.81) = 92.93 / 19.62 = 4.74 m [A1]

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Question 3: Forces and Equilibrium [6 marks]

(a) Draw a free-body diagram showing all forces acting on the plank. [2 marks]

Forces to show:

• Weight of plank: 200 N downward at centre (2.0 m from A) [B1]
• Weight of person: 600 N downward at x = 2.0 m from A
• Reaction at P (R_P): upward at A
• Reaction at Q (R_Q): upward at 3.0 m from A All forces correctly labelled with directions [B1]

(b) Determine the reaction force at Q when the person stands at the midpoint. [2 marks]

Taking moments about P: Clockwise moments = Anticlockwise moments (200 × 2.0) + (600 × 2.0) = R_Q × 3.0 [M1] 400 + 1200 = 3.0 R_Q R_Q = 1600 / 3.0 = 533 N [A1]

(c) Determine the maximum distance x from A that the person can stand without the plank tipping. [2 marks]

Plank tips when R_P = 0 (plank just lifts off support P) Taking moments about Q: 200 × 1.0 = 600 × (3.0 - x) [M1] 200 = 1800 - 600x 600x = 1600 x = 2.67 m [A1]

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Question 4: Momentum and Collisions [6 marks]

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

In a closed/isolated system, the total linear momentum remains constant provided no external forces act. [B1] OR: The total momentum before a collision equals the total momentum after the collision, in the absence of external forces. [B1]

(b) Calculate the velocity of trolley B after the collision. [2 marks]

m_A u_A + m_B u_B = m_A v_A + m_B v_B (2.0 × 3.0) + (1.0 × 0) = (2.0 × 1.0) + (1.0 × v_B) [M1] 6.0 = 2.0 + v_B v_B = 4.0 m s⁻¹ to the right [A1]

(c) Determine whether the collision is elastic or inelastic. [2 marks]

Initial KE = ½ × 2.0 × (3.0)² + 0 = 9.0 J [M1] Final KE = ½ × 2.0 × (1.0)² + ½ × 1.0 × (4.0)² = 1.0 + 8.0 = 9.0 J Initial KE = Final KE, therefore collision is elastic. [A1]

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Question 5: Work, Energy and Power [5 marks]

(a) Calculate the work done by the applied force. [1 mark]

W = F × d = 400 × 5.0 = 2000 J [A1]

(b) Calculate the gain in gravitational potential energy of the crate. [2 marks]

Vertical height gained: h = 5.0 sin 30° = 5.0 × 0.5 = 2.5 m [M1] GPE = mgh = 50 × 9.81 × 2.5 = 1226 J ≈ 1230 J [A1]

(c) Determine the frictional force acting on the crate. [2 marks]

Work done against friction = Work by applied force - Gain in GPE W_f = 2000 - 1226 = 774 J [M1] Frictional force f = W_f / d = 774 / 5.0 = 155 N [A1]

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Question 6: Current Electricity [5 marks]

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

Total resistance: R_total = R_ext + r = 5.5 + 0.50 = 6.0 Ω [M1] I = E / R_total = 12.0 / 6.0 = 2.0 A [A1]

(b) Calculate the terminal potential difference across the battery. [1 mark]

V = E - Ir = 12.0 - (2.0 × 0.50) = 11.0 V [A1] OR: V = IR_ext = 2.0 × 5.5 = 11.0 V

(c) Calculate the power dissipated in the external resistor. [2 marks]

P = I²R = (2.0)² × 5.5 [M1] P = 4.0 × 5.5 = 22.0 W [A1] OR: P = VI = 11.0 × 2.0 = 22.0 W

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Question 7: D.C. Circuits [6 marks]

(a) Draw the circuit diagram. [1 mark]

Correct diagram showing:

• 4.0 Ω and 12.0 Ω in parallel
• This combination in series with 2.0 Ω resistor
• 9.0 V battery connected across the whole circuit [B1]

(b) Calculate the total resistance of the circuit. [2 marks]

Parallel combination: 1/R_p = 1/4.0 + 1/12.0 = 3/12.0 + 1/12.0 = 4/12.0 [M1] R_p = 12.0/4 = 3.0 Ω R_total = R_p + 2.0 = 3.0 + 2.0 = 5.0 Ω [A1]

(c) Calculate the current through the 4.0 Ω resistor. [2 marks]

Total current: I_total = V / R_total = 9.0 / 5.0 = 1.8 A [M1] Voltage across parallel combination: V_p = I_total × R_p = 1.8 × 3.0 = 5.4 V Current through 4.0 Ω: I_4 = V_p / 4.0 = 5.4 / 4.0 = 1.35 A [A1]

(d) Calculate the potential difference across the 12.0 Ω resistor. [1 mark]

V_12 = V_p = 5.4 V [A1]

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Question 8: Potential Divider [4 marks]

(a) Calculate the output voltage across R₂. [2 marks]

V_out = [R₂ / (R₁ + R₂)] × V_supply [M1] V_out = [6.0 / (3.0 + 6.0)] × 9.0 = (6.0/9.0) × 9.0 = 6.0 V [A1]

(b) Calculate the new output voltage with load resistor. [2 marks]

R₂ parallel with 6.0 kΩ: 1/R_parallel = 1/6.0 + 1/6.0 = 2/6.0 R_parallel = 3.0 kΩ [M1] V_out = [3.0 / (3.0 + 3.0)] × 9.0 = (3.0/6.0) × 9.0 = 4.5 V [A1]

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Question 9: Nuclear Physics [4 marks]

(a) Write the nuclear equation for this decay. [2 marks]

¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ₑ [B2] (Award [B1] for correct daughter nucleus; [B1] for beta particle and antineutrino)

(b) Calculate the number of carbon-14 nuclei remaining after 17 190 years. [2 marks]

Number of half-lives: n = 17190 / 5730 = 3 [M1] N = N₀ × (½)ⁿ = 8.0 × 10¹⁰ × (½)³ = 8.0 × 10¹⁰ × ⅛ = 1.0 × 10¹⁰ [A1]

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Question 10: Radioactivity [5 marks]

(a) Determine the half-life of the source. [2 marks]

Activity halves from 800 to 400 Bq in 4.0 hours [M1] OR: Activity halves from 566 to 283 Bq in 4.0 hours (from t = 2.0 to t = 6.0) Half-life = 4.0 hours [A1]

(b) Calculate the decay constant λ. [1 mark]

λ = ln 2 / t₁/₂ = 0.693 / (4.0 × 3600) = 4.81 × 10⁻⁵ s⁻¹ [A1]

(c) Calculate the initial number of radioactive nuclei. [2 marks]

A₀ = λN₀ [M1] N₀ = A₀ / λ = 800 / (4.81 × 10⁻⁵) = 1.66 × 10⁷ [A1]

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Section B: Free Response Questions [30 marks]

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Question 11: Mechanics – Momentum and Energy [15 marks]

(a)(i) Calculate the speed of the ball just before it hits the ground. [2 marks]

Using v² = u² + 2as: v² = 0 + 2 × 9.81 × 20.0 [M1] v = √(392.4) = 19.8 m s⁻¹ [A1]

(a)(ii) Calculate the impulse exerted on the ball by the ground. [3 marks]

Taking upward as positive: Initial velocity before impact = -19.8 m s⁻¹ Final velocity after impact = +12.0 m s⁻¹ Impulse = Δp = m(v - u) = 0.50(12.0 - (-19.8)) [M1] = 0.50 × 31.8 [M1] = 15.9 N s upward [A1]

(a)(iii) Calculate the average force exerted on the ball by the ground. [2 marks]

F = Impulse / Δt = 15.9 / 0.080 [M1] = 199 N upward [A1]

(b)(i) Calculate the velocity of skater B after they push apart. [2 marks]

Conservation of momentum: 0 = m_A v_A + m_B v_B 0 = 60 × 2.0 + 40 × v_B [M1] v_B = -120/40 = -3.0 m s⁻¹ Velocity of B is 3.0 m s⁻¹ in opposite direction to A [A1]

(b)(ii) Calculate the total kinetic energy after they push apart. [2 marks]

KE_A = ½ × 60 × (2.0)² = 120 J [M1] KE_B = ½ × 40 × (3.0)² = 180 J Total KE = 120 + 180 = 300 J [A1]

(b)(iii) Explain where this kinetic energy comes from. [2 marks]

The kinetic energy comes from the chemical energy stored in the skaters' muscles. [B1] When they push, they do work on each other, converting chemical energy to kinetic energy. [B1]

(b)(iv) Calculate the distance skater A travels before coming to rest. [2 marks]

Work done by friction = Change in KE F × d = ½mv² [M1] 50 × d = ½ × 60 × (2.0)² = 120 d = 120/50 = 2.4 m [A1]

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Question 12: Electricity and Circuits [15 marks]

(a) Define electrical resistance and state Ohm's law. [2 marks]

Resistance = potential difference / current (R = V/I) [B1] Ohm's law: The current through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant. [B1]

(b)(i) Calculate the resistance of the lamp at rated voltage. [2 marks]

P = V²/R → R = V²/P [M1] R = (6.0)²/12.0 = 36.0/12.0 = 3.0 Ω [A1]

(b)(ii) Explain why resistance is lower when first switched on. [2 marks]

When first switched on, the filament is cold/at room temperature. [B1] As current flows, the filament heats up. The increased thermal vibrations of the metal ions cause more collisions with electrons, increasing resistance. Therefore, resistance is lower when cold. [B1]

(c)(i) Explain how to obtain I-V characteristic data. [2 marks]

Adjust the variable resistor to change the current in the circuit. [B1] For each setting, record the voltmeter reading (V across X) and ammeter reading (I through X). Repeat for a range of values. [B1]

(c)(ii) Plot graph of I against V. [3 marks]

Correct axes with labels and units [B1] All 7 points plotted correctly [B1] Smooth curve drawn through points [B1]

(c)(iii) Determine whether component X obeys Ohm's law. [2 marks]

The graph is not a straight line through the origin. [B1] Therefore, component X does NOT obey Ohm's law (current is not directly proportional to voltage). [B1]

(c)(iv) Calculate power dissipated when I = 0.50 A. [2 marks]

From graph, when I = 0.50 A, V ≈ 3.1 V [M1] P = VI = 3.1 × 0.50 = 1.55 W ≈ 1.6 W [A1] (Allow 1.5-1.6 W depending on graph reading)

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Question 13: Nuclear Physics and Applications [15 marks]

(a) Explain what is meant by "isotope." [2 marks]

Isotopes are atoms of the same element (same proton/atomic number) [B1] but with different numbers of neutrons (different mass/nucleon number). [B1]

(b)(i) Determine number of alpha and beta particles emitted. [3 marks]

Change in mass number: 238 - 206 = 32 Each alpha particle reduces mass number by 4: Number of alpha particles = 32/4 = 8 [M1] Change in atomic number due to 8 alpha particles: 8 × 2 = 16 Expected atomic number after alpha decays: 92 - 16 = 76 Actual atomic number: 82 Difference: 82 - 76 = 6 [M1] Each beta particle increases atomic number by 1: Number of beta particles = 6 [A1]

(b)(ii) Explain why alpha particles are more ionising than beta particles. [2 marks]

Alpha particles have a larger mass and charge (+2e) compared to beta particles (-1e). [B1] They travel more slowly and interact more strongly with atoms they pass, causing more ionisation per unit length of path. [B1]

(c)(i) Calculate the decay constant. [2 marks]

λ = ln 2 / t₁/₂ = 0.693 / (15 × 3600) [M1] = 0.693 / 54000 = 1.28 × 10⁻⁵ s⁻¹ [A1]

(c)(ii) Calculate activity after 45 hours. [2 marks]

Number of half-lives: n = 45/15 = 3 [M1] A = A₀ × (½)ⁿ = 1200 × (½)³ = 1200 × ⅛ = 150 Bq [A1]

(c)(iii) Calculate time for activity to decrease to 150 Bq. [2 marks]

From (c)(ii), activity = 150 Bq after 3 half-lives [M1] t = 3 × 15 = 45 hours [A1] OR: Using A = A₀e^(-λt): 150 = 1200e^(-λt) → e^(-λt) = 0.125 → λt = ln 8 → t = ln 8 / λ

(c)(iv) Suggest why measured activity may be higher than calculated. [2 marks]

Background radiation contributes to the measured count. [B1] The measured activity includes counts from both the source AND background radiation, making it appear higher than the source activity alone. [B1]

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END OF ANSWER KEY