A Level H2 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics H2 A-Level

Marking Scheme and Answers (Version 2)

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 event (collision/explosion) is equal to the total momentum after the event, provided no external resultant force acts on the system [1].
• Notes: Must mention "closed/isolated system" or "no external forces". "Momentum is conserved" alone is insufficient for full marks if the condition is omitted.

2. Power developed by car. [2]

• Answer:
  • Since speed is constant, driving force $F = \text{resistive force} = 800 \text{ N}$ [1].
  • Power $P = Fv = 800 \times 25 = 20,000 \text{ W}$ or $20 \text{ kW}$ [1].
• Notes: Accept $20 \text{ kW}$.

3. Define impulse. [1]

• Answer: Impulse is the product of the average force and the time interval during which the force acts ($I = F \Delta t$) [1].
• Alternative: Change in momentum.

4. Satellite acceleration. [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 changes, which means there is acceleration.

5. Free-body diagram on inclined plane. [2]

• Answer:
  • Weight ($mg$) acting vertically downwards [1].
  • Normal contact force ($N$) acting perpendicular to the plane [0.5].
  • Friction ($f$) acting up the slope (parallel to plane) [0.5].
• Notes: All three forces must be present and correctly directed for full marks. Labels must be clear.

Section B: Application and Analysis

6. Vertical projectile. (a) Max height. [2]

• Answer:
  • Using $v^2 = u^2 + 2as$: $0 = 15^2 + 2(-9.81)h$ [1].
  • $h = \frac{225}{19.62} \approx 11.5 \text{ m}$ [1].
• Notes: Accept $11.5 \text{ m}$ or $11.47 \text{ m}$.

(b) Time to return. [2]

• Answer:
  • Using $s = ut + \frac{1}{2}at^2$: $0 = 15t - 4.905t^2$ [1].
  • $t(15 - 4.905t) = 0 \Rightarrow t = \frac{15}{4.905} \approx 3.06 \text{ s}$ [1].
• Alternative: Time up = $v/a = 15/9.81 = 1.53 \text{ s}$. Total time = $2 \times 1.53 = 3.06 \text{ s}$.

7. Inelastic Collision. (a) Common velocity. [3]

• Answer:
  • Conservation of Momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$ [1].
  • $2.0(4.0) + 3.0(0) = (2.0 + 3.0)v$ [1].
  • $8.0 = 5.0v \Rightarrow v = 1.6 \text{ m s}^{-1}$ [1].

(b) Elastic or Inelastic? [3]

• Answer:
  • Calculate KE before: $KE_i = \frac{1}{2}(2.0)(4.0)^2 = 16 \text{ J}$ [1].
  • Calculate KE after: $KE_f = \frac{1}{2}(5.0)(1.6)^2 = 6.4 \text{ J}$ [1].
  • Since $KE_i \neq KE_f$ (KE is lost), the collision is inelastic [1].

8. Conical Pendulum. (a) Diagram. [2]

• Answer: Tension ($T$) along the string towards pivot [1]. Weight ($mg$) vertically downwards [1]. Resultant force (centripetal) horizontal towards center (optional but good practice).

(b) Tension. [2]

• Answer:
  • Resolve vertically: $T \cos(30^\circ) = mg$ [1].
  • $T = \frac{0.50 \times 9.81}{\cos(30^\circ)} = \frac{4.905}{0.866} \approx 5.66 \text{ N}$ [1].

(c) Speed. [3]

• Answer:
  • Resolve horizontally: $T \sin(30^\circ) = \frac{mv^2}{r}$ [1].
  • Radius $r = L \sin(30^\circ) = 1.2 \times 0.5 = 0.6 \text{ m}$ [1].
  • $5.66 \times 0.5 = \frac{0.50 \times v^2}{0.6} \Rightarrow 2.83 = \frac{0.5 v^2}{0.6}$.
  • $v^2 = \frac{2.83 \times 0.6}{0.5} = 3.396$.
  • $v \approx 1.84 \text{ m s}^{-1}$ [1].
• Alternative: $v = \sqrt{rg \tan \theta} = \sqrt{0.6 \times 9.81 \times \tan 30^\circ} \approx 1.84 \text{ m s}^{-1}$.

9. Cyclist on Hill. (a) Weight component. [2]

• Answer:
  • Component down slope $= mg \sin \theta$ [1].
  • $= 80 \times 9.81 \times \sin(5.0^\circ) \approx 68.4 \text{ N}$ [1].

(b) Power output. [3]

• Answer:
  • Since speed is constant, Driving Force $F_D = \text{Resistive Forces} + \text{Weight Component}$ [1].
  • $F_D = 40 + 68.4 = 108.4 \text{ N}$ [1].
  • Power $P = F_D v = 108.4 \times 6.0 \approx 650 \text{ W}$ (or $650.4 \text{ W}$) [1].

10. Spring. (a) Spring constant. [2]

• Answer:
  • $F = kx \Rightarrow k = F/x$ [1].
  • $k = 10 / 0.04 = 250 \text{ N m}^{-1}$ [1].

(b) Elastic Potential Energy. [2]

• Answer:
  • $EPE = \frac{1}{2}kx^2$ or $\frac{1}{2}Fx$ [1].
  • $EPE = 0.5 \times 250 \times (0.04)^2 = 0.2 \text{ J}$ [1].

Section C: Synthesis and Evaluation

11. Projectile from Cliff. (a) Components. [2]

• Answer:
  • $u_x = 20 \cos(30^\circ) = 17.32 \text{ m s}^{-1}$ [1].
  • $u_y = 20 \sin(30^\circ) = 10.0 \text{ m s}^{-1}$ [1].

(b) Time to hit ground. [3]

• Answer:
  • Take upward as positive. $s_y = -45 \text{ m}$, $u_y = 10 \text{ m s}^{-1}$, $a = -9.81 \text{ m s}^{-2}$.
  • $s = ut + \frac{1}{2}at^2 \Rightarrow -45 = 10t - 4.905t^2$ [1].
  • $4.905t^2 - 10t - 45 = 0$.
  • Using quadratic formula: $t = \frac{10 \pm \sqrt{100 - 4(4.905)(-45)}}{9.81}$ [1].
  • $t = \frac{10 \pm \sqrt{100 + 882.9}}{9.81} = \frac{10 \pm 31.35}{9.81}$.
  • Positive root: $t \approx 4.22 \text{ s}$ [1].

(c) Horizontal distance. [2]

• Answer:
  • $x = u_x t$ [1].
  • $x = 17.32 \times 4.22 \approx 73.1 \text{ m}$ [1].

12. Circular Motion on Flat/Banked Road. (a) Centripetal Force. [1]

• Answer: Friction between tyres and road [1].

(b) Max Speed. [3]

• Answer:
  • Max friction $F_{max} = \mu N = \mu mg$ [1].
  • $F_c = \frac{mv^2}{r} \Rightarrow \mu mg = \frac{mv^2}{r}$ [1].
  • $v = \sqrt{\mu gr} = \sqrt{0.80 \times 9.81 \times 50} = \sqrt{392.4} \approx 19.8 \text{ m s}^{-1}$ [1].

(c) Banking Explanation. [2]

• Answer:
  • The normal contact force from the road has a horizontal component ($N \sin \theta$) [1].
  • This component contributes to (or provides) the centripetal force, reducing the reliance on friction [1].

13. Elastic Collision of Pendulums. (a) Conditions. [2]

• Answer:
  • Total momentum is conserved [1].
  • Total kinetic energy is conserved [1].

(b) Equal Masses. [2]

• Answer:
  • Sphere X stops (velocity becomes zero) [1].
  • Sphere Y moves off with the initial velocity of X [1].

(c) Mass X > Mass Y. [2]

• Answer:
  • Sphere X continues to move forward (slower than before) [1].
  • Sphere Y moves forward with a higher velocity [1].

14. Rocket Propulsion. (a) Explanation. [3]

• Answer:
  • The rocket exerts a force on the gas to eject it downwards/backwards [1].
  • By Newton’s Third Law, the gas exerts an equal and opposite force on the rocket (upwards/forwards) [1].
  • This reaction force (thrust) causes the rocket to accelerate (Newton’s Second Law) [1].

(b) Thrust Expression. [2]

• Answer:
  • Force = Rate of change of momentum [1].
  • $F = \frac{d(mv)}{dt} = v_e \frac{dm}{dt}$ [1].

15. Force-Acceleration Graph. (a) Frictional Force. [1]

• Answer: $5 \text{ N}$ [1]. (This is the maximum static friction/threshold force).

(b) Mass from Gradient. [2]

• Answer:
  • Equation: $F - f = ma \Rightarrow a = \frac{1}{m}F - \frac{f}{m}$. Gradient is $1/m$ [1].
  • Using points $(5, 0)$ and $(15, 2)$: Gradient $= \frac{2-0}{15-5} = \frac{2}{10} = 0.2$ [1].
  • $m = 1/0.2 = 5 \text{ kg}$.
  • Note: The question asks to verify against given mass $2.0 \text{ kg}$. The calculated mass ($5 \text{ kg}$) does NOT match the given mass ($2.0 \text{ kg}$). This implies the graph data provided in the prompt description was hypothetical or there is a discrepancy. However, based strictly on the graph coordinates given in the prompt text:
  • Correction for Student Logic: If the mass is truly $2.0 \text{ kg}$, the gradient should be $0.5$. If the graph shows gradient $0.2$, the mass is $5 \text{ kg}$.
  • Marking: Award marks for correct calculation of gradient and inversion. If student notes the discrepancy, award full marks.

(c) Zero Acceleration. [2]

• Answer:
  • The applied force is less than or equal to the maximum static friction [1].
  • The net force is zero, so there is no acceleration [1].

16. Kinematics Calculus. (a) Velocity Expression. [1]

• Answer: $v = \frac{ds}{dt} = 6t - 6t^2$ [1].

(b) Time at Rest. [2]

• Answer:
  • $v = 0 \Rightarrow 6t - 6t^2 = 0$ [1].
  • $6t(1-t) = 0$. $t=0$ (start) or $t=1.0 \text{ s}$ [1].

(c) Acceleration at $t=0.5$. [2]

• Answer:
  • $a = \frac{dv}{dt} = 6 - 12t$ [1].
  • At $t=0.5$, $a = 6 - 12(0.5) = 0 \text{ m s}^{-2}$ [1].

17. Hinged Beam. (a) Tension. [4]

• Answer:
  • Take moments about the hinge [1].
  • Clockwise Moment (Weight): $200 \text{ N} \times 2.0 \text{ m}$ (acting at center) $= 400 \text{ Nm}$ [1].
  • Geometry: String length forms a 3-4-5 triangle? No, wall height 3m, beam 4m. Hypotenuse (string) $= \sqrt{3^2+4^2} = 5 \text{ m}$.
  • Vertical component of Tension $T_y = T \sin \theta$. $\sin \theta = \frac{3}{5} = 0.6$.
  • Anticlockwise Moment: $(T \sin \theta) \times 4.0 = T(0.6)(4.0) = 2.4 T$ [1].
  • Equilibrium: $2.4 T = 400 \Rightarrow T = \frac{400}{2.4} \approx 167 \text{ N}$ [1].

(b) Horizontal Hinge Force. [2]

• Answer:
  • Horizontal forces must balance. $H_x = T_x$ [1].
  • $T_x = T \cos \theta = 166.7 \times \frac{4}{5} = 166.7 \times 0.8 \approx 133 \text{ N}$ [1].

18. Orbital Mechanics. (a) Derive Speed. [3]

• Answer:
  • Gravitational force provides centripetal force [1].
  • $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1].
  • Cancel $m$ and one $r$: $\frac{GM}{r} = v^2 \Rightarrow v = \sqrt{\frac{GM}{r}}$ [1].

(b) Kepler’s Third Law. [2]

• Answer:
  • $v = \frac{2\pi r}{T}$ [1].
  • Substitute into (a): $\frac{2\pi r}{T} = \sqrt{\frac{GM}{r}}$. Square both sides: $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$.
  • $T^2 = \left( \frac{4\pi^2}{GM} \right) r^3$. Thus $T^2 \propto r^3$ [1].

19. Impulse and Rebound. (a) Speed before impact. [2]

• Answer:
  • $v^2 = u^2 + 2gh \Rightarrow v = \sqrt{2 \times 9.81 \times 2.0}$ [1].
  • $v \approx 6.26 \text{ m s}^{-1}$ (downwards) [1].

(b) Speed after impact. [2]

• Answer:
  • $v^2 = u^2 + 2gh$ (for upward motion to 1.5m, final v=0).
  • $u = \sqrt{2 \times 9.81 \times 1.5}$ [1].
  • $u \approx 5.42 \text{ m s}^{-1}$ (upwards) [1].

(c) Impulse. [3]

• Answer:
  • Impulse $= \Delta p = m(v_{final} - v_{initial})$ [1].
  • Take upward as positive: $v_{final} = +5.42$, $v_{initial} = -6.26$.
  • $I = 0.20 (5.42 - (-6.26)) = 0.20 (11.68)$ [1].
  • $I \approx 2.34 \text{ N s}$ [1].

20. Free Fall Experiment. (a) Equation. [1]

• Answer: $h = v_1 t + \frac{1}{2}gt^2$ [1].

(b) Graph Method. [2]

• Answer:
  • Rearrange to $\frac{h}{t} = v_1 + \frac{1}{2}gt$ [1].
  • Plot $\frac{h}{t}$ on y-axis against $t$ on x-axis. The gradient will be $\frac{1}{2}g$ [1].

(c) Systematic Error. [2]

• Answer:
  • Error: Air resistance acts on the ball, causing the measured acceleration to be less than $g$ [1].
  • Reduction: Use a heavier, denser sphere (like steel) to minimize the effect of air resistance relative to weight [1].
  • Alternative: Delay in electronic timer reaction. Use light gates with high sampling rate.