A Level H2 Physics Practice Paper 1

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

Version 1 - Marking Scheme

Section A: Kinematics and Dynamics

Question 1 (a) $v_{iy} = 40 \sin(35^\circ) = 22.94 \text{ m s}^{-1}$. $H = v_{iy}^2 / 2g = (22.94)^2 / (2 \times 9.81) = 26.7 \text{ m}$. [3] (b) $t_{flight} = 2 v_{iy} / g = 4.68 \text{ s}$. $v_{ix} = 40 \cos(35^\circ) = 32.77 \text{ m s}^{-1}$. Range $= 32.77 \times 4.68 = 153 \text{ m}$. [3] (c) There are no horizontal forces acting on the projectile (neglecting air resistance). According to Newton's First Law, the horizontal acceleration is zero, thus velocity remains constant. [2]

Question 2 (a) $F_N = mg \cos(30^\circ) = 2.5 \times 9.81 \times 0.866 = 21.2 \text{ N}$. [2] (b) Driving force $F_d = mg \sin(30^\circ) = 2.5 \times 9.81 \times 0.5 = 12.3 \text{ N}$. Max static friction $f_{s,max} = \mu F_N = 0.40 \times 21.2 = 8.5 \text{ N}$. Since $F_d > f_{s,max}$, the block will slide. [4] (c) $F_{applied} - mg \sin(30^\circ) - \mu mg \cos(30^\circ) = 0$. $F_{applied} = 12.3 + (0.20 \times 21.2)$ (assuming $\mu_k$ is same as $\mu_s$ or specified). $F_{applied} = 12.3 + 8.5 = 20.8 \text{ N}$. [4]

Question 3 (a) $v^2 = u^2 + 2as \implies 25^2 = 0 + 2(a)(150) \implies a = 625 / 300 = 2.08 \text{ m s}^{-2}$. [3] (b) $F_{net} = ma = 1200 \times 2.08 = 2500 \text{ N}$. [2] (c) $F_{net} = F_{drive} - F_{resistive} \implies 2500 = 3500 - F_r \implies F_r = 1000 \text{ N}$. [3]

Question 4 The rate of change of momentum of a body is directly proportional to the net force acting on it and takes place in the direction of the force. $\vec{F} = \frac{d\vec{p}}{dt}$. [3]

Section B: Energy, Momentum, and Circular Motion

Question 5 (a) In a closed system, the total linear momentum before an event is equal to the total linear momentum after the event, provided no external forces act. [2] (b) $m_1u_1 + m_2u_2 = (m_1+m_2)v \implies (1.0)(3.0) + (2.0)(-1.5) = (3.0)v \implies 3.0 - 3.0 = 3v \implies v = 0 \text{ m s}^{-1}$. [4] (c) $KE_{initial} = 0.5(1)(3^2) + 0.5(2)(1.5^2) = 4.5 + 2.25 = 6.75 \text{ J}$. $KE_{final} = 0$. Loss $= 6.75 \text{ J}$. [4]

Question 6 (a) $T=0$ at top: $mg = mv^2/r \implies v = \sqrt{gr} = \sqrt{9.81 \times 0.80} = 2.80 \text{ m s}^{-1}$. [3] (b) $T - mg = mv^2/r \implies T = m(g + v^2/r) = 0.15(9.81 + 2.8^2/0.8) = 0.15(9.81 + 9.8) = 2.94 \text{ N}$. [5] (c) At the bottom, the tension must provide the centripetal force AND overcome the weight of the sphere. At the top, weight assists in providing the centripetal force. [3]

Question 7 (a) $E_p = 0.5 k x^2 = 0.5 \times 200 \times (0.12)^2 = 1.44 \text{ J}$. [2] (b) Work done by friction = Initial Energy. $f_k \times d = E_p \implies (\mu mg) d = 1.44 \implies (0.20 \times 0.5 \times 9.81) d = 1.44 \implies 0.981 d = 1.44 \implies d = 1.47 \text{ m}$. [5]

Section C: Gravitational Fields

Question 8 (a) $F_c = F_g \implies mv^2/r = GMm/r^2$ where $r = R+h$. $v^2 = GM/(R+h)$. Since $v = 2\pi r / T$, then $(2\pi(R+h)/T)^2 = GM/(R+h) \implies T^2 = 4\pi^2(R+h)^3/GM \implies T = 2\pi \sqrt{(R+h)^3/GM}$. [6] (b) Orbital speed $v = \sqrt{GM/r}$. As $h$ increases, $r$ increases, so $v$ decreases. [3]

Question 9 (a) $g = GM/R^2 \implies M = gR^2/G = (4.5 \times (3.4 \times 10^6)^2) / (6.67 \times 10^{-11}) = 7.8 \times 10^{23} \text{ kg}$. [4] (b) $v_{esc} = \sqrt{2GM/R} = \sqrt{2 \times 6.67 \times 10^{-11} \times 7.8 \times 10^{23} / 3.4 \times 10^6} = 5520 \text{ m s}^{-1}$. [4] (c) The work done per unit mass in bringing a small test mass from infinity to that point. [3]