A Level H2 Physics Practice Paper 2

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

Version 2 — Mechanics Topic Set

Total Marks: 60

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

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

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