A Level H2 Physics Practice Paper 3

TuitionGoWhere Practice Paper - Physics H2 A-Level (Answers)

Version 3 of 5 - Mechanics

1. (a) The driving force equals the resistive force (net force is zero). [1] (b) At terminal speed, $F_D = F_R$. $2500 = k(40)$ $k = \frac{2500}{40} = 62.5 \text{ N s m}^{-1}$ (or $\text{kg s}^{-1}$) [2] (c) At rest, $v=0 \implies F_R = 0$. $F_{net} = F_D = 2500 \text{ N}$ $a = \frac{F}{m} = \frac{2500}{1200} = 2.08 \text{ m s}^{-2}$ [2]

2. (a) $v^2 = u^2 + 2as$. At max height, $v=0$. $0 = 15^2 + 2(-9.81)s$ $s = \frac{225}{19.62} = 11.5 \text{ m}$ [2] (b) $s = ut + \frac{1}{2}at^2$. Displacement $s=0$ for return. $0 = 15t - 4.905t^2$ $t(15 - 4.905t) = 0$ $t = 0$ (start) or $t = \frac{15}{4.905} = 3.06 \text{ s}$ [2] (c) Graph: Straight line with negative gradient. Y-intercept at $+15 \text{ m s}^{-1}$. X-intercept at $1.53 \text{ s}$ (time to max height). Ends at $t=3.06 \text{ s}$ with $v = -15 \text{ m s}^{-1}$. [2]

3. (a) In a closed system, the total momentum before collision equals the total momentum after collision, provided no external forces act. [1] (b) $m_A u_A + m_B u_B = (m_A + m_B)v$ $(2.0)(3.0) + 0 = (2.0 + 1.0)v$ $6.0 = 3.0v \implies v = 2.0 \text{ m s}^{-1}$ [2] (c) $KE_{initial} = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ $KE_{final} = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$ Since $KE_{initial} \neq KE_{final}$ (KE is lost), the collision is inelastic. [3]

4. (a) The only force acting on the satellite is gravity. It is constantly accelerating towards the Earth's center. Its tangential velocity ensures it misses the Earth, maintaining orbit. Thus, it is in free fall. [2] (b) $r = R + h = 6.37 \times 10^6 + 300 \times 10^3 = 6.67 \times 10^6 \text{ m}$ $F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)}{(6.67 \times 10^6)^2}$ $F = \frac{1.99 \times 10^{17}}{4.45 \times 10^{13}} = 4470 \text{ N}$ (approx) [3]

5. (a) Forces: Weight ($mg$) vertically down, Normal Reaction ($N$) perpendicular to plane, Friction ($f$) up the plane. [2] (b) Component of weight down slope: $W_{\parallel} = mg \sin 30^\circ = 5.0(9.81)(0.5) = 24.5 \text{ N}$ Max static friction: $f_{max} = \mu N = \mu mg \cos 30^\circ$ $f_{max} = 0.40(5.0)(9.81)(0.866) = 17.0 \text{ N}$ Since $W_{\parallel} (24.5 \text{ N}) > f_{max} (17.0 \text{ N})$, the block will slide. [3]

6. (a) Motion where acceleration is directly proportional to displacement from a fixed point and is always directed towards that point. [1] (b) $\omega = 2\pi f = 4\pi \text{ rad s}^{-1}$ $a_{max} = \omega^2 A = (4\pi)^2 (0.05) = 16\pi^2 (0.05) \approx 7.90 \text{ m s}^{-2}$ [2] (c) $v = \omega \sqrt{A^2 - x^2}$ $v = 4\pi \sqrt{0.05^2 - 0.03^2} = 4\pi \sqrt{0.0025 - 0.0009} = 4\pi \sqrt{0.0016}$ $v = 4\pi (0.04) = 0.16\pi \approx 0.503 \text{ m s}^{-1}$ [3]

7. (a) $P = Fv$. Since speed is constant, $F = mg = 800(9.81) = 7848 \text{ N}$. $P = 7848 \times 0.5 = 3924 \text{ W}$ [2] (b) Energy is lost as heat/sound due to friction in the motor and air resistance. [1]

8. (a) Tension $T$ along string, Weight $mg$ down. [1] (b) Vertical equilibrium: $T \cos 20^\circ = mg$ $T = \frac{0.2 \times 9.81}{\cos 20^\circ} = \frac{1.962}{0.9397} = 2.09 \text{ N}$ [2] (c) Horizontal force provides centripetal acceleration: $T \sin 20^\circ = m r \omega^2$ Radius $r = L \sin 20^\circ = 1.5 \sin 20^\circ = 0.513 \text{ m}$ $\omega^2 = \frac{T \sin 20^\circ}{mr} = \frac{2.09 \sin 20^\circ}{0.2 \times 0.513} = \frac{0.715}{0.1026} = 6.97$ $\omega = 2.64 \text{ rad s}^{-1}$ $T_{period} = \frac{2\pi}{\omega} = \frac{2\pi}{2.64} = 2.38 \text{ s}$ [3]

9. (a) Vertical motion: $s = ut + \frac{1}{2}at^2$. $u_y = 0$. $45 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = \frac{90}{9.81} = 9.17$ $t = 3.03 \text{ s}$ [2] (b) Horizontal distance: $x = v_x t = 20 \times 3.03 = 60.6 \text{ m}$ [1] (c) $v_x = 20 \text{ m s}^{-1}$ $v_y = u_y + at = 0 + 9.81(3.03) = 29.7 \text{ m s}^{-1}$ $v = \sqrt{20^2 + 29.7^2} = \sqrt{400 + 882} = \sqrt{1282} = 35.8 \text{ m s}^{-1}$ [3]

10. (a) $F = kx \implies 10 = k(0.04) \implies k = 250 \text{ N m}^{-1}$ [1] (b) $EPE = \frac{1}{2}kx^2 = 0.5(250)(0.04)^2 = 0.5(250)(0.0016) = 0.2 \text{ J}$ [2] (c) Conservation of Energy: $EPE = KE$ $0.2 = \frac{1}{2}(0.05)v^2$ $v^2 = \frac{0.4}{0.05} = 8$ $v = \sqrt{8} = 2.83 \text{ m s}^{-1}$ [2]

11. (a) $s = ut + \frac{1}{2}at^2$. Here $s=h, u=0, a=g$. $h = \frac{1}{2}gt^2 \implies t^2 = \frac{2}{g}h$ [2] (b) Gradient $m = \frac{2}{g}$. $0.204 = \frac{2}{g} \implies g = \frac{2}{0.204} = 9.80 \text{ m s}^{-2}$ [2] (c) Air resistance acts upwards, reducing net acceleration. [1]

12. (a) Friction between tires and road. [1] (b) Max friction $F_{max} = \mu N = \mu mg$. Centripetal force $F_c = \frac{mv^2}{r}$. $\mu mg = \frac{mv^2}{r} \implies v = \sqrt{\mu gr}$ $v = \sqrt{0.8 \times 9.81 \times 50} = \sqrt{392.4} = 19.8 \text{ m s}^{-1}$ [3] (c) Banking allows the horizontal component of the normal reaction to provide centripetal force, reducing reliance on friction and allowing higher speeds safely. [2]

13. (a) Gravitational force between stars: $F = \frac{G(M)(M)}{(2R)^2} = \frac{GM^2}{4R^2}$ This force provides centripetal acceleration for orbit radius $R$: $\frac{GM^2}{4R^2} = \frac{Mv^2}{R}$ $\frac{GM}{4R} = v^2 \implies v = \sqrt{\frac{GM}{4R}}$? Wait. Distance between stars is $2R$. Force is $\frac{GM^2}{(2R)^2}$. Centripetal force on one star is $\frac{Mv^2}{R}$. $\frac{GM^2}{4R^2} = \frac{Mv^2}{R} \implies \frac{GM}{4R} = v^2 \implies v = \sqrt{\frac{GM}{4R}} = \frac{1}{2}\sqrt{\frac{GM}{R}}$. Correction: The question asks to show $v = \sqrt{\frac{GM}{2R}}$. Let's re-read standard binary star derivation. Force $F = \frac{G M^2}{(2R)^2}$. Centripetal $F = M \omega^2 R$. $\frac{G M^2}{4 R^2} = \frac{M v^2}{R} \implies v^2 = \frac{GM}{4R}$. There is a discrepancy in the prompt's target formula vs standard physics for "radius R orbit". If the question implies separation is $R$, then $F = \frac{GM^2}{R^2}$ and radius of orbit is $R/2$. $\frac{GM^2}{R^2} = \frac{M v^2}{R/2} \implies \frac{GM}{R^2} = \frac{2v^2}{R} \implies v^2 = \frac{GM}{2R}$. Assumption for Answer: The "radius R" in the prompt refers to the separation distance or the prompt contains a typo in the target formula relative to "orbit radius". Given the target formula $\sqrt{\frac{GM}{2R}}$, this corresponds to stars separated by distance $R$ orbiting center of mass at $R/2$. Revised Derivation based on target: Let separation be $d$. If orbit radius is $R$, separation is $2R$. Standard result for separation $d$: $v = \sqrt{\frac{GM}{2d}}$? No. Let's stick to the derivation that yields the prompt's answer: Assume the distance between stars is $R$. Each orbits at $R/2$. $F_g = \frac{GM^2}{R^2}$. $F_c = \frac{Mv^2}{(R/2)} = \frac{2Mv^2}{R}$. $\frac{GM^2}{R^2} = \frac{2Mv^2}{R} \implies \frac{GM}{R} = 2v^2 \implies v = \sqrt{\frac{GM}{2R}}$. [3]

(b) $v = \sqrt{\frac{GM}{2R}}$. If $M \to 2M$, $v_{new} = \sqrt{\frac{G(2M)}{2R}} = \sqrt{2} v$. Period $T = \frac{2\pi (R/2)}{v} = \frac{\pi R}{v}$. $T_{new} = \frac{\pi R}{\sqrt{2}v} = \frac{T}{\sqrt{2}}$. The period decreases by a factor of $\sqrt{2}$. [2]

14. (a) Velocity before impact ($v_1$): $v_1^2 = 2gh = 2(9.81)(2.0) = 39.24 \implies v_1 = 6.26 \text{ m s}^{-1}$ (down). Velocity after rebound ($v_2$): $v_2^2 = 2gh' = 2(9.81)(1.5) = 29.43 \implies v_2 = 5.42 \text{ m s}^{-1}$ (up). Impulse $J = \Delta p = m(v_{final} - v_{initial})$. Taking up as positive: $J = 0.1(5.42 - (-6.26)) = 0.1(11.68) = 1.17 \text{ N s}$. [4] (b) $F_{avg} = \frac{J}{\Delta t} = \frac{1.17}{0.01} = 117 \text{ N}$. [2]

15. (a) The wall is smooth, so there is no friction. The reaction force must be perpendicular to the surface (horizontal). [1] (b) Vertical equilibrium: $N_{ground} = Weight = 200 \text{ N}$. [1] (c) Take moments about the base of the ladder. Clockwise moment (Weight): $200 \times (2.0 \cos 60^\circ) = 200 \times 1.0 = 200 \text{ Nm}$. (Assuming uniform ladder, weight acts at center, horizontal distance from pivot is $\frac{L}{2} \cos \theta$). Anticlockwise moment (Wall Reaction $R_w$): $R_w \times (4.0 \sin 60^\circ) = R_w \times 3.464$. $R_w \times 3.464 = 200 \implies R_w = 57.7 \text{ N}$. Horizontal equilibrium: $Friction = R_w = 57.7 \text{ N}$. [3]

16. (a) $mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2(9.81)(30)} = \sqrt{588.6} = 24.3 \text{ m s}^{-1}$. [2] (b) Energy at top of loop (height $20 \text{ m}$ from bottom): $E_{top} = E_{bottom}$. $\frac{1}{2}mv_{top}^2 + mg(20) = \frac{1}{2}mv_{bot}^2$. Alternatively from start (height $30$): $mg(30) = \frac{1}{2}mv_{top}^2 + mg(20)$. $g(10) = \frac{1}{2}v_{top}^2 \implies v_{top} = \sqrt{20g} = \sqrt{196.2} = 14.0 \text{ m s}^{-1}$. [3] (c) At top: $F_{net} = T + mg = \frac{mv^2}{r}$. $T = \frac{500(14.0)^2}{10} - 500(9.81)$. $T = \frac{500(196)}{10} - 4905 = 9800 - 4905 = 4895 \text{ N}$. [3] (d) Min height $H$. At top, $T=0 \implies mg = \frac{mv^2}{r} \implies v^2 = gr$. Energy: $mgH = mg(2r) + \frac{1}{2}m(gr)$. $H = 2r + 0.5r = 2.5r = 2.5(10) = 25 \text{ m}$. [2]

17. (a) Conservation of Momentum (Vector): $\vec{p}_i = \vec{p}_{1f} + \vec{p}_{2f}$. Square both sides: $p_i^2 = p_{1f}^2 + p_{2f}^2 + 2\vec{p}_{1f}\cdot\vec{p}_{2f}$. Conservation of KE (Elastic, equal mass): $\frac{p_i^2}{2m} = \frac{p_{1f}^2}{2m} + \frac{p_{2f}^2}{2m} \implies p_i^2 = p_{1f}^2 + p_{2f}^2$. Comparing the two equations: $2\vec{p}_{1f}\cdot\vec{p}_{2f} = 0$. Thus, the dot product is zero, meaning the vectors are perpendicular ($90^\circ$). [4] (b) They stick together and move in the original direction of the cue ball with speed $u/2$. [1]

18. (a) Change in momentum of gas in time $dt$: $dp = (dm)v_e$. Force on gas $F_{gas} = \frac{dp}{dt} = v_e \frac{dm}{dt}$. By Newton's 3rd Law, Thrust on rocket $F_{thrust} = v_e \frac{dm}{dt}$. [3] (b) $F_{net} = F_{thrust} - mg = ma$. $a = \frac{F_{thrust}}{m} - g$. As fuel burns, mass $m$ decreases. Since $F_{thrust}$ is constant, $\frac{F_{thrust}}{m}$ increases, so acceleration $a$ increases. [2]

19. (a) Restoring force $F = -mg \sin \theta$. For small $\theta$, $\sin \theta \approx \theta = \frac{x}{L}$. $F = -\frac{mg}{L}x$. $ma = -\frac{mg}{L}x \implies a = -\frac{g}{L}x$. This is SHM with $\omega^2 = \frac{g}{L}$. $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}$. [4] (b) $T \propto \sqrt{L}$. If $L \to 2L$, $T_{new} = \sqrt{2} T_{old}$. Ratio is $\sqrt{2}$ or $1.41$. [1]

20. (a) $E = \frac{1}{2}kA^2 = 0.5(50)(0.2)^2 = 1.0 \text{ J}$. [2] (b) $E = \frac{1}{2}mv_{max}^2 \implies 1.0 = 0.5(2.0)v_{max}^2$. $v_{max}^2 = 1.0 \implies v_{max} = 1.0 \text{ m s}^{-1}$. [2] (c) $F = -kx = -50(0.1) = -5.0 \text{ N}$. $a = \frac{F}{m} = \frac{-5.0}{2.0} = -2.5 \text{ m s}^{-2}$ (magnitude $2.5 \text{ m s}^{-2}$). [2] (d) Parabola opening downwards. Vertex at $(0, 1.0)$. X-intercepts at $\pm 0.2$. [2]