A Level H1 Physics Practice Paper 1

TuitionGoWhere Practice Paper — Answer Key (Version 1)

Subject: Physics H1 | Topic: Mechanics | Total Marks: 60

Section A Answers (21 marks)

Q1 [2]
Principle: In a closed (isolated) system, total linear momentum is conserved.
Marking: [B1] total momentum constant / $p_{\text{before}} = p_{\text{after}}$; [B1] no net external force / isolated system.
Teaching: Momentum is a vector; external forces like friction break conservation.

Q2 [2]
(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]
$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m} \Rightarrow m = \frac{p^2}{2E_k} = \frac{144}{48} = 3.0\ \text{kg}$ [M1]
$v = 12 / 3.0 = 4.0\ \text{m/s}$ [A1]
Marks: 3 allocated as shown.

Q4 [2]
Displacement is the shortest distance from initial to final position in a stated direction (vector). [1] Distance is total path length (scalar). [1]

Q5 [3]
$a = (v-u)/t = (20-0)/10 = 2.0\ \text{m/s}^2$ [M1]
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.0)(10^2) = 100\ \text{m}$ [M2]
Marks: 1 for accel, 2 for distance with working.

Q6 [3]
Graph: area = triangle (0–5s) + rectangle (5–10s) + triangle (10–15s).
Area = $\frac{1}{2}(5)(10) + (5)(10) + \frac{1}{2}(5)(10) = 25+50+25 = 100\ \text{m}$ [M3]
Teaching: Area under v-t graph = displacement; here motion is one-direction so distance = 100 m.

Q7 [2]
A body remains at rest or in uniform motion in a straight line unless acted upon by a net external force. [2]

Section B Answers (18 marks)

Q8 [3]
Forces: weight $80\ \text{N}$ downward at centre (2.0 m from A); hinge reaction at A (upward + horizontal components); tension $T$ upward at B. Label all. [3]

Q9 [3]
Take moments about A: $T \times 4.0 = 80 \times 2.0$ [M2]
$T = 160/4.0 = 40\ \text{N}$ [A1]
Teaching: Only weight produces moment about hinge.

Q10 [4]
Let $R_L$, $R_R$ be reactions.
$\Sigma M$ about right: $R_L \times 5.0 = 200 \times 2.5 + 100 \times 3.0 = 500 + 300 = 800$ [M2]
$R_L = 160\ \text{N}$ [A1]
Check: $R_R = 300-160 = 140\ \text{N}$ [M1]

Q11 [2]
For a body in equilibrium, sum of clockwise moments = sum of anticlockwise moments about any point. [2]

Q12 [3]
Balance: $5 \times 0.3 = 3 \times d$ (left side) → $d = 1.5/3 = 0.50\ \text{m}$ left of pivot. [M3]
(Using given 3 kg at unknown distance to balance 5 kg at 0.3 m right.)

Q13 [3]
A couple is two equal, opposite, parallel forces whose lines of action do not coincide, producing pure rotation. [2] Example: turning a steering wheel. [1]

Section C Answers (21 marks)

Q14 [4]
Vertical: $s = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2}(9.8)t^2 \Rightarrow t = \sqrt{90/9.8} = 3.03\ \text{s}$ [M2]
Horizontal: $R = vt = 15 \times 3.03 = 45.5\ \text{m}$ [M2]

Q15 [3]
$u_y = 25\sin30^\circ = 12.5\ \text{m/s}$ [M1]
$0 = u_y^2 - 2gh \Rightarrow h = (12.5)^2/(2\times9.8) = 7.97\ \text{m}$ [M2]

Q16 [4]
Momentum: $2.0\times4.0 = (3.0)v \Rightarrow v = 8/3 = 2.67\ \text{m/s}$ [M2]
KE before = $\frac{1}{2}(2)(16)=16\ \text{J}$; KE after = $\frac{1}{2}(3)(2.67^2)=10.7\ \text{J}$; lost = $5.3\ \text{J}$ [M2]

Q17 [2]
Elastic: KE conserved. Inelastic: KE not conserved (some to heat/deformation). [2]

Q18 [4]
$\omega = 2\pi f = 2\pi(3.0) = 18.85\ \text{rad/s}$ [M1]
$v = \omega r = 18.85\times1.2 = 22.62\ \text{m/s}$ [M1]
$F = mv^2/r = 0.50\times(22.62^2)/1.2 = 213\ \text{N}$ [M2]

Q19 [2]
Centripetal acceleration is acceleration directed towards centre of circular path, arising from change in velocity direction. [2]

Q20 [2]
Because velocity direction changes continuously; acceleration is centripetal, even if speed constant. [2]