A Level H2 Physics Practice Paper 2

TuitionGoWhere Exam Practice (AI) - Answer Key

Subject: Physics H2 | Paper: Practice Paper 2 (Version 2)

Section A

Question 1 (a) In a closed system (or isolated system), the total momentum before an event equals the total momentum after the event, provided no external forces act. [2] (b) $m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$ $(2.0 \times 4.0) + (3.0 \times 0) = (2.0 + 3.0)v_f$ $8.0 = 5.0v_f \Rightarrow v_f = 1.6 \text{ m s}^{-1}$ [3] (c) The collision is inelastic. Some kinetic energy is converted into internal energy (heat, sound, deformation of trolleys). [2]

Question 2 (a) $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.60} = 10.47 \text{ rad s}^{-1}$ [2] (b) $a_{\max} = \omega^2 X_0 = (10.47)^2 \times 0.08 = 8.77 \text{ m s}^{-2}$ [3] (c) $v_{\max} = \omega X_0 = 10.47 \times 0.08 = 0.838 \text{ m s}^{-1}$ [2] (d) At the equilibrium position (center of oscillation). [1]

Question 3 (a) $v^2 = 2gh = 2 \times 9.81 \times (1.2 \times (1 - \cos 15^\circ)) = 2 \times 9.81 \times 0.041 = 0.805 \text{ m}^2\text{s}^{-2}$ $v = 0.897 \text{ m s}^{-1}$ At bottom: $T - mg = \frac{mv^2}{r} \Rightarrow T = m(g + \frac{v^2}{r}) = 0.15(9.81 + \frac{0.805}{1.2}) = 0.15(9.81 + 0.671) = 1.57 \text{ N}$ [4] (b) $v = \sqrt{2 \times 9.81 \times 1.2(1 - \cos 15^\circ)} = 0.90 \text{ m s}^{-1}$ [3] (c) The sphere will move in a horizontal straight line (tangent to the arc) at the velocity it had at the bottom, subject to gravity (projectile motion). [2]

Question 4 (a) Diagram should show: Weight ($mg$) downwards, Normal reaction ($R$) perpendicular to incline, Applied force ($F$) up incline, Friction ($f$) down incline. [3] (b) Constant velocity $\Rightarrow \sum F = 0$. $F = mg \sin 30^\circ + f$ $f = \mu R = \mu mg \cos 30^\circ$ $F = (5.0 \times 9.81 \times 0.5) + (0.25 \times 5.0 \times 9.81 \times 0.866) = 24.53 + 10.61 = 35.14 \text{ N}$ [4] (c) $W_f = f \times d = 10.61 \times 4.0 = 42.44 \text{ J}$ (Work done by friction is negative, so $-42.4 \text{ J}$). [3]

Question 5 (a) The work done per unit mass in bringing a small test mass from infinity to that point. [2] (b) (i) $v = \sqrt{\frac{GM_E}{r}}$ where $r = 6.37 \times 10^6 + 36 \times 10^6 = 4.237 \times 10^7 \text{ m}$ $v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{4.237 \times 10^7}} = 3067 \text{ m s}^{-1}$ [3] (ii) $E_{\text{total}} = -\frac{GM_E m}{2r} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1200}{2 \times 4.237 \times 10^7} = -5.64 \times 10^9 \text{ J}$ [3] (iii) The negative sign indicates the satellite is bound to the Earth's gravitational field; work must be done to move it to infinity. [2]

Section B

Question 6 (a) Setup: Rigid support, string, small heavy bob, stopwatch, meter rule. Procedure: Measure $L$ from pivot to center of bob. Displace bob by small angle ($< 10^\circ$). Time for 20 oscillations to reduce random error. Repeat for different $L$. [6] (b) 1. Use a fiducial marker at the center to identify the exact point of oscillation for timing. 2. Ensure the angle of displacement is small to maintain SHM approximation. 3. Measure $L$ using a meter rule with mm graduations, ensuring the string is taut. [6] (c) $T^2 = \frac{4\pi^2}{g} L$. Plotting $T^2$ vs $L$ gives a straight line through origin. Gradient $m = \frac{4\pi^2}{g} \Rightarrow g = \frac{4\pi^2}{m}$. [4]

Question 7 (a) $P_{\text{total}} = (1.0 \times 5.0) + (2.0 \times -3.0) = 5.0 - 6.0 = -1.0 \text{ kg m s}^{-1}$ [2] (b) Conservation of momentum: $1v_A + 2v_B = -1$ Conservation of KE: $\frac{1}{2}(1)(5^2) + \frac{1}{2}(2)(3^2) = \frac{1}{2}(1)v_A^2 + \frac{1}{2}(2)v_B^2$ $12.5 + 9 = 0.5v_A^2 + v_B^2 \Rightarrow 21.5 = 0.5v_A^2 + v_B^2$ From (1): $v_A = -1 - 2v_B$ Substitute into (2): $21.5 = 0.5(-1 - 2v_B)^2 + v_B^2 = 0.5(1 + 4v_B + 4v_B^2) + v_B^2 = 0.5 + 2v_B + 3v_B^2$ $3v_B^2 + 2v_B - 21 = 0$. Using quadratic formula: $v_B = \frac{-2 \pm \sqrt{4 - 4(3)(-21)}}{6} = \frac{-2 \pm 16}{6}$ $v_B = 2.33 \text{ m s}^{-1}$ or $-3 \text{ m s}^{-1}$ (initial). So $v_B = 2.33 \text{ m s}^{-1}$, $v_A = -1 - 2(2.33) = -5.66 \text{ m s}^{-1}$ [6] (c) In the "stick together" case, the collision is perfectly inelastic. This results in the maximum possible loss of kinetic energy because the relative velocity of the particles becomes zero. [4]

Question 8 (a) At top, $T + mg = \frac{mv^2}{r}$. For minimum speed, $T \to 0$. $mg = \frac{mv_{\text{min}}^2}{r} \Rightarrow v_{\text{min}} = \sqrt{gr}$ [5] (b) Energy conservation: $E_{\text{top}} = E_{\text{bottom}}$ $\frac{1}{2}mv_{\text{top}}^2 + mgr = \frac{1}{2}mv_{\text{bottom}}^2$ $v_{\text{bottom}}^2 = v_{\text{top}}^2 + 2gr$ At bottom: $T - mg = \frac{mv_{\text{bottom}}^2}{r} = \frac{m(v_{\text{top}}^2 + 2gr)}{r}$ $T = mg + \frac{mv_{\text{top}}^2}{r} + 2mg = 3mg + \frac{mv_{\text{top}}^2}{r}$ [6] (c) Since $T = m(3g + \frac{v_{\text{top}}^2}{r})$, the tension is directly proportional to $m$. Increasing $m$ increases the tension linearly. [4]