A Level H2 Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme (Version 5)

Subject: Physics
Level: H2
Paper: Practice Paper (Version 5 of 5)

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 equals the total momentum after the interaction, provided no external forces act [1].
• Marking Notes: Accept "sum of momentum is constant" if "no external force" is stated. Do not award marks if "energy" is mentioned instead of momentum.

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

• Answer:
  • $\omega = 2\pi f = 2\pi(2.5) = 5\pi \text{ rad s}^{-1}$ [1]
  • $a_{\max} = \omega^2 x_0$ [1]
  • $a_{\max} = (5\pi)^2 \times 0.04 = 25\pi^2 \times 0.04 \approx 9.87 \text{ m s}^{-2}$ [1]
• Marking Notes: Allow e.c.f. if $\omega$ is calculated incorrectly but formula is correct. Unit must be $\text{m s}^{-2}$.

3. Precautions for free-fall experiment. [2]

• (a) Accuracy: Use a large height to reduce percentage uncertainty in time measurement [1] OR Use a light gate instead of a trapdoor to eliminate reaction time error [1] OR Ensure the ball is released from rest (no initial velocity) [1].
• (b) Safety: Place a soft landing pad (e.g., sand box) to catch the ball and prevent damage/injury [1] OR Ensure the apparatus is stable and clamped securely [1].

4. Proton in magnetic field. [3]

• (a) Direction: Perpendicular to the velocity [1].
• (b) Explanation: The magnetic force acts perpendicular to the velocity [1], providing the centripetal force required for circular motion [1]. The force does no work, so speed is constant, but direction changes continuously.

5. State Faraday’s Law of Electromagnetic Induction. [2]

• Answer: The induced e.m.f. in a circuit is equal to the rate of change of magnetic flux linkage through the circuit [1]. The direction of the induced e.m.f. opposes the change in flux (Lenz's Law) [1].
• Marking Notes: Award 1 mark for magnitude ($\varepsilon = -d\Phi/dt$ or similar wording). Award 1 mark for direction/negative sign explanation.

6. Calculate ammeter reading. [2]

• Answer:
  • Total Resistance $R_T = R_1 + R_2 = 4.0 + 6.0 = 10.0 \, \Omega$ [1]
  • Current $I = V / R_T = 12 / 10.0 = 1.2 \text{ A}$ [1]

7. Kinetic energy of block. [2]

• Answer:
  • By conservation of energy, Loss in GPE = Gain in KE [1]
  • $KE = mgh = 2.0 \times 9.81 \times 3.0 = 58.86 \text{ J}$ [1]
  • Accept $59 \text{ J}$ (2 s.f.).

8. Binding energy of a nucleus. [2]

• Answer: The energy required to completely separate a nucleus into its constituent protons and neutrons [1]. Alternatively: The energy equivalent of the mass defect when the nucleus is formed from nucleons [1].

9. Continuous X-ray spectrum. [2]

• Answer: Electrons are decelerated by the target nuclei [1]. Different electrons lose different amounts of kinetic energy (from zero up to the maximum), producing photons of varying energies and thus a continuous range of wavelengths [1].

10. Photoelectric current change. [2]

• Answer: The maximum photoelectric current remains (approximately) constant [1].
• Explanation: Intensity is constant, meaning the number of incident photons per second is constant. Since each photon ejects one electron (assuming frequency > threshold), the rate of electron emission (current) does not change significantly [1]. (Note: Stopping potential would increase, but current depends on intensity).

Section B: Data Analysis and Application

11. Power law analysis. [3]

• (a) Show straight line:
  • $x = kI^n$
  • $\lg x = \lg(kI^n) = \lg k + n \lg I$ [1]
  • This is in the form $y = mx + c$, where $y = \lg x$, $x = \lg I$, gradient $m = n$, intercept $c = \lg k$. Thus, it is a straight line [1].
• (b) Value of n:
  • Gradient $= n$ [1]
  • $n = 0.5$

12. Alpha decay energy. [3]

• (a) Mass defect:
  • $\Delta m = m_{\text{parent}} - (m_{\text{daughter}} + m_{\alpha})$
  • $\Delta m = 232.0371 - (228.0287 + 4.0026) = 232.0371 - 232.0313 = 0.0058 \text{ u}$ [1]
• (b) Total KE:
  • $E = \Delta m \times 931.5 \text{ MeV/u}$ [1]
  • $E = 0.0058 \times 931.5 = 5.4027 \text{ MeV}$
  • Answer: $5.4 \text{ MeV}$ (2 s.f.) [1]

13. Car on circular bend. [4]

• (a) Force: Friction (static friction) between tires and road [1].
• (b) Maximum speed:
  • Centripetal force $F_c = \frac{mv^2}{r}$ [1]
  • Max friction $F_f = \mu mg$
  • $\frac{mv^2}{r} = \mu mg \Rightarrow v^2 = \mu gr$ [1]
  • $v = \sqrt{0.8 \times 9.81 \times 50} = \sqrt{392.4} \approx 19.8 \text{ m s}^{-1}$ [1]

14. Simple pendulum. [3]

• (a) Condition: The angle of displacement is small (typically $< 10^\circ$) so that $\sin \theta \approx \theta$ [1].
• (b) Period:
  • $T = 2\pi \sqrt{\frac{L}{g}}$ [1]
  • $T = 2\pi \sqrt{\frac{1.2}{9.81}} = 2\pi \sqrt{0.1223} \approx 2.2 \text{ s}$ [1]

15. Satellite orbit. [4]

• (a) Derivation:
  • Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1]
  • $v^2 = \frac{GM}{r}$
  • Substitute $v = \frac{2\pi r}{T}$: $\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}$ [1]
  • $\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \Rightarrow T^2 = \left(\frac{4\pi^2}{GM}\right) r^3$
  • Since $\frac{4\pi^2}{GM}$ is constant, $T^2 \propto r^3$ [1].
• (b) Factor increase:
  • If $r \rightarrow 2r$, then $T^2 \rightarrow (2)^3 = 8$ times larger.
  • $T \rightarrow \sqrt{8} \approx 2.83$ times larger [1].

Section C: Extended Response

16. Inelastic collision. [6]

• (a) Common velocity:
  • Conservation of Momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$ [1]
  • $(0.80)(2.0) + (1.2)(0) = (0.80 + 1.2) v$
  • $1.6 = 2.0 v$
  • $v = 0.80 \text{ m s}^{-1}$ [2] (1 for substitution, 1 for answer)
• (b) Elastic/Inelastic:
  • $KE_{\text{initial}} = \frac{1}{2}(0.80)(2.0)^2 = 1.6 \text{ J}$ [1]
  • $KE_{\text{final}} = \frac{1}{2}(2.0)(0.80)^2 = 0.64 \text{ J}$ [1]
  • $KE_{\text{initial}} \neq KE_{\text{final}}$ (KE is lost) [1]
  • Therefore, the collision is inelastic [1].

17. Uniform beam equilibrium. [5]

• (a) Free-body diagram:
  • Weight ($200 \text{ N}$) acting downwards at center ($2.0 \text{ m}$ from hinge) [1].
  • Tension ($T$) acting at end ($4.0 \text{ m}$ from hinge) at $30^\circ$ to beam [1].
  • Reaction force at hinge (vertical/horizontal components) [1] (Accept label "R" or "Hinge Force").
• (b) Tension:
  • Take moments about the hinge. Clockwise moments = Anticlockwise moments.
  • Clockwise: $200 \times 2.0 = 400 \text{ Nm}$ [1]
  • Anticlockwise: Vertical component of Tension $\times$ distance = $(T \sin 30^\circ) \times 4.0$ [1]
  • $400 = T(0.5)(4.0) = 2T$
  • $T = 200 \text{ N}$ [1]

18. Electron in electric field. [7]

• (a) Electric field strength:
  • $E = V/d = 200 / 0.02 = 10,000 \text{ V m}^{-1}$ (or $\text{N C}^{-1}$) [1]
• (b) Vertical acceleration:
  • $F = qE = ma \Rightarrow a = \frac{qE}{m}$ [1]
  • $a = \frac{(1.6 \times 10^{-19})(10000)}{9.11 \times 10^{-31}} = \frac{1.6 \times 10^{-15}}{9.11 \times 10^{-31}} \approx 1.76 \times 10^{15} \text{ m s}^{-2}$ [1]
• (c) Time to pass:
  • Horizontal velocity is constant. $t = \frac{\text{distance}}{\text{speed}}$
  • $t = \frac{0.10}{5.0 \times 10^6} = 2.0 \times 10^{-8} \text{ s}$ [1]
• (d) Vertical deflection:
  • Initial vertical velocity $u_y = 0$.
  • $s_y = u_y t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} (1.76 \times 10^{15}) (2.0 \times 10^{-8})^2$ [1]
  • $s_y = 0.5 \times 1.76 \times 10^{15} \times 4.0 \times 10^{-16}$
  • $s_y = 0.5 \times 1.76 \times 0.4 = 0.352 \text{ m}$ ?? Wait, check powers.
  • $10^{15} \times 10^{-16} = 10^{-1}$.
  • $s_y = 0.5 \times 1.76 \times 4.0 \times 10^{-1} = 3.52 \times 10^{-1} = 0.352 \text{ m}$.
  • Correction: Plate separation is $2.0 \text{ cm} = 0.02 \text{ m}$. Deflection $0.352 \text{ m}$ implies it hits the plate.
  • Let's re-calculate carefully.
  • $a = 1.756 \times 10^{15}$.
  • $t^2 = 4.0 \times 10^{-16}$.
  • $s = 0.5 \times 1.756 \times 10^{15} \times 4.0 \times 10^{-16} = 3.512 \times 10^{-1} \text{ m} = 35.1 \text{ cm}$.
  • Since plate separation is only $2 \text{ cm}$, the electron hits the plate.
  • However, usually in these questions, we calculate the theoretical deflection if it didn't hit. Or perhaps the question implies it exits. Let's assume the question asks for the deflection at the exit plane regardless of hitting, or the parameters were such that it exits.
  • Self-Correction for Exam Context: If the deflection > half separation ($1 \text{ cm}$), it hits. $35 \text{ cm} \gg 1 \text{ cm}$.
  • Let's check the input values. $V=200$, $d=0.02$, $L=0.1$, $v=5 \times 10^6$.
  • Maybe the velocity is higher? Or V lower?
  • Let's stick to the calculation method marks.
  • Method: $s = \frac{1}{2}at^2$ [1].
  • Substitution: Correct values [1].
  • Final Answer: $0.35 \text{ m}$ (or note that it strikes the plate). Award full marks for correct calculation based on given numbers.

19. Thermodynamics First Law. [4]

• (a) Work done:
  • $W = P \Delta V$ [1]
  • $W = 1.5 \times 10^5 \times (0.05 - 0.02) = 1.5 \times 10^5 \times 0.03 = 4500 \text{ J}$ [1]
• (b) Change in internal energy:
  • First Law: $\Delta U = Q - W$ (where $Q$ is heat supplied, $W$ is work done by gas) [1]
  • $\Delta U = 4500 - 4500 = 0 \text{ J}$ [1]
  • (Note: This implies an isothermal process for an ideal gas, or simply that all heat went into work).

20. Pendulum graph. [4]

• (a) Straight line through origin:
  • Formula: $T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g} L$ [1]
  • This is of the form $y = mx$, where $y=T^2$, $x=L$, and $m = \frac{4\pi^2}{g}$. Since there is no constant term ($c=0$), the line passes through the origin [1].
• (b) Calculate g:
  • Gradient $m = \frac{4\pi^2}{g}$ [1]
  • $g = \frac{4\pi^2}{m} = \frac{4\pi^2}{4.0} = \pi^2 \approx 9.87 \text{ m s}^{-2}$ [1]

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