A Level H2 Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key - Version 5

Section A: Mechanics and Energy

Question 1 (a) $F_{net} = T - mg = mv^2/r \implies T = 0.20(5^2/1.5) + 0.20(9.81) = 3.33 + 1.96 = 5.29 \text{ N}$. [3] (b) At highest point, $T=0 \implies mg = mv^2/r \implies v = \sqrt{gr} = \sqrt{9.81 \times 1.5} = 3.84 \text{ m s}^{-1}$. [3] (c) $E_{bottom} = E_{top} \implies \frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mg(2r)$. $v_2^2 = v_1^2 - 4gr = 5^2 - 4(9.81)(1.5) = 25 - 58.86 = -33.86$. Correction/Note: The initial speed $5.0 \text{ m s}^{-1}$ is insufficient to reach the top. The sphere will oscillate or fall. (Student should identify that $v^2$ becomes negative, implying it doesn't reach the top). [4]

Question 2 (a) In a closed system, the total linear momentum remains constant provided no external forces act. [2] (b) $m_1u_1 + m_2u_2 = (m_1+m_2)v \implies (1.5)(3.0) + (2.5)(-2.0) = (4.0)v \implies 4.5 - 5.0 = 4v \implies v = -0.125 \text{ m s}^{-1}$ (opposite to A's initial direction). [3] (c) $KE_{init} = \frac{1}{2}(1.5)(3^2) + \frac{1}{2}(2.5)(2^2) = 6.75 + 5.0 = 11.75 \text{ J}$. $KE_{final} = \frac{1}{2}(4.0)(-0.125)^2 = 0.03125 \text{ J}$. Loss $= 11.75 - 0.03 = 11.72 \text{ J}$. [3]

Question 3 (a) $F_c = F_g \implies mv^2/R = GMm/R^2 \implies v^2 = GM/R$. $T = 2\pi R / v = 2\pi R / \sqrt{GM/R} = 2\pi \sqrt{R^3/GM}$. [4] (b) $T \propto R^{3/2}$. If $R \to 1.1R$, then $T \to (1.1)^{1.5} T \approx 1.1537 T$. Percentage increase $\approx 15.4\%$. [3]

Question 4 (a) Diagram should show: Weight ($mg$) downwards, Normal force ($N$) perpendicular to plane, Friction ($f$) up the plane. [2] (b) $F_{net} = mg\sin\theta - \mu mg\cos\theta = ma \implies a = g(\sin 30^\circ - 0.2\cos 30^\circ) = 9.81(0.5 - 0.173) = 3.21 \text{ m s}^{-2}$. [4] (c) $a_{up} = -g(\sin 30^\circ + 0.2\cos 30^\circ) = -9.81(0.5 + 0.173) = -6.60 \text{ m s}^{-2}$. $v^2 = u^2 + 2as \implies 0 = 2^2 + 2(-6.60)s \implies s = 4/13.2 = 0.303 \text{ m}$. [4]

Section B: Electricity and Magnetism

Question 5 (a) The magnitude of the induced EMF is equal to the rate of change of magnetic flux linkage. [2] (b) $\Phi = BA\cos(\omega t) \implies \varepsilon = -N \frac{d\Phi}{dt} = -NBA(-\omega \sin(\omega t)) = NBA\omega \sin(\omega t)$. [3] (c) $I_{max} = \varepsilon_{max}/R = (NBA\omega)/R = (50 \times 0.5 \times 0.02 \times 10) / 2.0 = 5/2 = 2.5 \text{ A}$. [3]

Question 6 (a) $R_{parallel} = (6 \times 3)/(6+3) = 2.0\ \Omega$. $R_{total} = r + R_1 + R_{parallel} = 1.0 + 4.0 + 2.0 = 7.0\ \Omega$. [3] (b) $I_{total} = \varepsilon/R_{total} = 12/7 = 1.71 \text{ A}$. Using current divider: $I_3 = I_{total} \times (R_2 / (R_2+R_3)) = 1.71 \times (6/9) = 1.14 \text{ A}$. [4] (c) $I_{total} = \varepsilon / (r + R_{ext})$. If $r$ increases, the denominator increases, so total current $I_{total}$ decreases. [3]

Section C: Modern Physics and Waves

Question 7 (a) $E_{photon} = hc/\lambda = (6.63\times 10^{-34} \times 3\times 10^8) / 300\times 10^{-9} = 6.63 \times 10^{-19} \text{ J} \approx 4.14 \text{ eV}$. $K_{max} = 4.14 - 2.2 = 1.94 \text{ eV}$. [3] (b) $eV_s = K_{max} \implies V_s = 1.94 \text{ V}$. [2] (c) $K_{max}$ remains constant because it depends on frequency, not intensity. Photoelectric current increases because more photons per second result in more emitted electrons. [4]

Question 8 (a) $A = A_0(1/2)^{t/T} \implies 1.5\times 10^3 = 1.2\times 10^4(1/2)^{24/T} \implies 0.125 = (1/2)^{24/T}$. $1/8 = (1/2)^3 \implies 24/T = 3 \implies T = 8 \text{ days}$. [4] (b) $\lambda = \ln 2 / T = 0.693 / (8 \times 24 \times 3600) = 9.95 \times 10^{-7} \text{ s}^{-1}$. [3] (c) $A = \lambda N$, where $A$ is activity, $\lambda$ is decay constant, and $N$ is number of undecayed nuclei. [3]