A Level H1 Physics Practice Paper 3

TuitionGoWhere Practice Paper — Physics H1 A-Level

Answer Key — Practice Paper: Mechanics

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Section A: Multiple Choice

1. B [1]

• At the highest point, the ball momentarily stops (velocity = 0), but gravity still acts downward, so acceleration = $9.8 \text{ m s}^{-2}$ downwards.
• Common mistake: Choosing A — students often assume zero velocity means zero acceleration. Acceleration is due to gravity and is constant throughout the motion.

2. C [1]

• Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(2.5)(8.0)^2 = \frac{1}{2}(2.5)(64) = 80$ m.
• Common mistake: Using $s = vt = (2.5 \times 8) = 20$ m (this gives distance only for constant velocity, not acceleration from rest).

3. B [1]

• Conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
• $(3.0)(4.0) + (6.0)(0) = (3.0 + 6.0)v$
• $12 = 9.0v$, so $v = 1.3$ m/s (to 2 s.f.).
• Common mistake: Forgetting that the second object is stationary (initial velocity = 0).

4. C [1]

• Momentum ($p = mv$) has both magnitude and direction, making it a vector. Kinetic energy, power, and work are all scalars.
• Teaching note: Vectors require both magnitude and direction to be fully described. Scalars are described by magnitude alone.

5. C [1]

• Work done: $W = F \times d = 12 \times 5.0 = 60$ J.
• Work is done when a force causes displacement in the direction of the force.

6. C [1]

• In projectile motion (ignoring air resistance), there is no horizontal force, so horizontal velocity remains constant. Vertically, the ball accelerates at $g = 9.8 \text{ m s}^{-2}$ downward.
• This is a fundamental principle of projectile motion: horizontal and vertical motions are independent.

7. C [1]

• Taking the initial direction as positive: $\Delta p = m(v - u) = 0.50(-8.0 - 10) = 0.50 \times (-18) = -9.0$ kg m/s.
• Magnitude of change in momentum = 9.0 kg m/s.
• Common mistake: Students may calculate $0.50 \times (10 - 8) = 1.0$ kg m/s, forgetting that the ball reverses direction (velocity sign changes).

8. C [1]

• Total downward force = weight of beam + load = $40 + 20 = 60$ N.
• By symmetry, each support carries half: $60 / 2 = 30$ N.
• Teaching note: For a uniform load at the midpoint of a symmetrically supported beam, reactions are equal.

9. B [1]

• The car is moving in a circle, so there must be a centripetal force: $F = \frac{mv^2}{r} = \frac{1200 \times 25^2}{50} = \frac{1200 \times 625}{50} = \frac{750\,000}{50} = 15\,000$ N.
• Common mistake: Choosing D — students may confuse "constant speed" with "no net force." A change in direction requires a net force (centripetal force).

10. C [1]

• Gain in GPE: $\Delta E_p = mgh = 2.0 \times 9.8 \times 3.0 = 58.8$ J.
• Rounded to 2 s.f. = 60 J (option D). However, 58.8 J is option C. Since the data is given to 2 s.f., the answer should be 59 J or 60 J. Option C (58.8 J) is the exact calculation and is the best answer here.
• Note: Accept C (58.8 J) as the precise answer.

11. B [1]

• Using conservation of energy: $mgh = \frac{1}{2}mv^2$, so $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.0} = \sqrt{39.2} = 6.26$ m/s ≈ 6.3 m/s.
• Common mistake: Using $v = gt$ without first finding $t$, or using $v = \sqrt{2gh}$ with incorrect values.

12. C [1]

• Horizontal component: $F_x = F\cos 60°$, so $F = \frac{F_x}{\cos 60°} = \frac{15}{0.5} = 30$ N.
• Teaching note: When a force is applied at an angle, use cosine for the component adjacent to the angle (horizontal) and sine for the component opposite the angle (vertical).

13. B [1]

• For perpendicular forces: $F_{\text{resultant}} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ N.
• This is a classic 3-4-5 triangle scaled by factor 2 (6-8-10).

14. C [1]

• Efficiency = $\frac{\text{useful output}}{\text{total input}} \times 100\%$, so useful output = $0.75 \times 800 = 600$ J.
• Common mistake: Calculating $800 \times 0.25 = 200$ J (this gives the wasted energy, not the useful output).

15. C [1]

• By conservation of energy: $KE_{\text{bottom}} = PE_{\text{top}} = mgh = 1.5 \times 9.8 \times 10 = 147$ J.
• Teaching note: When air resistance is ignored, all gravitational potential energy converts to kinetic energy.

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Section B: Structured Questions

16. (Kinematics) [5 marks]

(a) [2 marks]

• Using $v = u + at$: $v = 0 + 3.0 \times 10 = 30$ m/s.
• [B1] for correct formula/substitution, [B1] for correct answer with unit.

(b) [3 marks]

• Phase 1 (acceleration): $s_1 = \frac{1}{2}at^2 = \frac{1}{2}(3.0)(10)^2 = 150$ m.
• Phase 2 (constant velocity): $s_2 = vt = 30 \times 20 = 600$ m.
• Phase 3 (deceleration): Using average velocity: $s_3 = \frac{v + 0}{2} \times t = \frac{30}{2} \times 5.0 = 75$ m.
• Total distance: $s = 150 + 600 + 75 = 825$ m.
• [B1] for distance in phase 1, [B1] for distance in phase 2, [B1] for distance in phase 3 and total.
• Alternative for phase 3: $s = vt - \frac{1}{2}at^2$ where $a = 30/5.0 = 6.0 \text{ m s}^{-2}$, giving $s = 30(5.0) - \frac{1}{2}(6.0)(25) = 150 - 75 = 75$ m.

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17. (Dynamics — Newton's Second Law) [5 marks]

(a) [1 mark]

• Horizontal component: $F_x = T\cos 30° = 40 \times \cos 30° = 40 \times 0.866 = 34.6$ N ≈ 35 N.
• [B1] for correct answer.

(b) [1 mark]

• Net horizontal force: $F_{\text{net}} = F_x - f = 34.6 - 10 = 24.6$ N ≈ 25 N.
• [B1] for correct answer.

(c) [2 marks]

• Using Newton's second law: $F_{\text{net}} = ma$, so $a = \frac{F_{\text{net}}}{m} = \frac{24.6}{5.0} = 4.92$ m/s² ≈ 4.9 m/s².
• [B1] for correct formula/substitution, [B1] for correct answer.

(d) [1 mark]

• Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(4.92)(4.0)^2 = \frac{1}{2}(4.92)(16) = 39.4$ m ≈ 39 m.
• [B1] for correct answer.

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18. (Momentum — Conservation and Collisions) [6 marks]

(a) [1 mark]

• The principle of conservation of linear momentum states that the total momentum of a system remains constant (is conserved) provided no external forces act on the system (i.e., in a closed/isolated system).
• [B1] for a complete statement including the "no external forces" or "closed system" condition.

(b) [3 marks]

• Let $u$ be the initial speed of the bullet.
• Conservation of momentum: $m_{\text{bullet}} \times u + m_{\text{block}} \times 0 = (m_{\text{bullet}} + m_{\text{block}}) \times v$
• $0.020 \times u + 0 = (0.020 + 2.0) \times 0.50$
• $0.020u = 2.02 \times 0.50 = 1.01$
• $u = \frac{1.01}{0.020} = 50.5$ m/s ≈ 51 m/s (to 2 s.f.).
• [B1] for correct conservation of momentum equation, [B1] for correct substitution, [B1] for correct answer with unit.

(c) [2 marks]

• KE before collision: $\frac{1}{2}m_{\text{bullet}}u^2 = \frac{1}{2}(0.020)(50.5)^2 = 0.5 \times 0.020 \times 2550.25 = 25.5$ J.
• KE after collision: $\frac{1}{2}(m_{\text{bullet}} + m_{\text{block}})v^2 = \frac{1}{2}(2.02)(0.50)^2 = 0.5 \times 2.02 \times 0.25 = 0.253$ J.
• Since KE before (25.5 J) >> KE after (0.253 J), kinetic energy is not conserved, so the collision is inelastic.
• [B1] for calculating both kinetic energies, [B1] for correct conclusion with justification.
• Note: This is a perfectly inelastic collision (maximum KE loss) because the objects stick together.

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19. (Forces — Equilibrium) [6 marks]

(a) [2 marks]

• Free-body diagram should show:
  • Weight $W = mg$ acting vertically downward from the centre of the frame.
  • Two tension forces $T$, each acting along a string, directed upward and outward at 25° above the horizontal.
• [B1] for correct forces shown (weight + two tensions), [B1] for correct directions and angles.

(b) [2 marks]

• For vertical equilibrium, the net vertical force is zero:
  • Upward components: $2 \times T\sin\theta$ (the vertical component of each tension is $T\sin\theta$ since the angle is measured from the horizontal).
  • Downward force: $mg$.
  • Setting net force to zero: $2T\sin\theta = mg$.
  • Therefore: $T = \frac{mg}{2\sin\theta}$.
• [B1] for correct vertical force balance equation, [B1] for correct rearrangement to show the required expression.

(c) [2 marks]

• $T = \frac{mg}{2\sin\theta} = \frac{3.0 \times 9.8}{2 \times \sin 25°} = \frac{29.4}{2 \times 0.4226} = \frac{29.4}{0.8452} = 34.8$ N ≈ 35 N.
• [B1] for correct substitution, [B1] for correct answer with unit.
• Common mistake: Using $\cos 25°$ instead of $\sin 25°$ — since the angle is with the horizontal, the vertical component uses sine.

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20. (Work, Energy and Power) [8 marks]

(a) [2 marks]

• $\Delta E_p = mgh = 60 \times 9.8 \times 12 = 7056$ J ≈ 7100 J (to 2 s.f.) or 7.1 kJ.
• [B1] for correct formula/substitution, [B1] for correct answer with unit.

(b) [2 marks]

• Average power: $P = \frac{W}{t} = \frac{\Delta E_p}{t} = \frac{7056}{15} = 470.4$ W ≈ 470 W.
• [B1] for correct formula/substitution, [B1] for correct answer with unit.

(c) [2 marks]

• The student's muscles must also do work to:
  • Overcome friction between shoes and stairs.
  • Increase the kinetic energy of the student's limbs (internal energy changes).
  • Overcome air resistance (small effect).
  • Some energy is lost as heat in the muscles (metabolic inefficiency).
• [B1] for any two valid reasons, [B1] for clear explanation.
• Key concept: The total work done by muscles = gain in GPE + work against friction + heat losses + other energy transfers.

(d) [2 marks]

• Useful power output: $P_{\text{useful}} = \frac{mgh}{t} = \frac{100 \times 9.8 \times 12}{25} = \frac{11\,760}{25} = 470.4$ W.
• Efficiency = $\frac{P_{\text{useful}}}{P_{\text{input}}}$, so $P_{\text{input}} = \frac{P_{\text{useful}}}{\text{efficiency}} = \frac{470.4}{0.65} = 723.7$ W ≈ 720 W.
• [B1] for calculating useful power, [B1] for applying efficiency to find input power.

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Section C: Data-Based / Extended Response

21. (Projectile Motion) [7 marks]

(a) [2 marks]

• Vertical motion: $h = \frac{1}{2}gt^2$ (initial vertical velocity = 0).
• $45 = \frac{1}{2}(9.8)t^2$
• $t^2 = \frac{45 \times 2}{9.8} = \frac{90}{9.8} = 9.184$
• $t = 3.03$ s ≈ 3.0 s.
• [B1] for correct equation, [B1] for correct answer.

(b) [2 marks]

• Horizontal distance: $R = v_0 \times t = 20 \times 3.03 = 60.6$ m ≈ 61 m.
• [B1] for correct formula, [B1] for correct answer.

(c) [3 marks]

• Horizontal velocity remains constant: $v_x = 20$ m/s.
• Vertical velocity just before impact: $v_y = gt = 9.8 \times 3.03 = 29.7$ m/s.
• Resultant speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 29.7^2} = \sqrt{400 + 882.09} = \sqrt{1282.09} = 35.8$ m/s ≈ 36 m/s.
• [B1] for calculating vertical component, [B1] for using Pythagoras' theorem, [B1] for correct final answer.
• Teaching note: The final speed is found by combining horizontal and vertical components using Pythagoras, since they are perpendicular.

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22. (Energy and Momentum — Multi-Concept Synthesis) [8 marks]

(a) [3 marks]

• Conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
• $(4.0)(6.0) + (2.0)(0) = (4.0)(2.0) + (2.0)v_B$
• $24 = 8.0 + 2.0v_B$
• $2.0v_B = 16$
• $v_B = 8.0$ m/s.
• [B1] for correct conservation of momentum equation, [B1] for correct substitution, [B1] for correct answer with unit.

(b) [3 marks]

• KE before: $\frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 = \frac{1}{2}(4.0)(6.0)^2 + 0 = 0.5 \times 4.0 \times 36 = 72$ J.
• KE after: $\frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(4.0)(2.0)^2 + \frac{1}{2}(2.0)(8.0)^2 = 0.5 \times 4.0 \times 4.0 + 0.5 \times 2.0 \times 64 = 8.0 + 64 = 72$ J.
• [B1] for KE before, [B1] for KE after, [B1] for correct values.

(c) [1 mark]

• Since total kinetic energy before (72 J) = total kinetic energy after (72 J), the collision is elastic.
• [B1] for correct conclusion with justification.

(d) [1 mark]

• Using the impulse-momentum theorem on trolley B: $F \times \Delta t = \Delta p_B = m_B v_B - m_B u_B$
• $F \times 0.050 = 2.0 \times 8.0 - 0 = 16$
• $F = \frac{16}{0.050} = 320$ N.
• [B1] for correct answer with unit.
• Teaching note: The impulse-momentum theorem states that the impulse (force × time) equals the change in momentum. This is derived from Newton's second law.