A Level H2 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme

Subject: Physics H2 (9478) Paper: Practice Paper 2 (Structured Questions) Version: 2 of 5 Total Marks: 80

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Section A: Kinematics and Dynamics (Questions 1–5)

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Question 1 (4 marks)

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

Using v² = u² + 2as:

• u = 0, v = 25.0 m s⁻¹, s = 150 m
• a = v² / (2s) = (25.0)² / (2 × 150) = 625 / 300 = 2.0833... m s⁻²
• a = 2.08 m s⁻² (3 s.f.)

Marking:

• [1] Correct formula and substitution
• [1] Correct answer with units and appropriate s.f.

(b) Calculate the time taken to travel the 150 m. [2 marks]

Using v = u + at:

• t = v / a = 25.0 / 2.0833... = 12.0 s

Alternatively, using s = ut + ½at²:

• 150 = 0 + ½ × 2.0833... × t²
• t² = 300 / 2.0833... = 144
• t = 12.0 s

Marking:

• [1] Correct method
• [1] Correct answer: 12.0 s

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Question 2 (4 marks)

(a) Calculate the maximum height reached by the stone. [2 marks]

At maximum height, v = 0:

• v² = u² + 2as → 0 = (19.6)² + 2(−9.81)h
• h = (19.6)² / (2 × 9.81) = 384.16 / 19.62 = 19.58... m
• h = 19.6 m (3 s.f.)

Marking:

• [1] Correct equation with v = 0 and a = −g
• [1] Correct answer: 19.6 m

(b) Determine the time taken for the stone to return to ground level. [2 marks]

Time to reach maximum height: v = u + at → 0 = 19.6 − 9.81t → t = 2.00 s Total time = 2 × 2.00 = 4.00 s

Alternatively, using s = ut + ½at² with s = 0:

• 0 = 19.6t − ½(9.81)t²
• t(19.6 − 4.905t) = 0
• t = 0 or t = 19.6 / 4.905 = 4.00 s

Marking:

• [1] Correct method (symmetry or quadratic)
• [1] Correct answer: 4.00 s

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Question 3 (4 marks)

(a) Calculate the time taken for the ball to reach the ground. [2 marks]

Vertical motion: u_y = 0, s_y = 45.0 m, a = 9.81 m s⁻²

• s = ut + ½at² → 45.0 = 0 + ½(9.81)t²
• t² = 90.0 / 9.81 = 9.174...
• t = 3.03 s (3 s.f.)

Marking:

• [1] Correct use of vertical motion equation
• [1] Correct answer: 3.03 s

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

Horizontal motion: u_x = 12.0 m s⁻¹, t = 3.03 s

• s_x = u_x × t = 12.0 × 3.03 = 36.4 m (3 s.f.)

Marking:

• [1] Correct method using horizontal velocity
• [1] Correct answer: 36.4 m

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Question 4 (4 marks)

(a) Determine the acceleration at t = 2.0 s. [2 marks]

v = 4t² − 8t + 3 a = dv/dt = 8t − 8 At t = 2.0 s: a = 8(2.0) − 8 = 8.0 m s⁻²

Marking:

• [1] Correct differentiation
• [1] Correct substitution and answer: 8.0 m s⁻²

(b) Calculate the displacement from t = 0 to t = 3.0 s. [2 marks]

s = ∫v dt = ∫(4t² − 8t + 3) dt = (4/3)t³ − 4t² + 3t + C At t = 0, s = 0 → C = 0 At t = 3.0: s = (4/3)(27) − 4(9) + 3(3) = 36 − 36 + 9 = 9.0 m

Marking:

• [1] Correct integration
• [1] Correct answer: 9.0 m

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Question 5 (4 marks)

(a) Calculate the centripetal acceleration. [2 marks]

a_c = v² / r = (10.0)² / 50.0 = 100 / 50.0 = 2.00 m s⁻²

Marking:

• [1] Correct formula
• [1] Correct answer: 2.00 m s⁻²

(b) Calculate the centripetal force. [2 marks]

F_c = ma_c = 80.0 × 2.00 = 160 N

Marking:

• [1] Correct formula F = ma
• [1] Correct answer: 160 N

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Section B: Forces and Newton's Laws (Questions 6–10)

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Question 6 (4 marks)

(a) Calculate the maximum static friction force. [2 marks]

f_s,max = μ_s × N = μ_s × mg = 0.60 × 5.0 × 9.81 = 29.43 N f_s,max = 29.4 N (3 s.f.)

Marking:

• [1] Correct formula f = μN with N = mg
• [1] Correct answer: 29.4 N

(b) Determine the minimum value of F to set the block in motion. [2 marks]

F_min = f_s,max = 29.4 N

Marking:

• [1] Recognition that F must equal maximum static friction
• [1] Correct answer: 29.4 N

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Question 7 (4 marks)

(a) Draw free-body diagrams. [2 marks]

Block A (on table):

• Weight mg downward
• Normal reaction N upward
• Tension T to the right

Block B (hanging):

• Weight mg downward
• Tension T upward

Marking:

• [1] Correct forces on Block A (weight, normal, tension)
• [1] Correct forces on Block B (weight, tension)

(b) Calculate acceleration and tension. [2 marks]

For Block B: m_B g − T = m_B a → 2.0 × 9.81 − T = 2.0a → T = 19.62 − 2.0a For Block A: T = m_A a → T = 3.0a

Equating: 3.0a = 19.62 − 2.0a → 5.0a = 19.62 → a = 3.924 m s⁻² a = 3.92 m s⁻² (3 s.f.)

T = 3.0 × 3.924 = 11.77 N → T = 11.8 N (3 s.f.)

Marking:

• [1] Correct equations of motion for both blocks
• [1] Correct acceleration (3.92 m s⁻²) and tension (11.8 N)

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Question 8 (4 marks)

(a) Draw a diagram showing forces. [1 mark]

Forces:

• Weight mg vertically downward
• Normal reaction N perpendicular to road surface
• No friction (given)

Marking:

• [1] Correct diagram with weight and normal reaction shown

(b) Calculate the banking angle θ. [3 marks]

Resolving vertically: N cos θ = mg Resolving horizontally (centripetal): N sin θ = mv²/r

Dividing: tan θ = v²/(rg) = (20.0)² / (80.0 × 9.81) = 400 / 784.8 = 0.5097... θ = tan⁻¹(0.5097...) = 27.0° (3 s.f.)

Marking:

• [1] Correct resolution of forces
• [1] Correct derivation of tan θ = v²/rg
• [1] Correct answer: 27.0°

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Question 9 (4 marks)

(a) State the principle of conservation of linear momentum. [1 mark]

The total momentum of a closed system remains constant, provided no external forces act on the system.

Marking:

• [1] Correct statement including "closed system" or "no external forces"

(b) Calculate the initial thrust on the rocket. [3 marks]

Thrust = rate of change of momentum of exhaust gases = (dm/dt) × v_exhaust = 50.0 × 2000 = 100,000 N = 1.00 × 10⁵ N

Marking:

• [1] Correct formula: thrust = v(dm/dt)
• [1] Correct substitution
• [1] Correct answer: 1.00 × 10⁵ N

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Question 10 (4 marks)

(a) Calculate tension when accelerating upwards. [2 marks]

T − mg = ma T = m(g + a) = 800(9.81 + 1.50) = 800 × 11.31 = 9048 N T = 9.05 × 10³ N (3 s.f.)

Marking:

• [1] Correct equation T − mg = ma
• [1] Correct answer: 9.05 × 10³ N

(b) Determine tension at constant speed. [2 marks]

At constant speed, a = 0: T = mg = 800 × 9.81 = 7848 N T = 7.85 × 10³ N (3 s.f.)

Marking:

• [1] Recognition that a = 0
• [1] Correct answer: 7.85 × 10³ N

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Section C: Work, Energy and Power (Questions 11–15)

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Question 11 (4 marks)

(a) Calculate the work done by the force. [2 marks]

W = F · s = (3.0i + 4.0j) · (5.0i + 2.0j) = 3.0 × 5.0 + 4.0 × 2.0 = 15.0 + 8.0 = 23.0 J

Marking:

• [1] Correct dot product method
• [1] Correct answer: 23.0 J

(b) Calculate the speed at B. [2 marks]

Work-energy theorem: W = ΔKE = ½mv² − 0 23.0 = ½ × 2.0 × v² v² = 23.0 v = 4.80 m s⁻¹ (3 s.f.)

Marking:

• [1] Correct use of work-energy theorem
• [1] Correct answer: 4.80 m s⁻¹

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Question 12 (4 marks)

(a) Calculate elastic potential energy. [2 marks]

EPE = ½kx² = ½ × 200 × (0.15)² = 100 × 0.0225 = 2.25 J

Marking:

• [1] Correct formula
• [1] Correct answer: 2.25 J

(b) Calculate launch speed. [2 marks]

EPE → KE: ½kx² = ½mv² 2.25 = ½ × 0.050 × v² v² = 2.25 / 0.025 = 90 v = 9.49 m s⁻¹ (3 s.f.)

Marking:

• [1] Correct energy conversion
• [1] Correct answer: 9.49 m s⁻¹

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Question 13 (4 marks)

(a) Calculate power output. [2 marks]

P = work done per second = mgh/t = (dm/dt)gh P = 0.80 × 9.81 × 25.0 = 196.2 W P = 196 W (3 s.f.)

Marking:

• [1] Correct formula P = (dm/dt)gh
• [1] Correct answer: 196 W

(b) Calculate electrical power input. [2 marks]

Efficiency = P_out / P_in 0.70 = 196 / P_in P_in = 196 / 0.70 = 280 W

Marking:

• [1] Correct use of efficiency formula
• [1] Correct answer: 280 W

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Question 14 (4 marks)

(a) Calculate average power developed. [2 marks]

Acceleration: a = (v − u)/t = 30.0/8.0 = 3.75 m s⁻² Net force: F_net = ma = 1000 × 3.75 = 3750 N Engine force: F_engine = F_net + F_resistive = 3750 + 500 = 4250 N Distance: s = ½(u + v)t = ½(0 + 30.0) × 8.0 = 120 m Work done by engine: W = F_engine × s = 4250 × 120 = 510,000 J Average power: P = W/t = 510,000/8.0 = 63,750 W = 63.8 kW (3 s.f.)

Alternative: P_avg = F_engine × v_avg = 4250 × 15.0 = 63,750 W

Marking:

• [1] Correct method (either work/time or force × average velocity)
• [1] Correct answer: 63.8 kW

(b) Work done against resistive forces. [2 marks]

W_resistive = F_resistive × s = 500 × 120 = 60,000 J = 60.0 kJ

Marking:

• [1] Correct formula W = Fs
• [1] Correct answer: 60.0 kJ

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Question 15 (4 marks)

(a) Calculate work done against friction. [2 marks]

Normal reaction: N = mg cos 30° = 2.0 × 9.81 × cos 30° = 2.0 × 9.81 × 0.8660 = 16.99 N Friction force: f = μN = 0.25 × 16.99 = 4.248 N Work against friction: W_f = f × s = 4.248 × 3.0 = 12.74 J W_f = 12.7 J (3 s.f.)

Marking:

• [1] Correct calculation of friction force
• [1] Correct answer: 12.7 J

(b) Determine speed at bottom. [2 marks]

Loss in GPE = mgh = mgs sin 30° = 2.0 × 9.81 × 3.0 × 0.5 = 29.43 J Gain in KE = Loss in GPE − Work against friction ½mv² = 29.43 − 12.74 = 16.69 v² = 2 × 16.69 / 2.0 = 16.69 v = 4.09 m s⁻¹ (3 s.f.)

Marking:

• [1] Correct energy equation (GPE loss = KE gain + work against friction)
• [1] Correct answer: 4.09 m s⁻¹

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Section D: Momentum and Circular Motion (Questions 16–20)

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Question 16 (4 marks)

(a) State the principle of conservation of linear momentum. [1 mark]

The total momentum of a closed system remains constant, provided no external forces act on the system.

Marking:

• [1] Correct statement including "closed system" or "no external forces"

(b) Calculate velocity of 0.30 kg ball after collision. [3 marks]

Taking initial direction of 0.20 kg ball as positive: Initial momentum = 0.20 × 5.0 + 0.30 × 0 = 1.0 kg m s⁻¹ Final momentum = 0.20 × (−1.0) + 0.30 × v = −0.20 + 0.30v

Conservation of momentum: 1.0 = −0.20 + 0.30v 0.30v = 1.20 v = 4.0 m s⁻¹ (in the original direction of the 0.20 kg ball)

Marking:

• [1] Correct initial momentum calculation
• [1] Correct application of conservation of momentum
• [1] Correct answer: 4.0 m s⁻¹

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Question 17 (4 marks)

(a) Calculate gravitational force on satellite. [2 marks]

F = GMm/r² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 500) / (7.0 × 10⁶)² = (6.67 × 6.0 × 500 × 10¹³) / (49 × 10¹²) = (20,010 × 10¹³) / (49 × 10¹²) = 4.084... × 10³ N F = 4.08 × 10³ N (3 s.f.)

Marking:

• [1] Correct formula and substitution
• [1] Correct answer: 4.08 × 10³ N

(b) Calculate orbital speed. [2 marks]

Gravitational force provides centripetal force: GMm/r² = mv²/r v² = GM/r = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / (7.0 × 10⁶) = (40.02 × 10¹³) / (7.0 × 10⁶) = 5.717... × 10⁷ v = √(5.717... × 10⁷) = 7.56 × 10³ m s⁻¹

Marking:

• [1] Correct derivation v² = GM/r
• [1] Correct answer: 7.56 × 10³ m s⁻¹

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Question 18 (4 marks)

(a) Draw diagram showing forces on bob. [1 mark]

Forces:

• Weight mg vertically downward
• Tension T along the string towards the point of suspension

Marking:

• [1] Correct diagram with both forces labelled

(b) Calculate tension in the string. [3 marks]

Geometry: sin θ = r/L = 0.40/1.2 = 1/3 → θ = 19.47° cos θ = √(1 − sin²θ) = √(1 − 1/9) = √(8/9) = 0.9428

Vertically: T cos θ = mg T = mg / cos θ = (0.50 × 9.81) / 0.9428 = 4.905 / 0.9428 = 5.20 N (3 s.f.)

Marking:

• [1] Correct geometry (sin θ or cos θ)
• [1] Correct vertical resolution T cos θ = mg
• [1] Correct answer: 5.20 N

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Question 19 (4 marks)

(a) Calculate angular frequency. [1 mark]

ω = 2π/T = 2π/2.0 = π = 3.14 rad s⁻¹

Marking:

• [1] Correct answer: 3.14 rad s⁻¹

(b) Determine maximum acceleration. [1 mark]

a_max = ω²x₀ = (π)² × 0.080 = 9.8696 × 0.080 = 0.790 m s⁻² (3 s.f.)

Marking:

• [1] Correct answer: 0.790 m s⁻²

(c) Calculate total energy. [2 marks]

E_total = ½mω²x₀² = ½ × 0.10 × (π)² × (0.080)² = 0.05 × 9.8696 × 0.0064 = 0.05 × 0.06317 = 3.16 × 10⁻³ J (3 s.f.)

Marking:

• [1] Correct formula E = ½mω²x₀²
• [1] Correct answer: 3.16 × 10⁻³ J

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Question 20 (4 marks)

(a) Calculate centripetal force at top. [2 marks]

F_c = mv²/r = 0.15 × (4.0)² / 0.80 = 0.15 × 16 / 0.80 = 2.4 / 0.80 = 3.0 N

Marking:

• [1] Correct formula
• [1] Correct answer: 3.0 N

(b) Determine tension at top of circle. [2 marks]

At the top: T + mg = F_c T = F_c − mg = 3.0 − (0.15 × 9.81) = 3.0 − 1.4715 = 1.53 N (3 s.f.)

Marking:

• [1] Correct force equation at top (T + mg = mv²/r)
• [1] Correct answer: 1.53 N

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END OF ANSWER KEY

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Marking notes:

• Accept equivalent methods and alternative correct formulations.
• Deduct 1 mark for incorrect or missing units only once per question.
• Accept answers within ±1 in the last significant figure due to rounding variations.
• For definition questions, accept equivalent wording that captures the essential physics.