A Level H1 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme

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

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Section A: Structured Questions (60 marks)

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Question 1: Kinematics and Projectile Motion

(a) Time taken for stone to reach the sea [2]

Vertical motion: s = ut + ½at² 45.0 = 0 + ½(9.81)t² [M1] t² = (2 × 45.0) / 9.81 = 9.17 t = 3.03 s [A1]

(b) Horizontal distance [1]

Horizontal distance = horizontal velocity × time = 20.0 × 3.03 = 60.6 m [A1]

(c) Magnitude of velocity just before hitting sea [3]

Vertical component: v_y = u_y + at = 0 + 9.81 × 3.03 = 29.7 m s⁻¹ [M1] Horizontal component: v_x = 20.0 m s⁻¹ (constant) [M1] Resultant speed = √(v_x² + v_y²) = √(20.0² + 29.7²) = √(400 + 882) = √1282 = 35.8 m s⁻¹ [A1]

(d) Graph of vertical velocity vs. time [2]

• Straight line through origin with positive gradient [B1]
• Gradient = 9.81 m s⁻²
• Line extends from t = 0 to t = 3.03 s
• Final vertical velocity labelled as -29.7 m s⁻¹ (or 29.7 m s⁻¹ downward)
• Axes labelled correctly [B1]

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Question 2: Forces and Equilibrium

(a) Free-body diagram [3]

Forces to be shown:

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

All forces correctly labelled and positioned.

(b) Reaction force from wall [3]

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

(c) Frictional force [2]

Horizontal equilibrium: F = R_W [M1] F = 72.2 N [A1]

(d) Minimum coefficient of static friction [2]

Vertical equilibrium: R_G = 250 N [M1] μ_min = F / R_G = 72.2 / 250 = 0.289 [A1]

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Question 3: Work, Energy, and Power

(a) Acceleration [1]

a = (v - u) / t = (25.0 - 0) / 8.00 = 3.13 m s⁻² [A1]

(b) Distance travelled [2]

s = ut + ½at² = 0 + ½(3.13)(8.00)² [M1] s = 100 m [A1] (Alternative: s = ½(u + v)t = ½(0 + 25.0)(8.00) = 100 m)

(c) Work done by engine [3]

Resultant force = ma = 1200 × 3.13 = 3756 N [M1] Driving force = resultant force + resistive force = 3756 + 600 = 4356 N [M1] Work done = driving force × distance = 4356 × 100 = 4.36 × 10⁵ J [A1]

(d) Average power [2]

Average power = work done / time = 4.36 × 10⁵ / 8.00 [M1] = 5.45 × 10⁴ W [A1] (Alternative: P = Fv_avg = 4356 × 12.5 = 5.45 × 10⁴ W)

(e) Explanation [2]

As speed increases, air resistance (drag force) increases [B1]. The net force (driving force - resistive forces) decreases, so acceleration decreases. Eventually, when driving force equals total resistive forces, net force = 0 and the car reaches terminal velocity / constant maximum speed [B1].

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Question 4: Momentum and Collisions

(a) Principle of conservation of linear momentum [2]

The total momentum of a closed/isolated system remains constant [B1] provided no external resultant force acts on the system [B1].

(b) Velocity of trolley B after collision [2]

Total momentum before = total momentum after (2.00 × 3.00) + (1.00 × 0) = (2.00 × 1.00) + (1.00 × v_B) [M1] 6.00 = 2.00 + v_B v_B = 4.00 m s⁻¹ to the right [A1]

(c) Show collision is elastic [3]

Total KE before = ½(2.00)(3.00)² + 0 = 9.00 J [M1] Total KE after = ½(2.00)(1.00)² + ½(1.00)(4.00)² = 1.00 + 8.00 = 9.00 J [M1] Since total KE before = total KE after, the collision is elastic [A1].

(d) Conservation of momentum with magnetic repulsion [3]

Newton's Third Law: The force exerted by A on B is equal and opposite to the force exerted by B on A [B1]. The time of interaction is the same for both trolleys [B1]. Impulse on A = -Impulse on B (since FΔt is equal and opposite). Change in momentum of A = -Change in momentum of B. Therefore, total momentum remains constant [B1].

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Question 5: Current Electricity

(a) Current in circuit [2]

Total resistance = R + r = 5.50 + 0.500 = 6.00 Ω [M1] I = EMF / total resistance = 12.0 / 6.00 = 2.00 A [A1]

(b) Terminal potential difference [1]

V = EMF - Ir = 12.0 - (2.00 × 0.500) = 11.0 V [A1] (Alternative: V = IR = 2.00 × 5.50 = 11.0 V)

(c) Power dissipated in external resistor [2]

P = I²R = (2.00)² × 5.50 [M1] = 22.0 W [A1] (Alternative: P = VI = 11.0 × 2.00 = 22.0 W)

(d) Power wasted in internal resistance [1]

P = I²r = (2.00)² × 0.500 = 2.00 W [A1]

(e) Explanation using potential divider [2]

The internal resistance r and the external resistance R form a potential divider in series [B1]. The EMF is divided between r and R in proportion to their resistances. The terminal p.d. (voltage across R) = EMF × [R/(R + r)], which is less than the EMF because some voltage is dropped across the internal resistance [B1].

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Question 6: D.C. Circuits

(a) Circuit diagram [2]

• Battery symbol with 10.0 V labelled [B1]
• R₁ (4.00 kΩ) and R₂ (6.00 kΩ) in series
• Output voltage V_out labelled across R₂
• Correct circuit symbols and connections [B1]

(b) Output voltage across R₂ [2]

V_out = V_supply × [R₂/(R₁ + R₂)] [M1] = 10.0 × [6.00/(4.00 + 6.00)] = 10.0 × 0.600 = 6.00 V [A1]

(c) New output voltage with load resistor [4]

R₂ and load (3.00 kΩ) in parallel: 1/R_parallel = 1/6.00 + 1/3.00 = 1/6.00 + 2/6.00 = 3/6.00 [M1] R_parallel = 2.00 kΩ [A1]

New total resistance = R₁ + R_parallel = 4.00 + 2.00 = 6.00 kΩ [M1] New V_out = 10.0 × (2.00/6.00) = 3.33 V [A1]

(d) Explanation of voltage change [2]

The load resistor provides an additional path for current, reducing the effective resistance of the lower arm of the potential divider [B1]. This changes the ratio of resistances in the divider, reducing the fraction of the supply voltage appearing across the output [B1].

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Question 7: Nuclear Physics

(a) Nuclear equation for carbon-14 decay [2]

¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄ₑ [B1 for correct products, B1 for balanced atomic and mass numbers]

(b) Age of sample [3]

After n half-lives, fraction remaining = (½)ⁿ 0.25 = (½)ⁿ [M1] n = 2 (since ½² = ¼ = 0.25) [M1] Age = n × t₁/₂ = 2 × 5730 = 11,460 years [A1]

(c) Limitation of carbon-14 dating [2]

After about 50,000 years (approximately 9 half-lives), the activity of carbon-14 becomes too low to measure accurately [B1]. The count rate becomes comparable to background radiation, making reliable dating impossible [B1].

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Section B: Free Response Questions (20 marks)

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Question 8: Mechanics – Energy and Momentum

(a) Speed just before hitting ground [2]

Using v² = u² + 2as: v² = 0 + 2(9.81)(10.0) = 196.2 [M1] v = 14.0 m s⁻¹ [A1] (Alternative: using energy conservation: mgh = ½mv²)

(b) Impulse on ball [3]

Taking upward as positive: Initial velocity (just before impact) = -14.0 m s⁻¹ Final velocity (just after impact) = +8.00 m s⁻¹ [M1] Change in momentum = m(v - u) = 0.500(8.00 - (-14.0)) = 0.500 × 22.0 [M1] Impulse = 11.0 N s upward [A1]

(c) Average force [2]

F_avg = Impulse / time = 11.0 / 0.0500 [M1] = 220 N upward [A1]

(d) Maximum height after rebound [2]

Using v² = u² + 2as: 0 = (8.00)² + 2(-9.81)h [M1] h = 64.0 / (2 × 9.81) = 3.26 m [A1]

(e) Reason for not reaching original height [1]

Energy is lost during the impact (converted to thermal energy/sound/deformation of the ball and ground) [B1]. The collision is inelastic, so kinetic energy is not conserved.

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Question 9: Electricity – Circuit Analysis

(a) Resistance of thermistor at 300 K [2]

R_T = R₀ e^(k/θ) = 2.00 × e^(3000/300) [M1] = 2.00 × e¹⁰ = 2.00 × 22,026 = 4.41 × 10⁴ Ω [A1]

(b) Current at 300 K [2]

Total resistance = R_T + 100 = 44,100 + 100 = 44,200 Ω [M1] I = V / R_total = 6.00 / 44,200 = 1.36 × 10⁻⁴ A [A1]

(c) Potential difference across thermistor at 300 K [1]

V_T = I × R_T = 1.36 × 10⁻⁴ × 44,100 = 6.00 V [A1] (Effectively all the voltage as R_T >> 100 Ω)

(d) Effect of temperature increase to 350 K [3]

As temperature increases, the exponential term e^(k/θ) decreases because k/θ decreases [B1]. Therefore, R_T decreases significantly [B1]. With lower total resistance, the current in the circuit increases. Since R_T decreases relative to the fixed 100 Ω resistor, the potential difference across the thermistor decreases (potential divider effect) [B1].

(e) Practical application [2]

Temperature sensor / electronic thermometer [B1]. The thermistor is placed in one arm of a potential divider. As temperature changes, the resistance changes, altering the output voltage. This voltage can be calibrated to give a temperature reading [B1]. (Other valid applications: fire alarm, overheating protection in circuits, temperature control systems)

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Question 10: Waves and Photoelectric Effect

(a) Energy of photon in joules [2]

E = hf = hc/λ [M1] E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / (450 × 10⁻⁹) = (1.989 × 10⁻²⁵) / (4.50 × 10⁻⁷) = 4.42 × 10⁻¹⁹ J [A1]

(b) Work function in joules [1]

Φ = 2.00 eV × 1.60 × 10⁻¹⁹ J/eV = 3.20 × 10⁻¹⁹ J [A1]

(c) Maximum kinetic energy [2]

K_max = hf - Φ = 4.42 × 10⁻¹⁹ - 3.20 × 10⁻¹⁹ [M1] = 1.22 × 10⁻¹⁹ J [A1]

(d) Stopping potential [2]

eV_s = K_max [M1] V_s = K_max / e = 1.22 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 0.763 V [A1]

(e) Effect of doubling intensity [3]

(i) Maximum kinetic energy: No change [B1]. The energy of each photon depends only on frequency/wavelength (E = hf), not on intensity. Each photon still transfers the same energy, so K_max remains unchanged.

(ii) Photoelectric current: The current doubles [B1]. Doubling intensity means twice as many photons strike the surface per second. This releases twice as many photoelectrons per second, doubling the current [B1].

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END OF ANSWER KEY