A Level H1 Physics Practice Paper 5

TuitionGoWhere Exam Practice (AI) - Physics H1 A-Level

PRACTICE PAPER - VERSION 5 - ANSWER KEY

Subject: Physics
Level: H1 (8867)
Total Marks: 60

Section A: Structured Questions

1. State the principle of conservation of linear momentum. [2]

• Answer: In a closed system (or isolated system) [B1], the total momentum before interaction is equal to the total momentum after interaction (provided no external forces act) [B1].
• Note: Accept "Total momentum remains constant if net external force is zero."

2. Car motion. [3]

• (a) Resistive forces: Since speed is constant, acceleration is zero. By Newton's First Law, driving force = resistive force.
  • Answer: $800 \text{ N}$ [B1]
• (b) Power developed: $P = Fv$
  • Answer: $P = 800 \times 25 = 20,000 \text{ W}$ (or $20 \text{ kW}$) [M1, A1]

3. Vertical projectile. [4]

• (a) Max height: Using $v^2 = u^2 + 2as$. At max height, $v=0$. $u=15$, $a=-9.81$.
  • $0 = 15^2 + 2(-9.81)s$
  • $s = \frac{225}{19.62} = 11.47 \text{ m}$
  • Answer: $11.5 \text{ m}$ (3 s.f.) [M1, A1]
• (b) Graph: Straight line with negative gradient starting at $+15$ on y-axis, crossing t-axis, ending at $-15$.
  • Answer: Linear slope downwards [B1]; Intercepts/correct shape indicated [B1].

4. Inelastic collision. [3]

• Conservation of momentum: $m_A u_A + m_B u_B = (m_A + m_B)v$
  • $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0)v$
  • $6.0 = 3.0v$
  • $v = 2.0 \text{ m s}^{-1}$
  • Answer: $2.0 \text{ m s}^{-1}$ to the right [M1, M1, A1]

5. Plank equilibrium. [5]

• (a) Free-body diagram:
  • Weight of plank ($200 \text{ N}$) acting downwards at center ($2.0 \text{ m}$ from A). [B1]
  • Weight of boy ($400 \text{ N}$) acting downwards at distance $x$. [B1]
  • Reaction at X ($R_X$) upwards at A. [B1]
  • Reaction at Y ($R_Y$) upwards at $3.0 \text{ m}$ from A. [B1]
  • (Note: Award marks for correct labels and directions)
• (b) Max $x$ before tipping:
  • Condition for tipping: Reaction at X becomes zero ($R_X = 0$).
  • Take moments about support Y.
  • Clockwise moment = Anti-clockwise moment.
  • Weight of plank acts at $2.0 \text{ m}$ from A. Support Y is at $3.0 \text{ m}$ from A. Distance from Y to center = $1.0 \text{ m}$.
  • Moment of plank weight about Y: $200 \times 1.0 = 200 \text{ N m}$ (Anti-clockwise).
  • Moment of boy about Y: $400 \times (3.0 - x)$ (Clockwise, assuming boy is to the right of Y? No, boy is at $x$. If $x > 3$, moment is clockwise. If $x < 3$, moment is anti-clockwise. Wait. To tip over Y, the boy must be to the right of Y, creating a clockwise moment that overcomes the plank's weight moment? No. The plank weight creates a moment trying to rotate it back down. The boy creates a moment trying to tip it.
  • Let's check positions: A(0), Center(2), Y(3), B(4).
  • Pivot at Y.
  • Plank weight ($200 \text{ N}$) is at $2 \text{ m}$ from A. Distance to Y = $1 \text{ m}$ to the left. Moment = $200 \times 1 = 200 \text{ Nm}$ (Counter-Clockwise).
  • Boy ($400 \text{ N}$) is at $x$. To tip, boy must be to the right of Y ($x > 3$). Distance to Y = $x - 3$. Moment = $400(x - 3)$ (Clockwise).
  • Equilibrium limit: $200 = 400(x - 3)$
  • $0.5 = x - 3$
  • $x = 3.5 \text{ m}$
  • Answer: $3.5 \text{ m}$ [M1, M1, A1]

Section B: Data Interpretation and Problem Solving

6. Falling sphere in oil. [6]

• (a) Shape explanation:
  • Initially, velocity is zero, so drag is zero. Net force = weight. Acceleration is max ($g$). [B1]
  • As velocity increases, drag force increases. Net force ($W - D$) decreases, so acceleration decreases. [B1]
  • Eventually, drag equals weight. Net force is zero. Acceleration is zero. Velocity becomes constant (terminal velocity). [B1]
• (b) Drag at terminal velocity:
  • At terminal velocity, $D = W = mg$.
  • $D = 0.050 \times 9.81 = 0.4905 \text{ N}$.
  • Answer: $0.49 \text{ N}$ [M1, A1]
• (c) Initial acceleration:
  • At $t=0$, $v=0$, so Drag $= 0$.
  • $F_{net} = W$. $ma = mg \Rightarrow a = g$.
  • Answer: $9.81 \text{ m s}^{-2}$ [B1]

7. Block on inclined plane. [10]

• (a) Free-body diagram:
  • Weight ($mg$) vertically down. [B1]
  • Normal reaction ($N$) perpendicular to slope. [B1]
  • Friction ($f$) down the slope (opposing motion). [B1]
  • Applied force ($F$) up the slope. [B1]
• (b) Calculate $F$:
  • Resolve forces parallel to slope. Constant speed $\Rightarrow a=0$.
  • $F = mg \sin(30^\circ) + f$
  • $F = (5.0)(9.81)(0.5) + 12$
  • $F = 24.525 + 12 = 36.525 \text{ N}$
  • Answer: $36.5 \text{ N}$ (3 s.f.) [M1, M1, A1]
• (c) Work and Energy:
  • (i) Work done by $F$: $W = F \times d = 36.525 \times 4.0 = 146.1 \text{ J}$.
    • Answer: $146 \text{ J}$ [M1, A1]
  • (ii) Gain in GPE: $\Delta PE = mgh$. Height $h = d \sin(30^\circ) = 4.0 \times 0.5 = 2.0 \text{ m}$.
    • $\Delta PE = 5.0 \times 9.81 \times 2.0 = 98.1 \text{ J}$.
    • Answer: $98.1 \text{ J}$ [M1, A1]
  • (iii) Explanation: Work done by $F$ is used to increase GPE AND to do work against friction (dissipated as heat). [B1]

8. Projectile motion. [7]

• (a) Components:
  • $u_x = 40 \cos(30^\circ) = 34.64 \text{ m s}^{-1}$
  • $u_y = 40 \sin(30^\circ) = 20.0 \text{ m s}^{-1}$
  • Answer: $u_x = 34.6 \text{ m s}^{-1}$, $u_y = 20.0 \text{ m s}^{-1}$ [M1, A1]
• (b) Time of flight:
  • Consider vertical motion. Displacement $s_y = 0$ (returns to ground).
  • $s_y = u_y t + \frac{1}{2} a_y t^2$
  • $0 = 20t - 4.905t^2$
  • $t(20 - 4.905t) = 0$
  • $t = 0$ (start) or $t = \frac{20}{4.905} = 4.077 \text{ s}$
  • Answer: $4.08 \text{ s}$ [M1, M1, A1]
• (c) Horizontal range:
  • $R = u_x \times t = 34.64 \times 4.077 = 141.2 \text{ m}$
  • Answer: $141 \text{ m}$ [M1, A1]

9. Ice skaters. [7]

• (a) Velocity of Q:
  • Conservation of momentum. Initial $P = 0$.
  • $m_P v_P + m_Q v_Q = 0$
  • $(60)(-2.0) + (80)(v_Q) = 0$ (Taking right as positive, P moves left)
  • $-120 + 80 v_Q = 0$
  • $v_Q = \frac{120}{80} = 1.5 \text{ m s}^{-1}$
  • Answer: $1.5 \text{ m s}^{-1}$ to the right [M1, M1, A1]
• (b) Total KE:
  • $KE_P = \frac{1}{2}(60)(2.0)^2 = 120 \text{ J}$
  • $KE_Q = \frac{1}{2}(80)(1.5)^2 = 90 \text{ J}$
  • Total $KE = 120 + 90 = 210 \text{ J}$
  • Answer: $210 \text{ J}$ [M1, M1, A1]
• (c) Source of energy:
  • Chemical potential energy from the skaters' muscles / Internal energy. [B1]

10. Crane lifting load. [7]

• (a) Tension during acceleration:
  • $T - mg = ma$
  • $T = m(g + a) = 500(9.81 + 0.50) = 500(10.31)$
  • $T = 5155 \text{ N}$
  • Answer: $5160 \text{ N}$ (3 s.f.) [M1, M1, A1]
• (b) Height gained:
  • $v^2 = u^2 + 2as$
  • $4.0^2 = 0 + 2(0.50)s$
  • $16 = 1.0 s \Rightarrow s = 16 \text{ m}$
  • Answer: $16 \text{ m}$ [M1, A1]
• (c) Power at constant speed:
  • At constant speed, $T = mg = 500 \times 9.81 = 4905 \text{ N}$.
  • $P = Fv = T v = 4905 \times 4.0 = 19620 \text{ W}$
  • Answer: $19.6 \text{ kW}$ [M1, A1]

11. Spring energy. [7]

• (a) Spring constant:
  • $F = kx \Rightarrow 10 = k(0.05)$
  • $k = \frac{10}{0.05} = 200 \text{ N m}^{-1}$
  • Answer: $200 \text{ N m}^{-1}$ [M1, A1]
• (b) Elastic PE:
  • $EPE = \frac{1}{2} k x^2 = \frac{1}{2}(200)(0.05)^2$
  • $EPE = 100 \times 0.0025 = 0.25 \text{ J}$
  • Answer: $0.25 \text{ J}$ [M1, A1]
• (c) Launch speed:
  • $EPE = KE \Rightarrow 0.25 = \frac{1}{2} m v^2$
  • $0.25 = \frac{1}{2}(0.10)v^2$
  • $0.5 = 0.10 v^2 \Rightarrow v^2 = 5$
  • $v = \sqrt{5} = 2.236 \text{ m s}^{-1}$
  • Answer: $2.24 \text{ m s}^{-1}$ [M1, M1, A1]

12. Circular motion. [6]

• (a) Centripetal force source:
  • Friction between tires and road. [B1]
• (b) Max speed:
  • $F_c = \frac{mv^2}{r}$
  • $F_{friction} = \frac{mv^2}{r} \Rightarrow 8000 = \frac{1000 v^2}{50}$
  • $8000 = 20 v^2$
  • $v^2 = 400 \Rightarrow v = 20 \text{ m s}^{-1}$
  • Answer: $20 \text{ m s}^{-1}$ [M1, M1, A1]
• (c) Skidding explanation:
  • Required centripetal force exceeds maximum static friction. The car cannot maintain the circular path and moves in a straighter line (tangentially/outwards) relative to the curve. [B1, B1]

13. Impact momentum. [7]

• (a) Speed before impact:
  • $v^2 = u^2 + 2as = 0 + 2(9.81)(2.0) = 39.24$
  • $v = \sqrt{39.24} = 6.264 \text{ m s}^{-1}$
  • Answer: $6.26 \text{ m s}^{-1}$ [M1, A1]
• (b) Speed after impact:
  • Rebound height $1.5 \text{ m}$. At top, $v=0$.
  • $0 = u_{rebound}^2 - 2(9.81)(1.5)$
  • $u_{rebound} = \sqrt{29.43} = 5.425 \text{ m s}^{-1}$
  • Answer: $5.43 \text{ m s}^{-1}$ [M1, A1]
• (c) Change in momentum:
  • Take Up as positive.
  • $p_{initial} = 0.20 \times (-6.264) = -1.253 \text{ N s}$
  • $p_{final} = 0.20 \times (+5.425) = +1.085 \text{ N s}$
  • $\Delta p = p_f - p_i = 1.085 - (-1.253) = 2.338 \text{ N s}$
  • Answer: $2.34 \text{ N s}$ (upwards) [M1, M1, A1]

14. Ladder moments. [4]

• (a) Smooth wall:
  • Smooth surface cannot exert friction. Therefore, the reaction force is perpendicular to the surface (horizontal). [B1]
• (b) Derivation:
  • Take moments about the base (point of contact with ground).
  • Let $R_w$ be the horizontal force from the wall.
  • Moment of $R_w$: Force $\times$ perpendicular distance. Vertical height of contact = $L \sin \theta$.
  • Moment = $R_w (L \sin \theta)$ (Clockwise/Anti-clockwise depending on side, let's say CW).
  • Moment of Weight $W$: Acts at center ($L/2$). Perpendicular distance from base = $\frac{L}{2} \cos \theta$.
  • Moment = $W (\frac{L}{2} \cos \theta)$ (Opposite direction).
  • Equilibrium: $R_w L \sin \theta = W \frac{L}{2} \cos \theta$
  • $R_w = \frac{W \cos \theta}{2 \sin \theta} = \frac{W}{2 \tan \theta}$
  • Answer: $R_w = \frac{W}{2 \tan \theta}$ [M1, M1, A1]

15. Rocket launch. [5]

• (a) Initial acceleration:
  • $F_{net} = Thrust - Weight$
  • $F_{net} = 15000 - (1000)(9.81) = 15000 - 9810 = 5190 \text{ N}$
  • $a = \frac{F_{net}}{m} = \frac{5190}{1000} = 5.19 \text{ m s}^{-2}$
  • Answer: $5.19 \text{ m s}^{-2}$ [M1, M1, A1]
• (b) Effect of air resistance:
  • As speed increases, air resistance (drag) increases. [B1]
  • This reduces the net upward force ($Thrust - Weight - Drag$), so acceleration decreases. [B1]

16. Curved track and friction. [5]

• (a) Speed at bottom:
  • Conservation of Energy: $PE_{top} = KE_{bottom}$
  • $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$
  • $v = \sqrt{2(9.81)(3.0)} = \sqrt{58.86} = 7.672 \text{ m s}^{-1}$
  • Answer: $7.67 \text{ m s}^{-1}$ [M1, A1]
• (b) Distance on rough surface:
  • Work done by friction = Loss in KE
  • $f_k d = \frac{1}{2}mv^2$
  • Friction force $f_k = \mu N = \mu mg = 0.40 \times 2.0 \times 9.81 = 7.848 \text{ N}$
  • $KE_{bottom} = \frac{1}{2}(2.0)(7.672)^2 = 58.86 \text{ J}$ (Matches initial PE)
  • $7.848 d = 58.86$
  • $d = \frac{58.86}{7.848} = 7.5 \text{ m}$
  • Answer: $7.5 \text{ m}$ [M1, M1, A1]

17. Vector addition. [5]

• (a) Horizontal component:
  • $F_{1x} = 10 \text{ N}$
  • $F_{2x} = 10 \cos(60^\circ) = 5 \text{ N}$
  • $R_x = 10 + 5 = 15 \text{ N}$
  • Answer: $15 \text{ N}$ [M1, A1]
• (b) Vertical component:
  • $F_{1y} = 0$
  • $F_{2y} = 10 \sin(60^\circ) = 8.66 \text{ N}$
  • $R_y = 8.66 \text{ N}$
  • Answer: $8.66 \text{ N}$ [B1]
• (c) Resultant magnitude:
  • $R = \sqrt{R_x^2 + R_y^2} = \sqrt{15^2 + 8.66^2} = \sqrt{225 + 75} = \sqrt{300}$
  • $R = 17.32 \text{ N}$
  • Answer: $17.3 \text{ N}$ [M1, A1]

18. Satellite orbit. [3]

• (a) Direction of force:
  • Towards the center of the Earth. [B1]
• (b) Acceleration explanation:
  • Velocity is a vector (speed and direction). [B1]
  • Although speed is constant, the direction of motion is constantly changing. Therefore, velocity is changing, which means there is acceleration. [B1]

19. Free fall experiment. [4]

• (a) Equation:
  • $h = \frac{1}{2}gt^2$ [B1]
• (b) Gradient:
  • Graph of $h$ (y) vs $t^2$ (x). Equation $y = (\frac{1}{2}g)x$.
  • Gradient = $\frac{1}{2}g$ [B1]
• (c) Systematic error:
  • Example: Delay in timer starting (electromagnet release time). [B1]
  • Effect: Measured time $t$ is larger than actual fall time. Calculated $g$ will be smaller than actual $g$ (since $g \propto 1/t^2$). [B1]
  • Alternative: Air resistance. Effect: $g$ calculated is lower.

20. Velocity-time graph journey. [7]

• (a) Sketch:
  • 0-10s: Straight line from $(0,0)$ to $(10,20)$. [B1]
  • 10-30s: Horizontal line at $v=20$. [B1]
  • 30-35s: Straight line from $(30,20)$ to $(35,0)$. [B1] (Shape correct)
• (b) Total distance:
  • Area under graph.
  • Area 1 (Triangle): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
  • Area 2 (Rectangle): $20 \times 20 = 400 \text{ m}$
  • Area 3 (Triangle): $\frac{1}{2} \times 5 \times 20 = 50 \text{ m}$
  • Total = $100 + 400 + 50 = 550 \text{ m}$
  • Answer: $550 \text{ m}$ [M1, M1, A1]
• (c) Average speed:
  • Avg Speed = Total Distance / Total Time
  • Total Time = $10 + 20 + 5 = 35 \text{ s}$
  • Avg Speed = $\frac{550}{35} = 15.71 \text{ m s}^{-1}$
  • Answer: $15.7 \text{ m s}^{-1}$ [M1, A1]