A Level H2 Physics Practice Paper 3

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Practice Paper - Version 3 of 5 - Answer Key & Marking Scheme

Subject: Physics
Level: H2 A-Level
Topic: Mechanics

Section A: Structured Questions

1. State the Principle of Conservation of Linear Momentum. [2]

• Answer: In a closed system (or isolated system) [1], the total momentum before an interaction (collision/explosion) is equal to the total momentum after the interaction, provided no external resultant force acts on the system [1].
• Marking Notes: Accept "sum of momentum before = sum of momentum after". Must mention "closed/isolated system" or "no external forces".

2. Calculate the maximum acceleration of the ball. [3]

• Given: $m = 0.15 \text{ kg}$, $A = 4.0 \text{ cm} = 0.04 \text{ m}$, $f = 2.5 \text{ Hz}$.
• Formula: $a_{max} = \omega^2 A$ and $\omega = 2\pi f$.
• Working:
  • $\omega = 2 \pi (2.5) = 5\pi \approx 15.71 \text{ rad s}^{-1}$ [1]
  • $a_{max} = (15.71)^2 \times 0.04$ [1]
  • $a_{max} = 9.87 \text{ m s}^{-2}$ [1]
• Answer: $9.9 \text{ m s}^{-2}$ (2 s.f.)

3. Define gravitational field strength. [1]

• Answer: The gravitational force per unit mass acting on a small test mass placed at that point.
• Marking Notes: Must include "force per unit mass".

4. Explain why the satellite is accelerating. [2]

• Answer: Velocity is a vector quantity having both magnitude and direction [1]. Although the speed (magnitude) is constant, the direction of motion is continuously changing [1]. Therefore, the velocity is changing, which implies acceleration.

5. Block on inclined plane. (a) Free-body diagram. [2]

• Answer: Diagram must show:
  1. Weight ($mg$) acting vertically downwards. [1]
  2. Normal contact force ($N$) acting perpendicular to the plane. [1]
  3. Frictional force ($f$) acting up the slope (parallel to plane). [1]
  • Note: Award 2 marks for all three correct. 1 mark for two correct.

(b) Calculate frictional force. [2]

• Reasoning: Since velocity is constant, acceleration is zero. Forces are balanced.
• Working:
  • Component of weight down slope = $mg \sin \theta$ [1]
  • $f = mg \sin 30^\circ = 2.0 \times 9.81 \times 0.5$ [1]
  • $f = 9.81 \text{ N}$
• Answer: $9.8 \text{ N}$

6. State the relationship between impulse and change in momentum. [1]

• Answer: Impulse is equal to the change in momentum. ($I = \Delta p$)

7. Car on circular bend. (a) Calculate centripetal force. [2]

• Formula: $F_c = \frac{mv^2}{r}$
• Working: $F_c = \frac{1200 \times 20^2}{50}$ [1]
• Calculation: $F_c = \frac{1200 \times 400}{50} = 9600 \text{ N}$ [1]
• Answer: $9600 \text{ N}$

(b) Identify the force. [1]

• Answer: Friction (between tyres and road).

8. Elastic vs Inelastic collisions. [2]

• Answer:
  • In an elastic collision, total kinetic energy is conserved [1].
  • In an inelastic collision, total kinetic energy is not conserved (some is converted to heat/sound/deformation) [1].
  • Note: Momentum is conserved in both (if isolated).

9. Horizontal acceleration of projectile. [1]

• Answer: $0 \text{ m s}^{-2}$ (or zero).

10. Falling object velocity-time graph gradient. [3]

• Answer:
  • The gradient represents acceleration [1].
  • As speed increases, air resistance (drag) increases [1].
  • The resultant force ($Weight - Drag$) decreases, so acceleration decreases [1].

Section B: Data Analysis and Application

11. Simple Pendulum. (a) Theoretical relationship. [1]

• Answer: $T = 2\pi \sqrt{\frac{L}{g}}$ or $T^2 = \frac{4\pi^2}{g} L$.

(b) Calculate $g$. [3]

• Reasoning: From $T^2 = (\frac{4\pi^2}{g}) L$, the gradient $m = \frac{4\pi^2}{g}$.
• Working:
  • $g = \frac{4\pi^2}{\text{gradient}}$ [1]
  • $g = \frac{4 \pi^2}{4.02}$ [1]
  • $g = 9.817... \text{ m s}^{-2}$
• Answer: $9.82 \text{ m s}^{-2}$

12. Collision of trolleys. (a) Common velocity. [3]

• Principle: Conservation of Momentum.
• Working:
  • $m_A u_A + m_B u_B = (m_A + m_B) v$ [1]
  • $(0.50)(1.2) + (0.80)(0) = (0.50 + 0.80) v$
  • $0.60 = 1.30 v$ [1]
  • $v = 0.4615... \text{ m s}^{-1}$
• Answer: $0.46 \text{ m s}^{-1}$

(b) Loss in kinetic energy. [3]

• Working:
  • $KE_{initial} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.50 \times 1.2^2 = 0.36 \text{ J}$ [1]
  • $KE_{final} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 1.30 \times (0.4615)^2 = 0.1385 \text{ J}$ [1]
  • Loss $= 0.36 - 0.1385 = 0.2215 \text{ J}$
• Answer: $0.22 \text{ J}$

13. Crane lifting load. (a) Tension in cable. [3]

• Newton's 2nd Law: $T - mg = ma \Rightarrow T = m(g+a)$
• Working:
  • $T = 500 (9.81 + 0.50)$ [1]
  • $T = 500 (10.31)$ [1]
  • $T = 5155 \text{ N}$
• Answer: $5160 \text{ N}$ (3 s.f.)

(b) Work done by tension. [3]

• Working:
  • Distance $s = ut + \frac{1}{2}at^2 = 0 + 0.5(0.50)(4.0)^2 = 4.0 \text{ m}$ [1]
  • Work $W = F s \cos \theta$. Force and displacement are in same direction ($\theta=0$).
  • $W = T \times s = 5155 \times 4.0$ [1]
  • $W = 20620 \text{ J}$
• Answer: $2.06 \times 10^4 \text{ J}$

14. Vertical Circle. (a) Condition at highest point. [2]

• Answer: The tension in the string must be greater than or equal to zero ($T \ge 0$). For minimum speed, $T=0$ [1]. The weight provides the entire centripetal force [1].

(b) Derive $v_{min}$. [3]

• Working:
  • At top: $T + mg = \frac{mv^2}{r}$ [1]
  • For minimum speed, set $T=0$: $mg = \frac{mv_{min}^2}{r}$ [1]
  • $v_{min}^2 = gr \Rightarrow v_{min} = \sqrt{gr}$ [1]

15. Friction Experiment. (a) Static vs Kinetic friction. [2]

• Answer: The coefficient of static friction ($\mu_s$) is generally greater than the coefficient of kinetic friction ($\mu_k$) [1]. This is because interlocking between surface irregularities is stronger when surfaces are stationary relative to each other than when they are sliding [1].

(b) Precaution. [1]

• Answer: Use a force sensor/data logger to capture the peak force just before motion starts (rather than relying on human reaction time with a spring balance). OR Ensure the pulling force is strictly horizontal.

Section C: Long Structured Questions

16. Rocket Launch. (a) Acceleration increase. [3]

• Answer:
  • Newton's 2nd Law: $F_{net} = ma \Rightarrow a = \frac{F_{net}}{m}$ [1].
  • As fuel burns, the mass $m$ of the rocket decreases [1].
  • Assuming thrust is constant and drag/weight changes are secondary or thrust > weight, the decreasing denominator $m$ causes the acceleration $a$ to increase [1].

(b) Gravitational Potential Energy (GPE). [4]

• Answer:
  • GPE increases as the rocket moves away from Earth (work is done against gravity) [1].
  • The formula $\Delta E_p = mg\Delta h$ assumes $g$ is constant [1].
  • However, $g$ decreases with distance from the Earth's center ($g \propto \frac{1}{r^2}$) [1].
  • Therefore, for large $\Delta h$, the variation in $g$ is significant, and the general formula $E_p = -\frac{GMm}{r}$ must be used [1].

17. Coefficient of Restitution. (a) Show $e = \sqrt{\frac{h'}{h}}$. [4]

• Working:
  • Speed just before impact $u$: Using conservation of energy, $mgh = \frac{1}{2}mu^2 \Rightarrow u = \sqrt{2gh}$ [1].
  • Speed just after impact $v$: Using conservation of energy, $\frac{1}{2}mv^2 = mgh' \Rightarrow v = \sqrt{2gh'}$ [1].
  • Definition of $e$: $e = \frac{\text{speed of separation}}{\text{speed of approach}} = \frac{v}{u}$ (for impact with stationary ground) [1].
  • Substitute: $e = \frac{\sqrt{2gh'}}{\sqrt{2gh}} = \sqrt{\frac{h'}{h}}$ [1].

(b) Calculate $h'$. [2]

• Working:
  • $0.80 = \sqrt{\frac{h'}{2.0}}$
  • $0.80^2 = \frac{h'}{2.0}$
  • $0.64 = \frac{h'}{2.0} \Rightarrow h' = 1.28 \text{ m}$
• Answer: $1.3 \text{ m}$ (2 s.f.)

(c) Lost kinetic energy. [2]

• Answer: Converted into internal energy (heat) of the ball and surface [1], sound energy [1], and energy of deformation.

18. Conical Pendulum. (a) Free-body diagram. [2]

• Answer:
  1. Tension $T$ along the string, towards the pivot. [1]
  2. Weight $mg$ vertically downwards. [1]

(b) Derive Period $T$. [5]

• Working:
  • Resolve forces vertically: $T \cos \theta = mg$ --- (1) [1]
  • Resolve forces horizontally (provides centripetal force): $T \sin \theta = \frac{mv^2}{r}$ --- (2) [1]
  • Divide (2) by (1): $\tan \theta = \frac{v^2}{rg}$ [1]
  • Geometry: $r = L \sin \theta$. Also $v = \frac{2\pi r}{T_{period}}$.
  • Substitute $v$: $\tan \theta = \frac{(2\pi r / T_{period})^2}{rg} = \frac{4\pi^2 r}{T_{period}^2 g}$ [1]
  • Substitute $r = L \sin \theta$ and $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
  • $\frac{\sin \theta}{\cos \theta} = \frac{4\pi^2 L \sin \theta}{T_{period}^2 g}$
  • Cancel $\sin \theta$: $\frac{1}{\cos \theta} = \frac{4\pi^2 L}{T_{period}^2 g}$
  • $T_{period}^2 = \frac{4\pi^2 L \cos \theta}{g}$
  • $T_{period} = 2\pi \sqrt{\frac{L \cos \theta}{g}}$ [1]

19. Hump-backed Bridge. (a) Normal contact force < Weight. [3]

• Answer:
  • At the top of the bridge, the car undergoes circular motion, requiring a centripetal force directed downwards (towards the center of the circle) [1].
  • The resultant force is $W - N = \frac{mv^2}{R}$ [1].
  • Therefore, $N = W - \frac{mv^2}{R}$. Since $\frac{mv^2}{R} > 0$, $N < W$ [1].

(b) Maximum speed without losing contact. [3]

• Condition: Losing contact means $N = 0$.
• Working:
  • $0 = mg - \frac{mv^2}{R} \Rightarrow mg = \frac{mv^2}{R}$
  • $v^2 = gR$
  • $v = \sqrt{9.81 \times 40}$ [1]
  • $v = \sqrt{392.4} = 19.81 \text{ m s}^{-1}$
• Answer: $19.8 \text{ m s}^{-1}$

20. Conservation of Energy Experiment. (a) Expected relationship. [1]

• Answer: $v^2 = 2gh$ (so $v^2$ is directly proportional to $h$).

(b) Positive intercept on h-axis. [2]

• Answer:
  • This implies that a certain height $h$ is required before the trolley gains any measurable speed at the bottom, or more likely, there is a systematic error [1].
  • Reason: Work is done against friction/resistive forces. The trolley needs a minimum height to overcome static friction or the energy loss due to friction means $v^2$ is lower than expected for a given $h$. If the line intercepts the h-axis at $h > 0$ when $v=0$, it suggests that for small heights, the trolley does not reach the gate or friction prevents motion entirely until a threshold height is reached [1].
  • Alternative Acceptable Answer: The height $h$ was measured from the wrong reference point (e.g., top of trolley instead of center of mass, or not accounting for the length of the card interrupting the light gate).

(c) Determine $g$ from gradient. [2]

• Answer:
  • The equation is $v^2 = 2gh$. Comparing to $y = mx$, the gradient $m = 2g$ [1].
  • Therefore, $g = \frac{\text{gradient}}{2}$ [1].