A Level H2 Physics Practice Paper 2

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI) — Version 2

Subject: Physics H2
Level: A-Level
Paper: Practice Paper (Mechanics Topic Set)
Duration: 1 hour 15 minutes
Total Marks: 60
Name: ________________________
Class: ____________
Date: ____________

Instructions

• This practice paper contains 20 questions on the topic of Mechanics.
• Section A: Short Structured Questions (Q1–Q8, 2 marks each, total 16 marks)
• Section B: Calculation and Application (Q9–Q15, 4 marks each, total 28 marks)
• Section C: Extended Reasoning (Q16–Q20, 3–4 marks each, total 16 marks)
• Show all working clearly. Use SI units.
• The total marks (16 + 28 + 16) = 60 as stated.

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Section A: Short Structured Questions (16 marks)

1. State the principle of conservation of linear momentum. [2]

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2. A projectile is launched horizontally from a cliff. State the direction of its acceleration during flight, ignoring air resistance. [2]

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3. Define centripetal force in the context of circular motion. [2]

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4. For a body undergoing simple harmonic motion, write the equation linking acceleration $a$ and displacement $x$. [2]

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5. State Newton's second law of motion in terms of force and momentum. [2]

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6. A mass is attached to a spring obeying Hooke's law. State the formula for the elastic potential energy stored when extended by $x$. [2]

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7. Define gravitational field strength at a point. [2]

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8. In a perfectly inelastic collision, what quantity is conserved but kinetic energy is not? [2]

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Section B: Calculation and Application (28 marks)

9. A trolley of mass $1.2\ \text{kg}$ moving at $4.0\ \text{m s}^{-1}$ collides with a stationary trolley of mass $1.8\ \text{kg}$. After collision, the first trolley moves at $1.0\ \text{m s}^{-1}$ in the same direction. Calculate the velocity of the second trolley. [4]

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10. A mass of $0.50\ \text{kg}$ is attached to a spring of spring constant $200\ \text{N m}^{-1}$ and oscillates with amplitude $0.080\ \text{m}$. Calculate (a) the angular frequency $\omega$, (b) the maximum acceleration. [4]

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11. A stone is thrown horizontally from a height of $20\ \text{m}$ with speed $15\ \text{m s}^{-1}$. Calculate the time taken to reach the ground and the horizontal distance travelled. ($g = 9.8\ \text{m s}^{-2}$) [4]

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12. A car of mass $1000\ \text{kg}$ moves around a circular track of radius $50\ \text{m}$ at constant speed $20\ \text{m s}^{-1}$. Calculate the centripetal force required. [4]

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13. A satellite orbits Earth at a distance of $7.0 \times 10^{6}\ \text{m}$ from the centre. Given $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\text{kg}^{-2}$ and $M = 6.0 \times 10^{24}\ \text{kg}$, calculate the gravitational field strength $g$ at that point. [4]

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14. A pendulum has period $2.0\ \text{s}$. Calculate its length assuming $g = 9.8\ \text{m s}^{-2}$ and $T = 2\pi\sqrt{l/g}$. [4]

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15. A ball of mass $0.20\ \text{kg}$ is dropped from rest and hits the ground after $2.0\ \text{s}$. Calculate the impulse exerted by the ground if it rebounds at $5.0\ \text{m s}^{-1}$ upward. ($g = 9.8\ \text{m s}^{-2}$) [4]

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Section C: Extended Reasoning (16 marks)

16. Explain why a rider on a bicycle leaning into a turn is an application of centripetal force. [3]

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17. Describe the energy changes that occur during one cycle of undamped simple harmonic motion. [3]

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18. A student claims that in a collision between two ice hockey pucks on frictionless ice, both momentum and kinetic energy are always conserved. Evaluate this claim. [4]

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19. The diagram below shows a velocity-time graph for a projectile.

Image pending generation: graph for Q19.

Using the graph, explain how you would determine the time to reach maximum height and the horizontal range. [3]

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20. A geostationary satellite remains above the same point on Earth's equator. State two conditions necessary for this orbit and explain one application. [3]

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TuitionGoWhere Practice Paper - Physics H2 A-Level (Answers)

Version 2 — Mechanics Topic Set

Total Marks: 60

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Section A: Short Structured Questions

1. [2 marks] Principle: In a closed (isolated) system, total momentum before an event equals total momentum after, provided no net external force acts.

• 1 mark: system / closed system stated
• 1 mark: before = after / constancy with no external force

2. [2 marks] Direction: Vertically downward (due to gravity).

• 1 mark: downward
• 1 mark: due to gravity / $g$

3. [2 marks] Centripetal force is the resultant force acting toward the centre of the circular path, causing centripetal acceleration.

• 1 mark: toward centre
• 1 mark: resultant / net force causing circular motion

4. [2 marks] $a = -\omega^{2}x$

• 1 mark: negative sign
• 1 mark: $\omega^{2}x$ form

5. [2 marks] $F = \frac{dp}{dt}$ (rate of change of momentum)

• 1 mark: force = rate of change
• 1 mark: of momentum

6. [2 marks] $E = \frac{1}{2}kx^{2}$

• 1 mark: $\frac{1}{2}kx^{2}$
• 1 mark: identifies $k$ as spring constant

7. [2 marks] Gravitational field strength is force per unit mass at a point: $g = F/m$.

• 1 mark: force per unit mass
• 1 mark: $g = F/m$

8. [2 marks] Linear momentum is conserved; kinetic energy is not (converted to heat/sound/deformation).

• 1 mark: momentum
• 1 mark: KE not conserved stated

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Section B: Calculation and Application

9. [4 marks] Conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ $(1.2)(4.0) + (1.8)(0) = (1.2)(1.0) + (1.8)v_2$ $4.8 = 1.2 + 1.8v_2 \Rightarrow 1.8v_2 = 3.6 \Rightarrow v_2 = 2.0\ \text{m s}^{-1}$

• 1 mark: equation
• 1 mark: substitution
• 1 mark: correct algebra
• 1 mark: $2.0\ \text{m s}^{-1}$

10. [4 marks] (a) $\omega = \sqrt{k/m} = \sqrt{200/0.50} = \sqrt{400} = 20\ \text{rad s}^{-1}$ (b) $a_{\max} = \omega^{2}A = (20)^{2}(0.080) = 400 \times 0.080 = 32\ \text{m s}^{-2}$

• 1 mark: (a) formula
• 1 mark: (a) value 20
• 1 mark: (b) formula
• 1 mark: (b) 32 m/s²

11. [4 marks] Vertical: $h = \frac{1}{2}gt^{2} \Rightarrow 20 = 0.5(9.8)t^{2} \Rightarrow t^{2} = 4.08 \Rightarrow t = 2.02\ \text{s}$ Horizontal: $d = vt = 15 \times 2.02 = 30.3\ \text{m}$

• 1 mark: vertical eqn
• 1 mark: t = 2.02 s
• 1 mark: horizontal eqn
• 1 mark: 30.3 m

12. [4 marks] $F = mv^{2}/r = (1000)(20)^{2}/50 = 1000 \times 400 / 50 = 8000\ \text{N}$

• 1 mark: formula
• 1 mark: substitution
• 2 marks: 8000 N

13. [4 marks] $g = GM/r^{2} = (6.67\times10^{-11})(6.0\times10^{24}) / (7.0\times10^{6})^{2}$ $= (4.002\times10^{14}) / (4.9\times10^{13}) = 8.17\ \text{N kg}^{-1}$ (or m s⁻²)

• 1 mark: formula
• 1 mark: substitution
• 2 marks: 8.2 N/kg

14. [4 marks] $T = 2\pi\sqrt{l/g} \Rightarrow l = g(T/2\pi)^{2} = 9.8\times(2.0/6.283)^{2} = 9.8\times0.101 = 0.99\ \text{m}$

• 1 mark: rearrange
• 1 mark: sub
• 2 marks: 0.99 m

15. [4 marks] Impact speed: $v = gt = 9.8\times2 = 19.6\ \text{m s}^{-1}$ downward. Momentum before: $0.20\times(-19.6) = -3.92\ \text{kg m s}^{-1}$ After: $0.20\times5.0 = +1.0$ Impulse = $\Delta p = 1.0 - (-3.92) = 4.92\ \text{N s}$

• 1 mark: speed calc
• 1 mark: momenta
• 2 marks: 4.9 N s

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Section C: Extended Reasoning

16. [3 marks] Leaning shifts the normal force from ground so its horizontal component provides centripetal force toward turn centre; friction also acts. Without lean, bike would slip outward.

• 1 mark: centripetal needed
• 1 mark: lean provides component
• 1 mark: stability explanation

17. [3 marks] KE max at equilibrium, PE max at extremes; total E constant. Energy interchanges continuously: KE↔PE↔KE.

• 1 mark: KE at centre
• 1 mark: PE at extremes
• 1 mark: interchange constant total

18. [4 marks] Momentum conserved (frictionless). KE conserved only if elastic. If pucks stick or deform, KE lost. Claim false for inelastic.

• 1 mark: momentum yes
• 1 mark: KE only elastic
• 2 marks: evaluation with example

19. [3 marks] Max height when $v_y = 0$ → read t from graph (~1.5 s). Range = $v_x \times$ total t = 10 × 4 = 40 m.

• 1 mark: t from v_y=0
• 1 mark: use v_x constant
• 1 mark: multiply for range

20. [3 marks] Conditions: (1) period = 24 h, (2) above equator / circular. Application: weather/satellite comms.

• 1 mark: period
• 1 mark: equatorial
• 1 mark: application