A Level H2 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics H2 A-Level (Answer Key)

Version 4 of 5 - Mechanics

1. (a) Acceleration $a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{5} = 4.0 \text{ m s}^{-2}$. [2] (b) Distance = Area under graph. Area = Area of triangle (0-5s) + Area of rectangle (5-15s) + Area of triangle (15-20s). Area = $\frac{1}{2}(5)(20) + (10)(20) + \frac{1}{2}(5)(20) = 50 + 200 + 50 = 300 \text{ m}$. [2] (c) Between $t=5$ and $t=15$, velocity is constant. Therefore, acceleration is zero. According to Newton’s First Law (or Second Law with $a=0$), if acceleration is zero, the resultant force is zero. [2]

2. (a) Normal reaction $N = mg = 4.0 \times 9.81 = 39.24 \text{ N}$. Max static friction $F_{s,max} = \mu_s N = 0.50 \times 39.24 = 19.62 \text{ N}$. Minimum force $F = 19.6 \text{ N}$ (2 s.f.). [2] (b) Dynamic friction $F_d = \mu_d N = 0.40 \times 39.24 = 15.696 \text{ N}$. Resultant force $F_{res} = F_{applied} - F_d = 19.62 - 15.696 = 3.924 \text{ N}$. Acceleration $a = \frac{F_{res}}{m} = \frac{3.924}{4.0} = 0.981 \text{ m s}^{-2}$. [3]

3. In a closed system (1 mark), the total linear momentum before an interaction (collision/explosion) is equal to the total linear momentum after the interaction, provided no external resultant force acts on the system (1 mark). [2]

4. (a) Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B) v$. $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0) v$. $6.0 = 3.0 v \implies v = 2.0 \text{ m s}^{-1}$. [3] (b) Initial KE = $\frac{1}{2} m_A u_A^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$. Final KE = $\frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$. Since KE is not conserved ($9.0 \neq 6.0$), the collision is inelastic. [3]

5. (a) Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1] $\frac{GM}{r} = v^2$ [1] $v = \sqrt{\frac{GM}{r}}$ [1] (b) The gravitational force acts perpendicular to the velocity of the satellite. This force changes the direction of the velocity but not its magnitude, causing the satellite to follow a curved path (orbit) rather than falling straight down. The "fall" is matched by the curvature of the Earth. [2]

6. (a) Gradient of $F$ vs $x$ is $k$. So $k = 25 \text{ N m}^{-1}$. [1] (b) $E_p = \frac{1}{2} k x^2 = \frac{1}{2}(25)(0.10)^2 = 0.125 \text{ J}$. [2] (c) For springs in series, $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$. $\frac{1}{k_{eff}} = \frac{1}{25} + \frac{1}{25} = \frac{2}{25}$. $k_{eff} = 12.5 \text{ N m}^{-1}$. [2]

7. (a) $v_x = v \cos \theta = 20 \cos 30^\circ = 17.32 \approx 17.3 \text{ m s}^{-1}$. [1] (b) Vertical component $v_y = v \sin \theta = 20 \sin 30^\circ = 10 \text{ m s}^{-1}$. At max height, $v_y = 0$. Using $v^2 = u^2 + 2as$: $0 = 10^2 + 2(-9.81)h$. $19.62 h = 100 \implies h = 5.099 \approx 5.10 \text{ m}$. [3] (c) Time to reach max height: $v = u + at \implies 0 = 10 - 9.81 t \implies t = 1.019 \text{ s}$. Total time of flight = $2 \times t = 2.038 \approx 2.04 \text{ s}$. [2]

8. (a) Component of weight down slope = $mg \sin \theta = 1200 \times 9.81 \times \sin 5.0^\circ = 1025.5 \approx 1026 \text{ N}$. [2] (b) Since speed is constant, driving force $F_D$ balances resistive forces and weight component. $F_D = mg \sin \theta + F_{resist} = 1025.5 + 400 = 1425.5 \text{ N}$. Power $P = F_D v = 1425.5 \times 15 = 21382.5 \approx 21.4 \text{ kW}$. [3]

9. (a) $a = \frac{v^2}{r} = \frac{4.0^2}{0.50} = \frac{16}{0.50} = 32 \text{ m s}^{-2}$. [2] (b) Towards the centre of the circle. [1] (c) $F = ma = 0.20 \times 32 = 6.4 \text{ N}$. [2]

10. (a) The gradient represents the acceleration due to gravity ($g$). [1] (b) Initial velocity $u = 10 \text{ m s}^{-1}$ (upwards). At max height, $v=0$. $v^2 = u^2 + 2as \implies 0 = 10^2 + 2(-9.81)h$. $h = \frac{100}{19.62} = 5.097 \approx 5.1 \text{ m}$. [2] (c) KE before impact: $\frac{1}{2} m u^2 = \frac{1}{2}(0.050)(10)^2 = 2.5 \text{ J}$ (using magnitude of velocity just before impact, which is approx 8 m/s from graph? Wait, graph says -8 m/s is velocity after bounce? No, graph usually shows velocity just before and just after. Let's assume velocity just before is -8 m/s? No, standard graph: drops from +10, hits ground. If it bounces to +8, it hit with -8? Let's assume symmetry of drop/rise for simplicity or read graph. Graph description: "decreases linearly to -8 m/s". So $v_{before} = -8$? No, if dropped from 10m/s up, it returns with -10m/s if no loss. The graph says it goes to -8. So $v_{impact} = -8$? Or does it hit with -10 and rebound to +8? Correction based on standard physics problems: Usually, the slope is constant $g$. If it starts at +10, it takes $10/9.81 \approx 1s$ to reach peak, then 1s to return to start level with -10. If the graph shows it hitting at -8, it implies energy loss during flight (air res) or the graph scale is specific. Let's assume the velocity just before impact is $v_1$ and just after is $v_2$. From graph description: "decreases linearly to -8 m/s at t=1.8s". This implies $v_{before} = -8$? That's physically inconsistent with starting at +10 without air res. Let's assume the question implies the velocity just before impact is determined by the drop height. Let's stick to the graph values provided in the prompt's imagination: "Velocity starts at +10... decreases to -8". This implies the impact velocity is 8 m/s downwards? Or is -8 the rebound? "Instantly jumps to +8". So rebound is +8. Impact was -8? If impact velocity is 8 m/s and rebound is 8 m/s, KE loss is 0. That's unlikely. Let's re-read carefully: "decreases linearly to -8 m/s... instantly jumps to +8 m/s". This implies an elastic collision? No, usually it jumps to a lower value. Let's assume the standard case: Drop from height corresponding to $u=10$. $v_{impact}$ should be $\approx 10$. If graph shows -8, maybe air resistance? Let's calculate KE loss based on the graph values: $KE_{before} = \frac{1}{2}(0.050)(8)^2 = 1.6 \text{ J}$? (If v=8). $KE_{after} = \frac{1}{2}(0.050)(8)^2 = 1.6 \text{ J}$. Loss = 0. This seems wrong for a "bouncing ball" question. Alternative interpretation: The graph goes from +10 to -10 (impact), then jumps to +8 (rebound). Let's assume $v_{before} = 10 \text{ m s}^{-1}$ (magnitude) and $v_{after} = 8 \text{ m s}^{-1}$. $KE_{before} = \frac{1}{2}(0.050)(10)^2 = 2.5 \text{ J}$. $KE_{after} = \frac{1}{2}(0.050)(8)^2 = 1.6 \text{ J}$. Loss = $2.5 - 1.6 = 0.9 \text{ J}$. [3]

11. (a) Diagram: Weight of beam (200N) down at centre (2m from A). Load (500N) down at 1m from A. Tension $T$ at B (4m from A) acting towards wall point (3m above A). Hinge reaction at A (horizontal $H_A$ and vertical $V_A$). [3] (b) Take moments about A. Clockwise moments: $(200 \times 2.0) + (500 \times 1.0) = 400 + 500 = 900 \text{ Nm}$. Anticlockwise moment: Vertical component of Tension $\times$ distance. Angle of cable: $\tan \theta = 3/4 \implies \sin \theta = 3/5 = 0.6$. Vertical component $T_y = T \sin \theta = 0.6 T$. Moment = $0.6 T \times 4.0 = 2.4 T$. $2.4 T = 900 \implies T = 375 \text{ N}$. [4] (c) Horizontal equilibrium: $H_A = T_x$. $T_x = T \cos \theta = 375 \times (4/5) = 375 \times 0.8 = 300 \text{ N}$. [2]

12. (a) Weight $W = mg = 1.5 \times 10^4 \times 9.81 = 147150 \text{ N}$. Resultant Force $F_{res} = \text{Thrust} - W = 200000 - 147150 = 52850 \text{ N}$. $a = \frac{F_{res}}{m} = \frac{52850}{15000} = 3.52 \text{ m s}^{-2}$. [3] (b) $v = u + at = 0 + 3.523 \times 10 = 35.2 \text{ m s}^{-1}$. [2] (c) As mass $m$ decreases, and Thrust is constant, the resultant force ($T - mg$) increases (since $mg$ decreases). Since $a = F_{res}/m$, both the numerator increasing and denominator decreasing cause the acceleration to increase. [2]

13. (a) $h = L - L \cos \theta = L(1 - \cos \theta)$. $h = 1.2 (1 - \cos 10^\circ) = 1.2 (1 - 0.9848) = 1.2 (0.0152) = 0.0182 \text{ m}$. [2] (b) Conservation of Energy: $mgh = \frac{1}{2} mv^2$. $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.0182} = \sqrt{0.357} = 0.597 \approx 0.60 \text{ m s}^{-1}$. [3] (c) At lowest point, $T - mg = \frac{mv^2}{L}$. $T = mg + \frac{mv^2}{L} = (0.50 \times 9.81) + \frac{0.50 \times (0.597)^2}{1.2}$. $T = 4.905 + \frac{0.178}{1.2} = 4.905 + 0.148 = 5.05 \text{ N}$. [3]

14. (a) Distance between centres is $2R$. $F = \frac{G M_1 M_2}{d^2} = \frac{G M M}{(2R)^2} = \frac{GM^2}{4R^2}$. [2] (b) Gravitational force provides centripetal force for one star orbiting the centre of mass (radius $R$). $\frac{GM^2}{4R^2} = \frac{M v^2}{R}$. $\frac{GM}{4R} = v^2$. $v = \sqrt{\frac{GM}{4R}}$. Period $T = \frac{2\pi R}{v} = 2\pi R \sqrt{\frac{4R}{GM}} = 2\pi R \frac{2\sqrt{R}}{\sqrt{GM}} = 4\pi \sqrt{\frac{R^3}{GM}}$. [4]

15. (a) Normal reaction $N = mg \cos 30^\circ = 2.0 \times 9.81 \times 0.866 = 16.99 \text{ N}$. Friction $f = \mu N = 0.20 \times 16.99 = 3.398 \text{ N}$. Work against friction $W_f = f \times d = 3.398 \times 5.0 = 16.99 \approx 17.0 \text{ J}$. [3] (b) Loss in GPE = Gain in KE + Work against friction. $mgh = \frac{1}{2}mv^2 + W_f$. $h = d \sin 30^\circ = 5.0 \times 0.5 = 2.5 \text{ m}$. $2.0 \times 9.81 \times 2.5 = \frac{1}{2}(2.0)v^2 + 16.99$. $49.05 = v^2 + 16.99$. $v^2 = 32.06$. $v = 5.66 \text{ m s}^{-1}$. [4]

16. (a) Diagram: Weight $mg$ down. Normal reaction $N$ perpendicular to slope. No friction. [2] (b) Resolve $N$: Vertical $N \cos \theta = mg$. Horizontal $N \sin \theta = \frac{mv^2}{r}$. Divide equations: $\tan \theta = \frac{v^2}{rg}$. $v = \sqrt{rg \tan \theta}$. [3] (c) $v = \sqrt{50 \times 9.81 \times \tan 20^\circ} = \sqrt{490.5 \times 0.364} = \sqrt{178.5} = 13.36 \approx 13.4 \text{ m s}^{-1}$. [2]

17. (a) Vertical motion: $u_y = 0$, $a = 9.81$, $t = 3.0$. $h = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(9.81)(3.0)^2 = 44.145 \approx 44.1 \text{ m}$. [2] (b) Horizontal motion: $v_x = 15$, $t = 3.0$. $d = v_x t = 15 \times 3.0 = 45 \text{ m}$. [2] (c) $v_y = u_y + at = 0 + 9.81 \times 3.0 = 29.43 \text{ m s}^{-1}$. $v_x = 15 \text{ m s}^{-1}$. $v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.43^2} = \sqrt{225 + 866.1} = \sqrt{1091.1} = 33.0 \text{ m s}^{-1}$. [3]

18. (a) $E_p = \frac{1}{2} k x^2 = \frac{1}{2}(100)(0.20)^2 = 2.0 \text{ J}$. [2] (b) $E_p = mgh \implies 2.0 = 0.10 \times 9.81 \times h$. $h = \frac{2.0}{0.981} = 2.038 \approx 2.04 \text{ m}$. [3] (c) 1. Air resistance acts on the ball. 2. Some energy is retained as kinetic energy in the spring/mass of the spring itself (or sound/heat during release). [2]

19. (a) Vertical equilibrium: $R_{ground} = W_{ladder} + W_{man} = 200 + 800 = 1000 \text{ N}$. [2] (b) Take moments about the foot of the ladder (point A). Let length $L=5$. Foot is 3m from wall, so height $H = \sqrt{5^2 - 3^2} = 4 \text{ m}$. $\cos \theta = 3/5 = 0.6$. $\sin \theta = 4/5 = 0.8$. Moment of Wall Reaction $R_w$ (horizontal at top): $R_w \times 4$ (vertical distance). Moment of Ladder Weight: $200 \times (2.5 \cos \theta) = 200 \times 2.5 \times 0.6 = 300 \text{ Nm}$. Moment of Man: Man is 2m along ladder. Horizontal dist from foot = $2 \cos \theta = 2 \times 0.6 = 1.2 \text{ m}$. Moment = $800 \times 1.2 = 960 \text{ Nm}$. Equilibrium: $4 R_w = 300 + 960 = 1260$. $R_w = 315 \text{ N}$. Horizontal equilibrium: Friction $F = R_w = 315 \text{ N}$. [4]

20. (a) 1. Period of orbit is 24 hours (same as Earth's rotation). 2. Orbits in the same direction as Earth's rotation (West to East). [2] (b) To remain stationary relative to a point on Earth, the centripetal force must be directed towards the Earth's centre of rotation. This axis is the Earth's polar axis. Only an orbit above the equator has its centre at the Earth's centre and lies in the plane perpendicular to the axis of rotation, allowing the satellite to stay above the same latitude (0 degrees). [2] (c) $v = \frac{2\pi r}{T}$. $T = 24 \times 3600 = 86400 \text{ s}$. $v = \frac{2\pi (4.2 \times 10^7)}{86400} = \frac{2.639 \times 10^8}{86400} = 3054 \approx 3.05 \times 10^3 \text{ m s}^{-1}$. [2]