A Level H2 Physics Practice Paper 5

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key and Marking Scheme

Subject: Physics H2
Paper: Practice Paper – Version 5 (Mechanics Focus)

Section A: Structured Questions

1. (a) $s = \frac{1}{2}at^2$ [1] (b) Comparing $s = \frac{1}{2}at^2$ with $y = mx$, the gradient $m = \frac{1}{2}a$. Therefore, $a = 2 \times \text{gradient}$. [2] (c) The distance $s$ was measured from a point other than the starting position (e.g., the front of the trolley instead of the back, or a zero error in the ruler). [1]

2. Linear momentum is the product of mass and velocity. ($p = mv$) [1]

3. In a closed system (or isolated system), [1] the total momentum before an interaction equals the total momentum after the interaction, provided no external forces act. [1]

4. (a) Using $v^2 = u^2 + 2as$: $v^2 = 0 + 2(9.81)(1.2)$ $v = \sqrt{23.544} = 4.85$ m s⁻¹ [2] (b) Using $v^2 = u^2 + 2as$ for upward motion (final $v=0$ at max height): $0 = u^2 - 2(9.81)(0.80)$ $u = \sqrt{15.696} = 3.96$ m s⁻¹ [2] (c) Taking downward as positive: $p_{initial} = 0.15 \times 4.85 = 0.7275$ N s $p_{final} = 0.15 \times (-3.96) = -0.594$ N s $\Delta p = p_{final} - p_{initial} = -0.594 - 0.7275 = -1.3215$ N s Magnitude = 1.32 N s [2]

5. (a) At constant speed, Driving Force = Resistive Force. $2500 = k(30)^2$ $k = \frac{2500}{900} = 2.78$ kg m⁻¹ [2] (b) At rest ($v=0$), $F_R = 0$. $F_{net} = 2500$ N $a = \frac{F}{m} = \frac{2500}{1200} = 2.08$ m s⁻² [2] (c) As speed increases, resistive force $F_R = kv^2$ increases. [1] Since Driving Force is constant, the resultant force ($F_D - F_R$) decreases. [1] Therefore, acceleration decreases ($a = F_{net}/m$).

6. (a) Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1] $\frac{GM}{r} = v^2$ [1] $v = \sqrt{\frac{GM}{r}}$ [1] (b) Period $T = \frac{2\pi r}{v} = 2\pi r \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^3}{GM}}$. $T \propto r^{3/2}$ $\frac{T_B}{T_A} = (\frac{r_B}{r_A})^{3/2} = (\frac{4r}{r})^{3/2} = 4^{1.5} = 8$ [3]

7. (a) Normal reaction $R = mg = 2.0 \times 9.81 = 19.62$ N. Max static friction $F_{s,max} = \mu_s R = 0.40 \times 19.62 = 7.85$ N. Minimum force $F = 7.85$ N (or 7.8 N). [2] (b) Dynamic friction $F_d = \mu_d R = 0.30 \times 19.62 = 5.886$ N. Resultant force $F_{net} = F_{applied} - F_d = 7.848 - 5.886 = 1.962$ N. $a = \frac{F_{net}}{m} = \frac{1.962}{2.0} = 0.98$ m s⁻². [3]

8. (a) Vertical component $u_y = 20 \sin 30^\circ = 10$ m s⁻¹. At max height, $v_y = 0$. $0 = 10^2 - 2(9.81)h$ $h = \frac{100}{19.62} = 5.10$ m. [2] (b) Horizontal component $u_x = 20 \cos 30^\circ = 17.32$ m s⁻¹. Time of flight: $t = \frac{2u_y}{g} = \frac{20}{9.81} = 2.04$ s. Range $R = u_x t = 17.32 \times 2.04 = 35.3$ m. [2] (c) Velocity is horizontal only. Magnitude = $17.3$ m s⁻¹. Direction = Horizontal. [1]

9. (a) Diagram should show: Weight (downwards from center), Tension (along cable towards wall), Hinge reaction (horizontal and/or vertical components at hinge). [2] (b) Angle of cable with beam: $\tan \alpha = \frac{3}{4} \Rightarrow \alpha = 36.9^\circ$. Taking moments about hinge: Clockwise moment = Anticlockwise moment $W \times 2.0 = T \sin(36.9^\circ) \times 4.0$ $50 \times 2.0 = T \times 0.6 \times 4.0$ $100 = 2.4 T$ $T = 41.7$ N. [3]

10. (a) $k = \frac{F}{x} = \frac{10}{0.04} = 250$ N m⁻¹. [1] (b) $EPE = \frac{1}{2}kx^2 = 0.5 \times 250 \times (0.04)^2 = 0.20$ J. [2] (c) New extension for 20 N: $x_2 = \frac{20}{250} = 0.08$ m. Work done = Change in EPE = $\frac{1}{2}k(x_2^2 - x_1^2)$ $= 0.5 \times 250 \times (0.08^2 - 0.04^2)$ $= 125 \times (0.0064 - 0.0016) = 125 \times 0.0048 = 0.60$ J. [2]

Section B: Data-Based and Contextual Questions

11. (a) Squaring the equation: $T^2 = \frac{4\pi^2}{g} L$. This is in the form $y = mx$. The gradient $m = \frac{4\pi^2}{g}$. Thus $g = \frac{4\pi^2}{\text{gradient}}$. [2] (b) Gradient calculation from points (e.g., (0,0) and (1.00, 4.00)): $m = \frac{4.00 - 0}{1.00 - 0} = 4.00$ s² m⁻¹. (Accept range 3.9–4.1 based on best fit). [2] (c) $g = \frac{4\pi^2}{4.00} = \pi^2 \approx 9.87$ m s⁻². [2] (d) No effect. [1] The period of a simple pendulum is independent of mass ($T = 2\pi \sqrt{L/g}$ does not contain $m$). [1]

12. (a) Frictional force between tires and road. [1] (b) Max friction provides centripetal force: $\mu mg = \frac{mv^2}{r}$ $v^2 = \mu gr = 0.80 \times 9.81 \times 50 = 392.4$ $v = \sqrt{392.4} = 19.8$ m s⁻¹. [3] (c) Banking allows the normal contact force to have a horizontal component. [1] This horizontal component contributes to the centripetal force, reducing the reliance on friction and allowing higher speeds before skidding occurs. [1]

13. (a) Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B) v$ $0.50(0.80) + 0 = (0.50 + 0.30) v$ $0.40 = 0.80 v$ $v = 0.50$ m s⁻¹. [3] (b) Initial KE = $\frac{1}{2}(0.50)(0.80)^2 = 0.16$ J. Final KE = $\frac{1}{2}(0.80)(0.50)^2 = 0.10$ J. Loss = $0.16 - 0.10 = 0.06$ J. [3] (c) Inelastic collision. [1]

14. (a) Weight $W = mg = 5.0 \times 10^4 \times 9.81 = 4.905 \times 10^5$ N. [1] (b) Resultant Force $F_{net} = \text{Thrust} - \text{Weight}$ $F_{net} = 8.0 \times 10^5 - 4.905 \times 10^5 = 3.095 \times 10^5$ N. [2] (c) $a = \frac{F_{net}}{m} = \frac{3.095 \times 10^5}{5.0 \times 10^4} = 6.19$ m s⁻². [2] (d) As the rocket burns fuel, its mass $m$ decreases. [1] Since $F_{net}$ is roughly constant (or increases as drag decreases/thrust stays same), $a = F_{net}/m$ increases as $m$ decreases. [1]

15. (a) At top: $T + mg = \frac{mv^2}{r}$ $T = \frac{0.20(4.0)^2}{0.80} - 0.20(9.81)$ $T = \frac{3.2}{0.8} - 1.962 = 4.0 - 1.962 = 2.04$ N. [3] (b) Minimum speed when $T=0$: $mg = \frac{mv_{min}^2}{r} \Rightarrow v_{min} = \sqrt{gr}$ $v_{min} = \sqrt{9.81 \times 0.80} = \sqrt{7.848} = 2.80$ m s⁻¹. [2] (c) At bottom, Tension acts upwards, Weight downwards. $T - mg = \frac{mv^2}{r} \Rightarrow T = \frac{mv^2}{r} + mg$. At top, $T + mg = \frac{mv^2}{r} \Rightarrow T = \frac{mv^2}{r} - mg$. Thus, tension is greater at the bottom by $2mg$ (assuming constant $v$). [2]

Section C: Long Structured Questions

16. (a) Diagram: Weight (vertical down), Normal Reaction (perpendicular to slope), Friction (up the slope). [2] (b) Component down slope = $mg \sin \theta = 70 \times 9.81 \times \sin 20^\circ = 234.6$ N. [2] (c) Normal Reaction $R = mg \cos \theta = 70 \times 9.81 \times \cos 20^\circ = 643.8$ N. Friction $F = \mu R = 0.10 \times 643.8 = 64.4$ N. [2] (d) Resultant force down slope = $234.6 - 64.4 = 170.2$ N. $a = \frac{170.2}{70} = 2.43$ m s⁻². [3] (e) $v^2 = u^2 + 2as = 0 + 2(2.43)(100) = 486$. $v = \sqrt{486} = 22.0$ m s⁻¹. [2] (f) On horizontal: Friction $F = \mu mg = 0.10 \times 70 \times 9.81 = 68.67$ N. Deceleration $a = -\frac{68.67}{70} = -0.981$ m s⁻². $0 = v^2 + 2as \Rightarrow 0 = 22.0^2 + 2(-0.981)s$. $s = \frac{484}{1.962} = 247$ m. [3]

17. (a) Total mass $M = 1000$ kg. (i) $T = Mg = 1000 \times 9.81 = 9810$ N. [2] (ii) $T - Mg = Ma \Rightarrow T = M(g+a) = 1000(9.81 + 1.5) = 11310$ N. [3] (iii) Moving down with deceleration means acceleration is upwards. Same as (ii). $T = 11310$ N. [3] (b) For passenger: $R - mg = ma$. $R = m(g+a) = 80(9.81 + 1.5) = 80(11.31) = 904.8$ N. [3] (c) If cable breaks, $a = g$ downwards. Normal reaction $R = m(g-a) = m(g-g) = 0$. [1] The passenger exerts no force on the scale/floor, creating the sensation of weightlessness. [1]

18. (a) Upward journey: Gravity acts down, Air resistance acts down (opposing motion). Resultant force $F = mg + bv$. [1] As $v$ decreases, air resistance decreases, so resultant force decreases. Acceleration decreases (but is always $> g$ initially and approaches $g$ as $v \to 0$). [1] (b) Time up < Time down. [1] Energy is lost to air resistance throughout the flight. [1] Therefore, at any given height, the speed on the way down is less than the speed on the way up. Since average speed is lower on the way down for the same distance, time taken is longer. [1] (c) Graph: Starts at $+15$. Curve concave up (decreasing gradient magnitude) to $v=0$. Then curve becomes negative, asymptotic to terminal velocity (straight line slope $g$ if $v$ small, but generally curved). Final $v$ magnitude $< 15$. [3] (d) Work is done against air resistance. [1] Mechanical energy is dissipated as heat. [1] Thus, final KE < initial KE, so final speed < initial speed.

19. (a) The wall is smooth, so there is no friction. [1] The only force is normal to the surface. (b) Let $N_w$ be normal force from wall. Ladder length $L=5$. Weight acts at $L/2 = 2.5$ m from foot. Horizontal distance of CG from foot = $2.5 \cos 60^\circ = 1.25$ m. Vertical height of top from foot = $5 \sin 60^\circ = 4.33$ m. Moments about foot: Clockwise (Weight): $200 \times 1.25 = 250$ Nm. Anticlockwise (Wall Normal): $N_w \times 4.33$. $N_w \times 4.33 = 250 \Rightarrow N_w = 57.7$ N. [4] (c) Horizontal equilibrium: Friction $F = N_w = 57.7$ N. [2] (d) Vertical equilibrium: Normal reaction from ground $N_g = \text{Weight} = 200$ N. Limiting friction $F_{max} = \mu N_g$. To prevent slipping, $F \le \mu N_g$. $57.7 \le \mu (200)$. $\mu \ge \frac{57.7}{200} = 0.29$. [3]

20. (a) Consider velocity vectors $\vec{v}_1$ and $\vec{v}_2$ at times $t$ and $t+\Delta t$. Angle between radii is $\Delta \theta$. Angle between velocity vectors is also $\Delta \theta$. Change in velocity $\Delta v \approx v \Delta \theta$ (for small $\theta$). Acceleration $a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} = v \frac{d\theta}{dt}$. Since $v = r \omega$ and $\omega = \frac{d\theta}{dt}$, then $\frac{d\theta}{dt} = \frac{v}{r}$. $a = v (\frac{v}{r}) = \frac{v^2}{r}$. [3] (b) Tension provides the centripetal force. $T = F_c = ma = \frac{mv^2}{r}$. [1] (c) $T_1 = \frac{mv^2}{r}$. $T_2 = \frac{m(2v)^2}{r/2} = \frac{m(4v^2)}{r/2} = \frac{8mv^2}{r} = 8 T_1$. Tension increases by a factor of 8. [2] (d) The string has mass, so each segment of the string requires centripetal force. [1] The tension must support the centripetal force for the particle PLUS the centripetal force for the portion of the string further out. Thus, tension is maximum at the fixed point and decreases towards the particle. [1]