A Level H2 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 4
Total Marks: 80

Section A: Structured Questions

Question 1: Kinematics and Projectile Motion [8 marks]

(a) [2 marks]

• Horizontal component: $v_x = 25.0 \cos 40.0^\circ = 19.2 \text{ m s}^{-1}$ [1]
• Vertical component: $v_y = 25.0 \sin 40.0^\circ = 16.1 \text{ m s}^{-1}$ [1]
• Award [1] for each correct calculation with correct units.

(b) [2 marks]

• At maximum height, $v_y = 0$ [1]
• Using $v = u + at$: $0 = 16.1 - 9.81t$
• $t = \frac{16.1}{9.81} = 1.64 \text{ s}$ [1]
• Award [1] for method, [1] for correct answer with units.

(c) [2 marks]

• Using $v^2 = u^2 + 2as$: $0 = (16.1)^2 + 2(-9.81)h$ [1]
• $h = \frac{(16.1)^2}{2 \times 9.81} = 13.2 \text{ m}$ [1]
• Award [1] for correct equation, [1] for correct answer with units.

(d) [2 marks]

• Total time of flight: $t_{\text{total}} = 2 \times 1.64 = 3.28 \text{ s}$ [1]
• Horizontal distance: $x = v_x \times t_{\text{total}} = 19.2 \times 3.28 = 63.0 \text{ m}$ [1]
• Award [1] for total time, [1] for correct distance with units.

Question 2: Forces and Equilibrium [8 marks]

(a) [2 marks]

• Diagram showing weight $W = mg = 12.0 \times 9.81 = 117.7 \text{ N}$ acting downwards [1]
• Tension forces $T_1$ and $T_2$ acting at $30.0^\circ$ and $45.0^\circ$ above horizontal respectively [1]
• All forces clearly labelled with directions.

(b) [2 marks]

• Vertical equilibrium: $T_1 \sin 30.0^\circ + T_2 \sin 45.0^\circ = 117.7$ [1]
• Horizontal equilibrium: $T_1 \cos 30.0^\circ = T_2 \cos 45.0^\circ$ [1]
• Award [1] for each correct equation.

(c) [3 marks]

• From horizontal: $T_1 \cos 30.0^\circ = T_2 \cos 45.0^\circ$ → $T_2 = T_1 \frac{\cos 30.0^\circ}{\cos 45.0^\circ} = 1.225 T_1$ [1]
• Substitute into vertical: $T_1 \sin 30.0^\circ + 1.225 T_1 \sin 45.0^\circ = 117.7$ [1]
• $T_1(0.500 + 0.866) = 117.7$ → $T_1 = \frac{117.7}{1.366} = 86.2 \text{ N}$ [1]
• Award [1] for substitution, [1] for correct algebra, [1] for correct answer with units.

(d) [1 mark]

• $T_2 = 1.225 \times 86.2 = 106 \text{ N}$ [1]
• Accept $105.6 \text{ N}$ or $106 \text{ N}$.

Question 3: Work, Energy, and Power [10 marks]

(a) [2 marks]

• Horizontal component of force: $F_x = 30.0 \cos 25.0^\circ = 27.2 \text{ N}$ [1]
• Work done: $W = F_x \times d = 27.2 \times 4.00 = 109 \text{ J}$ [1]
• Award [1] for horizontal component, [1] for correct work with units.

(b) [3 marks]

• Vertical forces: $N + F \sin 25.0^\circ - mg = 0$ [1]
• $N = mg - F \sin 25.0^\circ$ [1]
• $N = (5.00 \times 9.81) - (30.0 \times \sin 25.0^\circ) = 49.05 - 12.68 = 36.4 \text{ N}$ [1]
• Award [1] for equilibrium statement, [1] for correct expression, [1] for correct answer with units.

(c) [2 marks]

• Frictional force: $f = \mu_k N = 0.200 \times 36.4 = 7.28 \text{ N}$ [1]
• Work against friction: $W_f = f \times d = 7.28 \times 4.00 = 29.1 \text{ J}$ [1]
• Award [1] for friction calculation, [1] for correct work with units.

(d) [3 marks]

• Net work: $W_{\text{net}} = 109 - 29.1 = 79.9 \text{ J}$ [1]
• Work-energy theorem: $W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - 0$ [1]
• $v = \sqrt{\frac{2 \times 79.9}{5.00}} = \sqrt{31.96} = 5.65 \text{ m s}^{-1}$ [1]
• Award [1] for net work, [1] for correct application of theorem, [1] for correct answer with units.

Question 4: Circular Motion [6 marks]

(a) [1 mark]

• The frictional force between the tyres and the road provides the centripetal force. [1]
• Accept: static friction.

(b) [3 marks]

• Centripetal force: $f = \frac{mv^2}{r}$ [1]
• Maximum friction: $f_{\max} = \mu_s N = \mu_s mg$ [1]
• Equating: $\mu_s mg = \frac{mv^2}{r}$ → $v = \sqrt{\mu_s g r} = \sqrt{0.600 \times 9.81 \times 50.0} = 17.2 \text{ m s}^{-1}$ [1]
• Award [1] for each step; deduct [1] if mass not cancelled correctly.

(c) [2 marks]

• On a wet road, the coefficient of friction is reduced. [1]
• Since $v_{\max} = \sqrt{\mu g r}$, a smaller $\mu$ results in a lower maximum safe speed. [1]
• Award [1] for identifying reduced friction, [1] for linking to equation.

Question 5: Momentum and Collisions [8 marks]

(a) [1 mark]

• The total momentum of a closed system remains constant provided no external forces act. [1]
• Accept: In the absence of external forces, total momentum before collision equals total momentum after collision.

(b) [3 marks]

• Conservation of momentum: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$ [1]
• $(2.00 \times 4.00) + (3.00 \times 0) = (2.00 \times 0.800) + (3.00 \times v_B)$ [1]
• $8.00 = 1.60 + 3.00 v_B$ → $v_B = \frac{6.40}{3.00} = 2.13 \text{ m s}^{-1}$ [1]
• Award [1] for equation, [1] for substitution, [1] for correct answer with units.

(c) [3 marks]

• Initial KE: $KE_i = \frac{1}{2} \times 2.00 \times (4.00)^2 = 16.0 \text{ J}$ [1]
• Final KE: $KE_f = \frac{1}{2} \times 2.00 \times (0.800)^2 + \frac{1}{2} \times 3.00 \times (2.13)^2 = 0.640 + 6.81 = 7.45 \text{ J}$ [1]
• Since $KE_f < KE_i$, the collision is inelastic. [1]
• Award [1] for each KE calculation, [1] for correct conclusion with justification.

(d) [1 mark]

• The track is frictionless / no external forces act on the system. [1]
• Accept any valid assumption.

Section B: Long Structured Questions

Question 6: Gravitational Fields and Satellite Motion [13 marks]

(a) [2 marks]

• A geostationary orbit is one in which the satellite remains above a fixed point on the Earth's equator. [1]
• Condition: The orbital period must be 24 hours (or equal to Earth's rotational period) / The orbit must be equatorial. [1]
• Award [1] for definition, [1] for one correct condition.

(b) [1 mark]

• Orbital radius: $r = R_E + h = 6.37 \times 10^6 + 3.58 \times 10^7 = 4.22 \times 10^7 \text{ m}$ [1]
• Award [1] for correct calculation.

(c) [2 marks]

• $F = \frac{GM_E m}{r^2}$ [1]
• $F = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(850)}{(4.22 \times 10^7)^2} = 190 \text{ N}$ [1]
• Award [1] for formula, [1] for correct answer with units.

(d) [3 marks]

• Gravitational force provides centripetal force: $\frac{GM_E m}{r^2} = \frac{mv^2}{r}$ [1]
• $v = \sqrt{\frac{GM_E}{r}}$ [1]
• $v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{4.22 \times 10^7}} = 3.07 \times 10^3 \text{ m s}^{-1}$ [1]
• Award [1] for equating forces, [1] for derived expression, [1] for correct answer with units.

(e) [3 marks]

• Period: $T = \frac{2\pi r}{v} = \frac{2\pi \times 4.22 \times 10^7}{3.07 \times 10^3}$ [1]
• $T = 8.64 \times 10^4 \text{ s} = 24.0 \text{ hours}$ [1]
• The period is 24 hours, so the satellite is in a geostationary orbit (provided it is also in an equatorial orbit). [1]
• Award [1] for formula, [1] for correct period, [1] for correct comment.

(f) [2 marks]

• In orbit, the gravitational force provides the necessary centripetal force for circular motion. [1]
• There is no air resistance or other dissipative forces in space, so no energy is lost and the orbit is maintained without fuel. [1]
• Award [1] for force balance explanation, [1] for absence of dissipative forces.

Question 7: Simple Harmonic Motion [15 marks]

(a) [2 marks]

• Restoring force: $F = -kx$ [1]
• Since $F = ma$, we have $ma = -kx$ → $a = -\frac{k}{m}x$, which is of the form $a = -\omega^2 x$, indicating SHM. [1]
• Award [1] for stating Hooke's law, [1] for showing acceleration proportional to negative displacement.

(b) [2 marks]

• $\omega = \sqrt{\frac{k}{m}}$ [1]
• $\omega = \sqrt{\frac{40.0}{0.250}} = 12.6 \text{ rad s}^{-1}$ [1]
• Award [1] for formula, [1] for correct answer with units.

(c) [2 marks]

• $v_{\max} = \omega x_0$ [1]
• $v_{\max} = 12.6 \times 0.0600 = 0.756 \text{ m s}^{-1}$ [1]
• Award [1] for formula, [1] for correct answer with units.

(d) [2 marks]

• $a_{\max} = \omega^2 x_0$ [1]
• $a_{\max} = (12.6)^2 \times 0.0600 = 9.53 \text{ m s}^{-2}$ [1]
• Award [1] for formula, [1] for correct answer with units.

(e) [2 marks]

• General form: $x = x_0 \cos(\omega t)$ (since $x = x_0$ at $t = 0$) [1]
• $x = 0.0600 \cos(12.6 t)$ where $x$ is in metres and $t$ in seconds [1]
• Award [1] for correct form, [1] for correct substitution of values.

(f) [3 marks]

• Graph of KE vs. displacement should be parabolic, with maximum at $x = 0$ and zero at $x = \pm x_0$. [1]
• Maximum KE: $KE_{\max} = \frac{1}{2}mv_{\max}^2 = \frac{1}{2} \times 0.250 \times (0.756)^2 = 0.0714 \text{ J}$ [1]
• Axes labelled: $x$-axis from $-0.0600 \text{ m}$ to $+0.0600 \text{ m}$; $y$-axis from $0$ to $0.0714 \text{ J}$ (or appropriate scale). [1]
• Award [1] for correct shape, [1] for correct maximum value, [1] for correct axis labels.

(g) [2 marks]

• Precaution: Ensure the spring oscillates vertically (or horizontally without friction) to minimise energy loss / Use small amplitude oscillations to ensure the spring obeys Hooke's law. [1]
• Explanation: This ensures the motion approximates ideal SHM, making the period independent of amplitude and consistent with the theoretical formula. [1]
• Award [1] for a valid precaution, [1] for a clear explanation linking to improved accuracy.

Question 8: Dynamics and Connected Bodies [12 marks]

(a) [3 marks]

• Block P: Weight $m_P g$ downwards, normal reaction $N$ upwards, tension $T$ to the right, friction $f$ to the left. [1.5]
• Block Q: Weight $m_Q g$ downwards, tension $T$ upwards. [1.5]
• Award [1.5] for each correct free-body diagram with all forces labelled.

(b) [2 marks]

• For block P (horizontal): $T - f = m_P a$ [1]
• Where $f = \mu_k N = \mu_k m_P g$ [1]
• Award [1] for equation of motion, [1] for friction expression.

(c) [2 marks]

• For block Q (vertical): $m_Q g - T = m_Q a$ [1]
• Award [1] for correct equation, [1] for correct sign convention (consistent with part (b)).

(d) [4 marks]

• From (b): $T - \mu_k m_P g = m_P a$ → $T - (0.250 \times 3.00 \times 9.81) = 3.00a$ → $T - 7.36 = 3.00a$ [1]
• From (c): $m_Q g - T = m_Q a$ → $(2.00 \times 9.81) - T = 2.00a$ → $19.62 - T = 2.00a$ [1]
• Adding equations: $19.62 - 7.36 = 5.00a$ → $a = \frac{12.26}{5.00} = 2.45 \text{ m s}^{-2}$ [1]
• Tension: $T = 19.62 - 2.00(2.45) = 19.62 - 4.90 = 14.7 \text{ N}$ [1]
• Award [1] for each equation, [1] for correct acceleration, [1] for correct tension.

(e) [1 mark]

• Assumption: The string is inextensible and massless / The pulley is frictionless. [1]
• Explanation: In reality, a string with mass or a pulley with friction would affect the tension and acceleration, reducing accuracy. [1]
• Award [1] for a valid assumption with brief explanation.

END OF ANSWER KEY

Marking notes: Award marks for correct method even if final answer has minor arithmetic errors. Deduct marks for missing units only once per question. Accept equivalent correct expressions and alternative valid approaches.