A Level H1 Physics Practice Paper 2

A-Level Physics H1 Practice Paper - Mechanics (Version 2 of 5)

Marking Scheme and Answer Key

Total Marks: 60

Section A

1. Principle of Conservation of Linear Momentum

• Answer: In a closed/isolated system [B1], the total linear momentum remains constant (or is conserved) provided no external resultant force acts on the system [B1].
• Marks: 2

2. Ball Impact

• (a) Change in Momentum
  • Take upward as positive.
  • Initial velocity $u = -8.0 \text{ m s}^{-1}$, Final velocity $v = +6.0 \text{ m s}^{-1}$.
  • $\Delta p = m(v - u)$ [M1]
  • $\Delta p = 0.15(6.0 - (-8.0)) = 0.15(14.0)$ [M1]
  • $\Delta p = 2.1 \text{ N s}$ (or $\text{kg m s}^{-1}$) [A1]
  • Marks: 3
• (b) Average Resultant Force
  • $F_{avg} = \frac{\Delta p}{\Delta t}$ [M1]
  • $F_{avg} = \frac{2.1}{0.050} = 42 \text{ N}$ [A1]
  • Marks: 2

3. Plank Equilibrium

• (a) Free-Body Diagram
  • Weight of plank ($200 \text{ N}$) acting downwards at center ($2.0 \text{ m}$ from A). [B1]
  • Weight of student ($600 \text{ N}$) acting downwards at distance $x$. [B1]
  • Reaction force at P ($R_P$) acting upwards at A. [B1]
  • Reaction force at Q ($R_Q$) acting upwards at $3.0 \text{ m}$ from A. [B1] (Max 3 marks, accept clear labels)
  • Marks: 3
• (b) Maximum Distance $x$
  • Condition for tipping: The plank is on the verge of tipping about support Q. Reaction at P becomes zero ($R_P = 0$). [M1]
  • Take moments about Q.
  • Clockwise moment = Anticlockwise moment.
  • Weight of student creates clockwise moment: $600 \times (x - 3.0)$? No, student is at $x$ from A. Q is at $3.0$ m from A.
  • Distance of student from Q = $(3.0 - x)$ if $x < 3$, or $(x - 3.0)$ if $x > 3$.
  • Wait, let's look at the geometry. A is at 0. Q is at 3.0 m (1m from B, length 4m). Center of mass is at 2.0 m.
  • Moments about Q:
    • Plank weight ($200 \text{ N}$) acts at 2.0 m. Distance from Q = $3.0 - 2.0 = 1.0 \text{ m}$. This creates an anticlockwise moment (tending to rotate A down). Moment = $200 \times 1.0 = 200 \text{ Nm}$.
    • Student weight ($600 \text{ N}$) acts at $x$. To tip, student must be to the right of Q? No, if student moves right of Q, they create a clockwise moment. The plank weight creates an anticlockwise moment.
    • For equilibrium limit: Moment of Student = Moment of Plank Weight.
    • $600 \times (x - 3.0) = 200 \times 1.0$ [M1]
    • $600(x - 3.0) = 200$
    • $x - 3.0 = \frac{200}{600} = \frac{1}{3}$
    • $x = 3.0 + 0.333 = 3.33 \text{ m}$ [A1]
  • Alternatively, if the question implies tipping about P (student moves left), but student starts at A? "Maximum value of x" usually implies moving towards the overhang or far support. Here Q is the pivot for tipping if he goes too far right.
  • Answer: $3.33 \text{ m}$ [A1]
  • Marks: 4

4. Car Dynamics

• (a) Acceleration Decrease
  • Resultant force $F_{res} = F_{drive} - F_R = F_{drive} - kv^2$. [M1]
  • As $v$ increases, $F_R$ increases, so $F_{res}$ decreases. Since $a = F_{res}/m$, acceleration decreases. [A1]
  • Marks: 2
• (b) Constant $k$
  • At max speed, $a = 0$, so $F_{drive} = F_R$. [M1]
  • $2400 = k(40)^2$
  • $k = \frac{2400}{1600} = 1.5 \text{ kg m}^{-1}$ (or $\text{N s}^2 \text{m}^{-2}$) [A1]
  • Marks: 3 (1 for condition, 1 for substitution, 1 for answer)
• (c) Power
  • $P = F v$ [M1]
  • $P = 2400 \times 40 = 96,000 \text{ W}$ or $96 \text{ kW}$ [A1]
  • Marks: 2

5. Inelastic Collision

• (a) Common Velocity
  • Conservation of Momentum: $m_X u_X + m_Y u_Y = (m_X + m_Y)v$ [M1]
  • Taking right as positive: $(2.0)(3.0) + (1.0)(-1.0) = (2.0 + 1.0)v$
  • $6.0 - 1.0 = 3.0v$
  • $5.0 = 3.0v \Rightarrow v = 1.67 \text{ m s}^{-1}$ (to the right) [A1]
  • Marks: 3
• (b) Inelastic Proof
  • $KE_{initial} = \frac{1}{2}(2.0)(3.0)^2 + \frac{1}{2}(1.0)(-1.0)^2 = 9.0 + 0.5 = 9.5 \text{ J}$ [M1]
  • $KE_{final} = \frac{1}{2}(3.0)(1.667)^2 = 4.17 \text{ J}$ [M1]
  • $KE_{initial} \neq KE_{final}$ (KE is lost), so collision is inelastic. [A1]
  • Marks: 3

Section B

6. Skydiver Graph

• (a) Motion Description
  • Initially, weight > air resistance, so there is a resultant downward force and acceleration. [B1]
  • As speed increases, air resistance increases, reducing the resultant force and thus acceleration (gradient of graph decreases). [B1]
  • Eventually, air resistance equals weight. Resultant force is zero, acceleration is zero, and terminal velocity is reached. [B1]
  • Marks: 3
• (b) Distance Fallen
  • Distance = Area under v-t graph. [M1]
  • Approximate area (counting squares or trapezium estimate). Graph goes to 50 m/s in 20s. If linear, area = $0.5 \times 20 \times 50 = 500$. But it's curved (concave down). Area is greater than triangle.
  • Let's assume standard shape. Area $\approx 0.7 \times 20 \times 50$? Or count squares.
  • Accept reasonable estimate between $600 \text{ m}$ and $800 \text{ m}$ depending on curve shape provided in visual. Let's assume approx $700 \text{ m}$. [A1]
  • Marks: 2
• (c) Parachute Opening
  • Opening parachute greatly increases surface area, causing a large increase in air resistance. [B1]
  • Air resistance becomes much larger than weight, creating a large upward resultant force (deceleration). [B1]
  • Marks: 2

7. Inclined Plane

• (a) Free-Body Diagram
  • Weight ($mg$) vertically down. [B1]
  • Normal reaction ($N$) perpendicular to slope. [B1]
  • Friction ($F_f$) down the slope (opposing motion up). [B1] (Wait, motion is UP, so friction is DOWN).
  • Applied Force ($F$) up the slope. [B1]
  • Marks: 2 (Accept 2 correct forces for 1 mark, all 4 for 2)
• (b) Force F
  • Constant speed $\Rightarrow$ Equilibrium. Forces up slope = Forces down slope.
  • $F = mg \sin(30^\circ) + F_f$ [M1]
  • $F = (5.0)(9.81)(0.5) + 10$
  • $F = 24.525 + 10 = 34.5 \text{ N}$ [A1]
  • Marks: 3
• (c) Power
  • $P = F v$ [M1]
  • $P = 34.525 \times 2.0 = 69.0 \text{ W}$ (or $69.1 \text{ W}$) [A1]
  • Marks: 2

8. Spring

• (a) Spring Constant
  • $F = kx \Rightarrow k = F/x$ [M1]
  • $k = 10 / 0.04 = 250 \text{ N m}^{-1}$ [A1]
  • Marks: 2
• (b) Elastic Potential Energy
  • $E = \frac{1}{2}kx^2$ or $\frac{1}{2}Fx$ [M1]
  • $E = 0.5 \times 10 \times 0.04 = 0.20 \text{ J}$ [A1]
  • Marks: 2
• (c) Energy Changes
  • At lowest point: Max Elastic PE, Min KE, Min Gravitational PE (relative to equilibrium).
  • Moving up: Elastic PE converts to KE and Gravitational PE.
  • At equilibrium: Max KE.
  • At highest point: Max Gravitational PE, Min Elastic PE (depending on reference), Zero KE.
  • Key point: Continuous interchange between Kinetic Energy, Gravitational Potential Energy, and Elastic Potential Energy. Total energy is conserved. [B1 for each valid transition mention, max 3]
  • Marks: 3

9. Projectile

• (a) Horizontal Component
  • $u_x = 20 \cos(30^\circ) = 17.3 \text{ m s}^{-1}$ [A1]
  • Marks: 1
• (b) Max Height
  • $u_y = 20 \sin(30^\circ) = 10 \text{ m s}^{-1}$. At max height, $v_y = 0$.
  • $v^2 = u^2 + 2as \Rightarrow 0 = 10^2 + 2(-9.81)h$ [M1]
  • $19.62 h = 100 \Rightarrow h = 5.10 \text{ m}$ [A1]
  • Marks: 3
• (c) Time of Flight
  • Time to max height: $v = u + at \Rightarrow 0 = 10 - 9.81t \Rightarrow t = 1.02 \text{ s}$.
  • Total time = $2 \times 1.02 = 2.04 \text{ s}$ [A1]
  • Marks: 2

10. Satellite

• (a) Centripetal Force
  • Gravitational force (or Weight) [B1]
  • Marks: 1
• (b) Free Fall
  • The only force acting on the satellite is gravity. [B1]
  • It is constantly accelerating towards the Earth (changing direction), which is the definition of free fall, even though its tangential velocity keeps it in orbit. [B1]
  • Marks: 2

Section C

11. Car Braking

• (a) Braking Force
  • Work-Energy Principle: Work done by brakes = Change in KE.
  • $F d = \frac{1}{2}mv^2$ [M1]
  • $F(40) = 0.5(1000)(20)^2 = 200,000$
  • $F = \frac{200,000}{40} = 5000 \text{ N}$ [A1]
  • Marks: 3
• (b) New Stopping Distance
  • $F d' = \frac{1}{2}m(v')^2$
  • $5000 d' = 0.5(1000)(40)^2 = 800,000$
  • $d' = \frac{800,000}{5000} = 160 \text{ m}$ [A1]
  • Marks: 2
• (c) Relationship
  • KE is proportional to $v^2$. [B1]
  • Since stopping force is constant, stopping distance is proportional to KE, and thus proportional to $v^2$. Doubling speed quadruples the distance. [B1]
  • Marks: 2

12. Elastic Collision Spheres

• (a) Speed before impact
  • Conservation of Energy: $mgh = \frac{1}{2}mv^2$ [M1]
  • $v = \sqrt{2gh}$ [A1]
  • Marks: 2
• (b) Elastic Conditions
  • Total kinetic energy is conserved. [B1]
  • Relative speed of approach equals relative speed of separation. [B1]
  • Marks: 2
• (c) Rebound Explanation
  • Conservation of Momentum: $m u_A = m v_A + 2m v_B$.
  • Conservation of KE (or relative speed): $u_A = v_B - v_A$ (since B is at rest).
  • Solving these shows $v_A$ is negative (rebound) because $m_A < m_B$. Specifically, for elastic collision with target at rest, $v_A = u_A \frac{m_A - m_B}{m_A + m_B}$. Since $m_A < m_B$, the numerator is negative. [M1 for logic, A1 for conclusion]
  • Marks: 3

13. Lift

• (a) Tension
  • Total mass $M = 500 + 70 = 570 \text{ kg}$.
  • $T - Mg = Ma \Rightarrow T = M(g + a)$ [M1]
  • $T = 570(9.81 + 1.5) = 570(11.31) = 6447 \text{ N}$ [A1]
  • Marks: 3
• (b) Normal Reaction
  • For passenger: $R - mg = ma \Rightarrow R = m(g + a)$ [M1]
  • $R = 70(11.31) = 792 \text{ N}$ [A1]
  • Marks: 2
• (c) Constant Velocity
  • Acceleration is zero. [B1]
  • $R = mg$. The normal reaction decreases to equal the passenger's weight ($70 \times 9.81 = 687 \text{ N}$). [B1]
  • Marks: 2

14. River Boat

• (a) Resultant Velocity
  • $v_{res} = \sqrt{3.0^2 + 5.0^2} = \sqrt{9 + 25} = \sqrt{34} = 5.83 \text{ m s}^{-1}$ [M1]
  • Direction: $\theta = \tan^{-1}(3/5) = 31.0^\circ$ East of North. [A1]
  • Marks: 3
• (b) Time to Cross
  • Time depends on velocity component perpendicular to bank ($5.0 \text{ m s}^{-1}$).
  • $t = \frac{100}{5.0} = 20 \text{ s}$ [A1]
  • Marks: 2
• (c) Distance Downstream
  • $d = v_{river} \times t = 3.0 \times 20 = 60 \text{ m}$ [A1]
  • Marks: 1

15. Friction Experiment

• (a) Graph Sketch
  • Straight line through origin. Positive gradient. [B1]
  • Marks: 1
• (b) Friction-Compensated
  • The track is tilted slightly so that the component of weight down the slope balances friction. [B1]
  • This ensures that the resultant force on the trolley is equal to the applied force only, allowing $F=ma$ to be tested directly without subtracting friction. [B1]
  • Marks: 2
• (c) Uncompensated Graph
  • Line does not pass through origin. [B1]
  • It has an x-intercept (force required to overcome static friction before acceleration starts). Gradient remains same (mass unchanged). [B1]
  • Marks: 2

16. Ice Block

• (a) Speed at Bottom
  • $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh}$ [M1]
  • $v = \sqrt{2(9.81)(5.0)} = \sqrt{98.1} = 9.90 \text{ m s}^{-1}$ [A1]
  • Marks: 2
• (b) Frictional Force
  • Work done by friction = Loss in KE.
  • $F_f d = \frac{1}{2}mv^2$ [M1]
  • $F_f (10) = 0.5(2.0)(9.90)^2 = 98.1 \text{ J}$
  • $F_f = 9.81 \text{ N}$ [A1]
  • Marks: 3

17. Impulse Definition

• Answer: Impulse is the product of the average force and the time interval over which it acts ($F \Delta t$). Alternatively, it is the change in momentum. [B1]
• Marks: 1

18. Tennis Ball

• (a) Impulse
  • Impulse = Area under Force-Time graph.
  • Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$ [M1]
  • $I = 0.5 \times 0.01 \times 200 = 1.0 \text{ N s}$ [A1]
  • Marks: 2
• (b) Final Speed
  • $I = \Delta p = m(v - u)$
  • $1.0 = 0.06(v - 0)$ [M1]
  • $v = \frac{1.0}{0.06} = 16.7 \text{ m s}^{-1}$ [A1]
  • Marks: 2

19. Ladder

• (a) No Friction at Wall
  • The wall is smooth. [B1]
  • Marks: 1
• (b) Friction Direction at Ground
  • Towards the wall (to oppose the tendency of the ladder to slip outwards). [B1]
  • Marks: 1
• (c) Person Climbing
  • As the person climbs up, the moment of their weight about the base increases. [M1]
  • To maintain rotational equilibrium, the normal reaction from the wall must increase. Consequently, the horizontal frictional force at the ground (which balances the wall's normal reaction) must increase. [A1]
  • Marks: 2

20. Kinematics Calculus

• (a) Velocity at t=3
  • $v = \frac{ds}{dt} = 4t - 8$ [M1]
  • At $t=3$, $v = 4(3) - 8 = 4 \text{ m s}^{-1}$ [A1]
  • Marks: 2
• (b) Instantaneously at Rest
  • $v = 0 \Rightarrow 4t - 8 = 0$ [M1]
  • $t = 2.0 \text{ s}$ [A1]
  • Marks: 2
• (c) Total Distance
  • Particle changes direction at $t=2$.
  • Distance 1 ($t=0$ to $2$): $s(0) = 5$, $s(2) = 2(4) - 16 + 5 = -3$. Distance = $|-3 - 5| = 8 \text{ m}$. [M1]
  • Distance 2 ($t=2$ to $4$): $s(4) = 2(16) - 32 + 5 = 5$. Distance = $|5 - (-3)| = 8 \text{ m}$. [M1]
  • Total Distance = $8 + 8 = 16 \text{ m}$ [A1]
  • Marks: 3