A Level H2 Physics Practice Paper 4

Answer Key - TuitionGoWhere Practice Paper (AI) Version 4

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) (i) x-axis: $(0.5 \times 2.0) = (0.5 \times 0.5 \cos 30^\circ) + (0.8 \times v_{2x}) \implies 1.0 = 0.2165 + 0.8v_{2x} \implies v_{2x} = 0.98\text{ m s}^{-1}$ y-axis: $0 = (0.5 \times 0.5 \sin 30^\circ) + (0.8 \times v_{2y}) \implies 0 = 0.125 + 0.8v_{2y} \implies v_{2y} = -0.156\text{ m s}^{-1}$ $v_2 = \sqrt{0.98^2 + (-0.156)^2} = 0.99\text{ m s}^{-1}$ [4] (ii) $KE_{initial} = 0.5 \times 0.5 \times 2^2 = 1.0\text{ J}$ $KE_{final} = (0.5 \times 0.5 \times 0.5^2) + (0.5 \times 0.8 \times 0.99^2) = 0.0625 + 0.392 = 0.45\text{ J}$ $KE_{initial} \neq KE_{final}$, therefore the collision is inelastic. [3]

Question 2 (a) Centripetal force = Gravitational force: $mv^2/(R+h) = GMm/(R+h)^2 \implies v^2 = GM/(R+h)$. $T = 2\pi(R+h)/v = 2\pi(R+h) / \sqrt{GM/(R+h)} = 2\pi \sqrt{(R+h)^3/GM}$. [4] (b) Orbital speed $v \propto 1/\sqrt{R+h}$, so speed decreases. Period $T \propto (R+h)^{3/2}$, so period increases. [2] (c) $KE + PE = 0 \implies 0.5mv^2 - GMm/R = 0 \implies v = \sqrt{2GM/R}$. [3]

Question 3 (a) $\omega = \sqrt{k/m} = \sqrt{80/0.2} = \sqrt{400} = 20\text{ rad s}^{-1}$. [2] (b) $a_{\max} = \omega^2 X_0 = 20^2 \times 0.05 = 400 \times 0.05 = 20\text{ m s}^{-2}$. [2] (c) $E = 0.5 k X_0^2$. [2]

Section B

Question 4 (a) $KE = 0.5 k X_0^2 = 0.5 \times 500 \times 0.1^2 = 2.5\text{ J}$. [2] (b) Work done by friction = Initial Energy $\implies \mu mg d = 2.5 \implies 0.2 \times 2 \times 9.81 \times d = 2.5 \implies 3.924d = 2.5 \implies d = 0.64\text{ m}$. [4] (c) Lubrication reduces $\mu$, therefore the distance $d$ increases as less energy is dissipated per unit distance. [2]

Question 5 (a) Forces: Tension $T$ (right), Friction $f$ (left), Weight $m_Ag$ (down), Normal force $N$ (up). [2] (b) For A: $T - \mu m_Ag = m_Aa \implies T - (0.3 \times 3 \times 9.81) = 3a \implies T - 8.829 = 3a$ For B: $m_Bg - T = m_Ba \implies (2 \times 9.81) - T = 2a \implies 19.62 - T = 2a$ Adding: $19.62 - 8.829 = 5a \implies 10.791 = 5a \implies a = 2.16\text{ m s}^{-2}$. [4] (c) $T = 3(2.16) + 8.829 = 15.31\text{ N}$. [2]

Question 6 (a) $F_{net} = T - mg = mv^2/r \implies T = m(g + v^2/r) = m(9.81 + 25/1.5) = m(9.81 + 16.67) = 26.48m$. (Since $m$ not given, answer as $26.5m$ or assume $m=1$ for $26.5\text{ N}$). [3] (b) At top: $T + mg = mv^2/r$. For min velocity, $T=0 \implies mg = mv_{top}^2/r \implies v_{top} = \sqrt{gr} = \sqrt{9.81 \times 1.5} = 3.84\text{ m s}^{-1}$. Using energy: $0.5mv_{bot}^2 - mg(2r) = 0.5mv_{top}^2 \implies v_{bot}^2 = v_{top}^2 + 4gr = gr + 4gr = 5gr$. $v_{bot} = \sqrt{5 \times 9.81 \times 1.5} = \sqrt{73.575} = 8.58\text{ m s}^{-1}$. [4]

Section C

Question 7 (a) Use a set square to ensure the ruler is vertical; use a fiducial marker for the start/end points. [3] (b) Use a denser, smaller object (e.g., steel ball bearing) to increase the ratio of weight to surface area. [2] (c) $t_{measured} = t_{actual} + 0.01$. Since $g = 2s/t^2$, a larger $t$ results in a smaller calculated $g$. [3]

Question 8 (a) $v_y = u \sin\theta - gt$. At max height $v_y = 0 \implies t = (u \sin\theta)/g$. $H = (u \sin\theta)t - 0.5gt^2 = (u^2 \sin^2\theta)/g - 0.5(u^2 \sin^2\theta)/g = (u^2 \sin^2\theta)/(2g)$. [3] (b) $R = (u^2 \sin 2\theta)/g$. $R$ is max when $\sin 2\theta = 1 \implies 2\theta = 90^\circ \implies \theta = 45^\circ$. [3] (c) $t = (2u \sin\theta)/g = (2 \times 20 \times \sin 30^\circ)/9.81 = 20/9.81 = 2.04\text{ s}$. [3]

Question 9 (a) $m_1u_1 + m_2u_2 = (m_1+m_2)v \implies (1 \times 4) + (2 \times -2) = 3v \implies 4 - 4 = 3v \implies v = 0\text{ m s}^{-1}$. [3] (b) $KE_{initial} = 0.5(1)(4^2) + 0.5(2)(2^2) = 8 + 4 = 12\text{ J}$. $KE_{final} = 0$. Loss = $12\text{ J}$. [3] (c) Momentum is conserved because there are no external forces. KE is not conserved because the collision is perfectly inelastic; energy is converted to heat/sound/deformation. [2]

Question 10 (a) Vertical: $T \cos\theta = mg \implies T = mg / \cos\theta$. [3] (b) Horizontal: $T \sin\theta = mv^2/r \implies (mg/\cos\theta)\sin\theta = mv^2/r \implies g \tan\theta = v^2/r$. $v = \sqrt{gr \tan\theta}$. $T_{period} = 2\pi r / v = 2\pi r / \sqrt{gr \tan\theta} = 2\pi \sqrt{r / (g \tan\theta)}$. [4] (c) No change. The expression for $T_{period}$ is independent of mass $m$. [2]