A Level H1 Physics Practice Paper 3

TuitionGoWhere Practice Paper — Answer Key (Version 3)

Subject: Physics H1
Level: A-Level
Topic: Mechanics
Total Marks: 60

Section A: Foundations of Mechanics (Q1–7) — 21 marks

Q1. [2]
Teaching note: Conservation of linear momentum applies to a closed system.
Answer: In a closed (isolated) system, the total linear momentum remains constant (momentum before = momentum after) provided no external net force acts. [B1 for constant total momentum, B1 for no external force / closed system]

Q2. [2 total: 1+1]
(a) $p = mv$ [1]
(b) $E_k = \frac{1}{2}mv^2$ [1]
Common trap: Omitting $\frac{1}{2}$ in kinetic energy.

Q3. [3]
Given $p = 12\ \text{N·s}$, $E_k = 24\ \text{J}$.
$p = mv \Rightarrow v = p/m$ [M1]
$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m} \Rightarrow m = \frac{p^2}{2E_k}$ [M1]
$m = \frac{12^2}{2 \times 24} = \frac{144}{48} = 3.0\ \text{kg}$ [A1]
$v = 12 / 3.0 = 4.0\ \text{m s}^{-1}$ [A1]
Marks: M1+M1+A1+A1 (3 counted as method+answer).

Q4. [3]
Horizontal component of pull: $F_x = 20\cos30^\circ = 17.3\ \text{N}$ [M1]
Net force: $F_{net} = 17.3 - 5.0 = 12.3\ \text{N}$ [M1]
$a = F_{net}/m = 12.3 / 4.0 = 3.1\ \text{m s}^{-2}$ [A1]

Q5. [2]
Work done by a constant force = product of force component in direction of displacement and the displacement. [B1] $W = F s \cos\theta$ or $F \times d$ in direction. [B1]

Q6. [4 total: 2+2]
(a) Acceleration = gradient = $12/4 = 3.0\ \text{m s}^{-2}$ [2]
(b) Distance = area: triangle (½×4×12=24) + rect (4×12=48) + triangle (½×2×12=12) = 84 m [2]

Q7. [3]
Forces: weight of plank $300\ \text{N}$ at centre (2.5 m from A), boy $500\ \text{N}$ at 2.0 m from A, upward reactions at A and B. [B1 each force correctly placed, B1 for labels] (Diagram expected from placeholder.)

Section B: Dynamics and Equilibrium (Q8–14) — 21 marks

Q8. [4]
Take right as positive: $p_i = 2.0(4.0) + 3.0(-2.0) = 8 - 6 = 2.0\ \text{kg m s}^{-1}$ [M2]
$(2.0+3.0)v = 2.0 \Rightarrow v = 0.40\ \text{m s}^{-1}$ right [A2]

Q9. [5]
Moments about left support (at 1.0 m): $R_2 \times 2.0 = 200(1.0) + 150(0.5)$ [M2]
$R_2 = (200+75)/2 = 137.5\ \text{N}$ [A1]
$\Sigma F_y=0: R_1 + 137.5 = 350 \Rightarrow R_1 = 212.5\ \text{N}$ [M1+A1]

Q10. [2]
Dynamic equilibrium: object moves at constant velocity. Newton's first law: if net force zero, velocity constant. [B1] Hence acceleration zero, net force zero. [B1]

Q11. [4]
$u_y = 20\sin30^\circ = 10\ \text{m s}^{-1}$ [M1]
At max height $v_y=0$: $0 = 10^2 - 2(9.8)h$ [M1]
$h = 100 / 19.6 = 5.1\ \text{m}$ [A2]

Q12. [4]
Vertical: $45 = \frac{1}{2}(9.8)t^2 \Rightarrow t = \sqrt{90/9.8} = 3.03\ \text{s}$ [M2]
Range $= 15 \times 3.03 = 45.5\ \text{m}$ [A2]

Q13. [4]
$W = mg = 19.6\ \text{N}$; component down plane $= 19.6\sin25^\circ = 8.28\ \text{N}$ [M1]
Normal $N = 19.6\cos25^\circ = 17.8\ \text{N}$ [M1]
Max friction $= \mu N = 0.30 \times 17.8 = 5.34\ \text{N}$ [M1]
Since $8.28 > 5.34$, block moves down. [A1]

Q14. [2]
Rate of change of momentum proportional to net force: $\vec{F} = m\vec{a}$ or $F = dp/dt$. [B1+B1]

Section C: Energy and Circular Motion (Q15–20) — 18 marks

Q15. [3]
$mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ [M1]
$v = \sqrt{2 \times 9.8 \times 8.0} = \sqrt{156.8} = 12.5\ \text{m s}^{-1}$ [A2]

Q16. [2]
$P = Fv \Rightarrow F = P/v = 40000/20 = 2000\ \text{N}$ [2]

Q17. [4]
$\omega = 2\pi f = 2\pi \times 4.0 = 25.1\ \text{rad s}^{-1}$ [M1]
$v = \omega r = 25.1 \times 0.80 = 20.1\ \text{m s}^{-1}$ [M1]
$F = mv^2/r = 1.2 \times 20.1^2 / 0.80 = 605\ \text{N}$ [A2]

Q18. [2]
$a = v^2/r = (7.5\times10^3)^2 / (7.0\times10^6) = 8.0\ \text{m s}^{-2}$ [2]

Q19. [2]
Elastic: total KE conserved. [B1] Inelastic: total KE not conserved (some lost as heat/sound). [B1]

Q20. [5]
Initial KE: $\frac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$ [M1]
Final $v$: $2.0(3.0) = 3.5v \Rightarrow v = 1.71\ \text{m s}^{-1}$ [M1]
Final KE: $\frac{1}{2}(3.5)(1.71)^2 = 5.1\ \text{J}$ [M1]
$9.0 \neq 5.1$, so not conserved. [A2]