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

TuitionGoWhere Exam Practice (AI)

Subject: Physics H2 (9478) Level: A-Level Paper: Practice Paper 1 (Mechanics) Version: 1 of 5 Duration: 1 hour 30 minutes Total Marks: 60

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of 20 questions on the topic of Mechanics.
Answer all questions in the spaces provided.
Show all working for calculation questions. Marks are awarded for correct method and final answer.
You may use a scientific calculator.
The number of marks is given in brackets [ ] at the end of each question or part question.
Take g = 9.81 m s⁻² unless otherwise stated.

Section A: Structured Response (Questions 1–8)

Answer all questions in this section.

1. State the principle of conservation of linear momentum.

[2 marks]

2. A ball of mass 0.50 kg undergoes simple harmonic motion with amplitude 0.040 m and angular frequency 2.5 rad s⁻¹.

Calculate the maximum acceleration of the ball.

[2 marks]

3. A student performs an experiment to verify the principle of conservation of momentum using two gliders on a linear air track. One glider of mass 0.30 kg moves with velocity 0.80 m s⁻¹ and collides with a stationary glider of mass 0.50 kg. After the collision, the gliders stick together.

(a) Calculate the velocity of the combined gliders after the collision.

[2 marks]

(b) State one precaution the student should take to improve the accuracy of this experiment.

[1 mark]

4. A car of mass 1200 kg accelerates uniformly from rest to 25 m s⁻¹ in 8.0 s on a straight, level road.

Calculate the average power developed by the engine during this acceleration. Ignore resistive forces.

[3 marks]

5. A stone is projected horizontally from the top of a vertical cliff of height 45 m. The stone lands in the sea at a horizontal distance of 60 m from the base of the cliff.

Calculate the initial speed of projection.

[3 marks]

6. A particle of mass 0.20 kg moves in a horizontal circle of radius 0.50 m on a smooth table. The particle is attached to a string that passes through a hole in the centre of the table and is connected to a hanging mass of 0.30 kg, as shown in Fig. 6.1.

The system is in equilibrium with the hanging mass stationary.

(a) State the force that provides the centripetal force for the particle on the table.

[1 mark]

(b) Calculate the angular speed of the particle.

[3 marks]

7. A satellite of mass 500 kg orbits the Earth in a circular orbit of radius 7.0 × 10⁶ m. The mass of the Earth is 6.0 × 10²⁴ kg and the gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².

(a) Calculate the gravitational force acting on the satellite.

[2 marks]

(b) Hence, calculate the orbital speed of the satellite.

[2 marks]

8. A block of mass 2.0 kg slides down a rough inclined plane from rest. The plane is inclined at 30° to the horizontal and has a length of 3.0 m. The coefficient of kinetic friction between the block and the plane is 0.25.

Calculate the speed of the block at the bottom of the incline.

[4 marks]

Section B: Calculation and Application (Questions 9–15)

Answer all questions in this section.

9. A ball of mass 0.15 kg is thrown vertically upwards with an initial speed of 12 m s⁻¹ from a point 1.5 m above the ground. Air resistance is negligible.

(a) Calculate the maximum height reached by the ball above the ground.

[3 marks]

(b) Calculate the speed of the ball when it hits the ground.

[2 marks]

10. A lorry of mass 8000 kg travelling at 15 m s⁻¹ collides with a stationary car of mass 1200 kg. After the collision, the lorry and car move together.

(a) Calculate the common velocity immediately after the collision.

[2 marks]

(b) Calculate the kinetic energy lost in the collision.

[2 marks]

[1 mark]

11. A conical pendulum consists of a bob of mass 0.25 kg attached to a string of length 0.80 m. The bob moves in a horizontal circle such that the string makes an angle of 30° with the vertical.

(a) Draw a free-body diagram showing the forces acting on the bob.

[2 marks]

(b) Calculate the tension in the string.

[2 marks]

[3 marks]

12. A spring of spring constant 200 N m⁻¹ is fixed at one end. A mass of 0.50 kg is attached to the free end and pulled horizontally, extending the spring by 0.10 m from its equilibrium position. The mass is then released from rest on a smooth horizontal surface.

(a) Calculate the initial acceleration of the mass.

[2 marks]

(b) Calculate the maximum speed of the mass during its subsequent motion.

[2 marks]

13. Two particles A and B of masses 0.40 kg and 0.60 kg respectively are connected by a light inextensible string that passes over a smooth pulley. Particle A rests on a smooth horizontal table, while particle B hangs freely, as shown in Fig. 13.1.

When the system is released from rest, calculate:

(a) the acceleration of the particles;

[3 marks]

(b) the tension in the string.

[2 marks]

14. A projectile is fired from ground level with an initial speed of 50 m s⁻¹ at an angle of 40° above the horizontal. Air resistance is negligible.

(a) Calculate the time of flight of the projectile.

[2 marks]

(b) Calculate the maximum height reached.

[2 marks]

15. A car of mass 1500 kg travels around a banked circular track of radius 80 m. The track is banked at an angle of 20° to the horizontal. The coefficient of static friction between the tyres and the road is 0.40.

Calculate the maximum speed at which the car can travel around the track without slipping.

[4 marks]

Section C: Data Analysis and Extended Response (Questions 16–20)

Answer all questions in this section.

16. A student investigates the motion of a falling mass using a motion sensor. The mass of 0.50 kg is dropped from rest. The velocity-time graph obtained is shown in Fig. 16.1 (not reproduced here). The graph shows that the velocity increases from 0 to 8.0 m s⁻¹ in the first 1.5 s, then remains constant at 8.0 m s⁻¹.

(a) Explain why the velocity becomes constant after 1.5 s.

[2 marks]

(b) Calculate the acceleration of the mass during the first 1.5 s.

[2 marks]

17. A student performs an experiment to determine the acceleration of free fall, g, using a simple pendulum. The student measures the period T for different lengths L of the pendulum. The relationship is given by:

$T = 2\pi\sqrt{\frac{L}{g}}$

The student plots a graph of T² against L and obtains a straight line with gradient 4.05 s² m⁻¹.

(a) Show that the gradient of the T² against L graph is equal to $\frac{4\pi^2}{g}$ .

[2 marks]

(b) Use the gradient to calculate a value for g.

[2 marks]

(c) The accepted value of g is 9.81 m s⁻². Calculate the percentage difference between the student's value and the accepted value.

[1 mark]

(d) Suggest one reason for the difference, and state how the experiment could be improved.

[2 marks]

18. A particle of mass m moves in a circular path of radius r with constant speed v. The centripetal force F acting on the particle is given by:

$F = \frac{mv^2}{r}$

A student suggests that the centripetal force could also be expressed as $F = km^\alpha v^\beta r^\gamma$ , where k, α, β, and γ are dimensionless constants.

Use dimensional analysis to determine the values of α, β, and γ.

[4 marks]

19. A block of mass 2.0 kg is placed on a rough horizontal surface. A horizontal force F is applied to the block. The variation of the frictional force f with the applied force F is shown in Fig. 19.1 (not reproduced here). The graph shows that f increases linearly with F up to a maximum value of 5.9 N, after which f remains constant at 5.9 N as F increases further.

(a) Explain the shape of the graph.

[3 marks]

(b) Determine the coefficient of static friction between the block and the surface.

[2 marks]

20. A rocket of initial mass 5000 kg, including 3000 kg of fuel, is launched vertically upwards from rest. The exhaust gases are ejected at a constant speed of 2500 m s⁻¹ relative to the rocket, and the fuel is consumed at a constant rate of 50 kg s⁻¹. Assume g remains constant at 9.81 m s⁻² and air resistance is negligible.

(a) State Newton's second law of motion in terms of momentum.

[1 mark]

(b) Calculate the initial thrust provided by the rocket engine.

[2 marks]

(d) Explain, without calculation, how the acceleration of the rocket changes as the fuel is consumed, assuming the thrust remains constant.

[2 marks]

END OF PAPER

Check your work carefully. Ensure all answers are in the spaces provided.

Answers

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

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Mechanics) Version: 1 of 5 Total Marks: 60

Section A: Structured Response (Questions 1–8)

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

Answer: The total momentum of a closed system (or isolated system) remains constant, provided no net external force acts on the system. OR: In the absence of external forces, the total momentum before an interaction equals the total momentum after the interaction.

Marking:

1 mark: Reference to "closed system" or "isolated system" or "no external forces"
1 mark: Statement that total momentum remains constant / total momentum before = total momentum after

2. Calculate the maximum acceleration of the ball. [2 marks]

Answer: $a_{\max} = \omega^2 X_0$ $a_{\max} = (2.5)^2 \times 0.040$ $a_{\max} = 6.25 \times 0.040$ $a_{\max} = 0.25$ m s⁻²

Marking:

1 mark: Correct formula $a_{\max} = \omega^2 X_0$
1 mark: Correct answer 0.25 m s⁻² (accept 0.250 m s⁻²)

3. (a) Calculate the velocity of the combined gliders after the collision. [2 marks]

Answer: By conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(0.30 \times 0.80) + (0.50 \times 0) = (0.30 + 0.50)v$ $0.24 = 0.80v$ $v = 0.30$ m s⁻¹

Marking:

1 mark: Correct application of conservation of momentum
1 mark: Correct answer 0.30 m s⁻¹

(b) State one precaution the student should take to improve the accuracy of this experiment. [1 mark]

Answer (any one):

Ensure the air track is level to eliminate gravitational effects along the track.
Use light gates or motion sensors to measure velocities accurately instead of manual timing.
Ensure the gliders move freely with minimal friction by maintaining clean air holes.
Minimise parallax error when reading measuring instruments.

Marking:

1 mark: Any valid, specific precaution linked to accuracy

4. Calculate the average power developed by the engine. [3 marks]

Answer: Acceleration: $a = \frac{v - u}{t} = \frac{25 - 0}{8.0} = 3.125$ m s⁻² Force: $F = ma = 1200 \times 3.125 = 3750$ N Distance travelled: $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 25) \times 8.0 = 100$ m Work done: $W = Fs = 3750 \times 100 = 375,000$ J Average power: $P = \frac{W}{t} = \frac{375,000}{8.0} = 46,875$ W ≈ 47 kW

Alternative method: $P = Fv_{\text{avg}} = 3750 \times 12.5 = 46,875$ W

Marking:

1 mark: Correct calculation of acceleration or force
1 mark: Correct calculation of work done or use of average velocity
1 mark: Correct answer 4.69 × 10⁴ W or 46.9 kW (accept 47 kW)

5. Calculate the initial speed of projection. [3 marks]

Answer: Vertical motion: $s_y = u_y t + \frac{1}{2}a_y t^2$ $45 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = \frac{45 \times 2}{9.81} = 9.174$ $t = 3.03$ s

Horizontal motion: $s_x = u_x t$ $60 = u_x \times 3.03$ $u_x = 19.8$ m s⁻¹

Marking:

1 mark: Correct calculation of time of flight
1 mark: Correct use of horizontal motion equation
1 mark: Correct answer 19.8 m s⁻¹ (accept 19.8–20.0 depending on rounding)

6. (a) State the force that provides the centripetal force for the particle on the table. [1 mark]

Answer: The tension in the string (which equals the weight of the hanging mass).

Marking:

1 mark: Tension (in the string)

(b) Calculate the angular speed of the particle. [3 marks]

Answer: Tension = weight of hanging mass = $0.30 \times 9.81 = 2.943$ N Centripetal force: $F = m\omega^2 r$ $2.943 = 0.20 \times \omega^2 \times 0.50$ $\omega^2 = \frac{2.943}{0.10} = 29.43$ $\omega = 5.42$ rad s⁻¹

Marking:

1 mark: Correct calculation of tension (2.94 N)
1 mark: Correct use of $F = m\omega^2 r$
1 mark: Correct answer 5.42 rad s⁻¹

7. (a) Calculate the gravitational force acting on the satellite. [2 marks]

Answer: $F = \frac{GMm}{r^2}$ $F = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(500)}{(7.0 \times 10^6)^2}$ $F = \frac{2.001 \times 10^{17}}{4.9 \times 10^{13}}$ $F = 4.08 \times 10^3$ N ≈ 4080 N

Marking:

1 mark: Correct substitution into formula
1 mark: Correct answer 4.08 × 10³ N (or 4080 N)

(b) Hence, calculate the orbital speed of the satellite. [2 marks]

Answer: Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^6}}$ $v = \sqrt{5.717 \times 10^7}$ $v = 7.56 \times 10^3$ m s⁻¹ = 7560 m s⁻¹

Marking:

1 mark: Equating gravitational force to centripetal force or using $v = \sqrt{GM/r}$
1 mark: Correct answer 7.56 × 10³ m s⁻¹

8. Calculate the speed of the block at the bottom of the incline. [4 marks]

Answer: Component of weight down slope: $mg\sin\theta = 2.0 \times 9.81 \times \sin 30° = 9.81$ N Normal reaction: $N = mg\cos\theta = 2.0 \times 9.81 \times \cos 30° = 16.99$ N Frictional force: $f = \mu N = 0.25 \times 16.99 = 4.25$ N Net force down slope: $F_{\text{net}} = 9.81 - 4.25 = 5.56$ N Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{5.56}{2.0} = 2.78$ m s⁻² Using $v^2 = u^2 + 2as$ : $v^2 = 0 + 2 \times 2.78 \times 3.0 = 16.68$ $v = 4.08$ m s⁻¹

Alternative (energy method): Loss in GPE = $mgh = 2.0 \times 9.81 \times (3.0 \times \sin 30°) = 2.0 \times 9.81 \times 1.5 = 29.43$ J Work done against friction = $f \times s = (\mu mg\cos\theta) \times 3.0 = 0.25 \times 2.0 \times 9.81 \times \cos 30° \times 3.0 = 12.74$ J Gain in KE = $29.43 - 12.74 = 16.69$ J $\frac{1}{2}mv^2 = 16.69$ $v = \sqrt{\frac{2 \times 16.69}{2.0}} = 4.09$ m s⁻¹

Marking:

1 mark: Correct calculation of component of weight down slope or GPE loss
1 mark: Correct calculation of frictional force or work done against friction
1 mark: Correct calculation of net force/acceleration or net energy
1 mark: Correct answer 4.08–4.09 m s⁻¹

Section B: Calculation and Application (Questions 9–15)

9. (a) Calculate the maximum height reached by the ball above the ground. [3 marks]

Answer: Using $v^2 = u^2 + 2as$ : $0 = 12^2 + 2(-9.81)s$ $s = \frac{144}{19.62} = 7.34$ m (height above launch point) Maximum height above ground = $7.34 + 1.5 = 8.84$ m

Marking:

1 mark: Correct use of $v^2 = u^2 + 2as$ with v = 0
1 mark: Correct calculation of height above launch point (7.34 m)
1 mark: Adding initial height for correct answer 8.84 m

(b) Calculate the speed of the ball when it hits the ground. [2 marks]

Answer: Using conservation of energy or $v^2 = u^2 + 2as$ : $v^2 = 12^2 + 2(9.81)(1.5)$ (considering downward displacement from launch to ground) $v^2 = 144 + 29.43 = 173.43$ $v = 13.2$ m s⁻¹

Alternative: From max height: $v^2 = 0 + 2(9.81)(8.84) = 173.4$ $v = 13.2$ m s⁻¹

Marking:

1 mark: Correct method (energy or kinematics)
1 mark: Correct answer 13.2 m s⁻¹

10. (a) Calculate the common velocity immediately after the collision. [2 marks]

Answer: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(8000 \times 15) + (1200 \times 0) = (8000 + 1200)v$ $120,000 = 9200v$ $v = 13.0$ m s⁻¹

Marking:

1 mark: Correct conservation of momentum equation
1 mark: Correct answer 13.0 m s⁻¹

(b) Calculate the kinetic energy lost in the collision. [2 marks]

Answer: Initial KE = $\frac{1}{2} \times 8000 \times 15^2 = 900,000$ J Final KE = $\frac{1}{2} \times 9200 \times 13.04^2 = 782,609$ J KE lost = $900,000 - 782,609 = 117,391$ J ≈ $1.17 \times 10^5$ J

Marking:

1 mark: Correct calculation of initial and final KE
1 mark: Correct answer 1.17 × 10⁵ J (or 117 kJ)

Answer: The lost kinetic energy is converted to internal energy (heat and sound) and is used in deforming the vehicles during the collision.

Marking:

1 mark: Reference to heat/sound/internal energy/deformation

11. (a) Draw a free-body diagram showing the forces acting on the bob. [2 marks]

Answer: Two forces:

Weight (mg) acting vertically downwards
Tension (T) acting along the string at 30° to the vertical

Marking:

1 mark: Weight correctly shown (vertical, downwards)
1 mark: Tension correctly shown (along string direction)

(b) Calculate the tension in the string. [2 marks]

Answer: Resolving vertically: $T\cos 30° = mg$ $T\cos 30° = 0.25 \times 9.81 = 2.4525$ $T = \frac{2.4525}{\cos 30°} = \frac{2.4525}{0.8660} = 2.83$ N

Marking:

1 mark: Correct vertical resolution $T\cos\theta = mg$
1 mark: Correct answer 2.83 N

Answer: Radius of circle: $r = L\sin 30° = 0.80 \times 0.5 = 0.40$ m Horizontal component of tension provides centripetal force: $T\sin 30° = m\omega^2 r$ $2.83 \times 0.5 = 0.25 \times \omega^2 \times 0.40$ $1.415 = 0.10\omega^2$ $\omega^2 = 14.15$ $\omega = 3.76$ rad s⁻¹

Marking:

1 mark: Correct radius (0.40 m)
1 mark: Correct equation $T\sin\theta = m\omega^2 r$
1 mark: Correct answer 3.76 rad s⁻¹

12. (a) Calculate the initial acceleration of the mass. [2 marks]

Answer: Spring force: $F = kx = 200 \times 0.10 = 20$ N Acceleration: $a = \frac{F}{m} = \frac{20}{0.50} = 40$ m s⁻²

Marking:

1 mark: Correct calculation of spring force (20 N)
1 mark: Correct answer 40 m s⁻²

(b) Calculate the maximum speed of the mass during its subsequent motion. [2 marks]

Answer: By conservation of energy: Initial elastic PE = Maximum KE $\frac{1}{2}kx^2 = \frac{1}{2}mv_{\max}^2$ $\frac{1}{2} \times 200 \times (0.10)^2 = \frac{1}{2} \times 0.50 \times v_{\max}^2$ $1.0 = 0.25 v_{\max}^2$ $v_{\max} = 2.0$ m s⁻¹

Marking:

1 mark: Correct energy conversion (elastic PE to KE)
1 mark: Correct answer 2.0 m s⁻¹

13. (a) Calculate the acceleration of the particles. [3 marks]

Answer: For particle B (hanging): $m_B g - T = m_B a$ $0.60 \times 9.81 - T = 0.60a$ ... (1)

For particle A (on table): $T = m_A a$ $T = 0.40a$ ... (2)

Substituting (2) into (1): $5.886 - 0.40a = 0.60a$ $5.886 = 1.00a$ $a = 5.89$ m s⁻²

Marking:

1 mark: Correct equation for particle B
1 mark: Correct equation for particle A
1 mark: Correct answer 5.89 m s⁻²

(b) Calculate the tension in the string. [2 marks]

Answer: $T = m_A a = 0.40 \times 5.886 = 2.35$ N

Marking:

1 mark: Correct substitution
1 mark: Correct answer 2.35 N

14. (a) Calculate the time of flight of the projectile. [2 marks]

Answer: Initial vertical velocity: $u_y = 50 \sin 40° = 32.14$ m s⁻¹ Time to reach max height: $t_{\text{up}} = \frac{u_y}{g} = \frac{32.14}{9.81} = 3.28$ s Total time of flight: $T = 2 \times 3.28 = 6.55$ s

Alternative: $s_y = u_y t - \frac{1}{2}gt^2$ , with $s_y = 0$ : $0 = 32.14t - 4.905t^2$ $t(32.14 - 4.905t) = 0$ $t = 0$ or $t = 6.55$ s

Marking:

1 mark: Correct calculation of vertical component or correct equation
1 mark: Correct answer 6.55 s

(b) Calculate the maximum height reached. [2 marks]

Answer: $v_y^2 = u_y^2 + 2a_y s_y$ $0 = (32.14)^2 + 2(-9.81)H$ $H = \frac{(32.14)^2}{19.62} = 52.6$ m

Marking:

1 mark: Correct method
1 mark: Correct answer 52.6 m

Answer: Horizontal velocity: $u_x = 50 \cos 40° = 38.30$ m s⁻¹ Range: $R = u_x \times T = 38.30 \times 6.55 = 251$ m

Marking:

1 mark: Correct horizontal velocity
1 mark: Correct answer 251 m

15. Calculate the maximum speed at which the car can travel around the track without slipping. [4 marks]

Answer: For banked track with friction, resolving forces:

Vertically: $N\cos\theta - f\sin\theta = mg$ Horizontally: $N\sin\theta + f\cos\theta = \frac{mv^2}{r}$

At maximum speed, friction is limiting: $f = \mu N$

From vertical equation: $N\cos 20° - \mu N\sin 20° = mg$ $N(\cos 20° - 0.40 \sin 20°) = mg$ $N(0.9397 - 0.1368) = mg$ $N(0.8029) = mg$ $N = \frac{mg}{0.8029}$

From horizontal equation: $N\sin 20° + \mu N\cos 20° = \frac{mv^2}{r}$ $N(\sin 20° + 0.40 \cos 20°) = \frac{mv^2}{r}$ $N(0.3420 + 0.3759) = \frac{mv^2}{r}$ $N(0.7179) = \frac{mv^2}{r}$

Substituting N: $\frac{mg}{0.8029} \times 0.7179 = \frac{mv^2}{r}$ $\frac{0.7179}{0.8029} \times g = \frac{v^2}{r}$ $0.8941 \times 9.81 = \frac{v^2}{80}$ $8.771 = \frac{v^2}{80}$ $v^2 = 701.7$ $v = 26.5$ m s⁻¹

Marking:

1 mark: Correct resolution of forces (vertical and horizontal)
1 mark: Correct use of limiting friction $f = \mu N$
1 mark: Correct algebraic manipulation
1 mark: Correct answer 26.5 m s⁻¹

Section C: Data Analysis and Extended Response (Questions 16–20)

16. (a) Explain why the velocity becomes constant after 1.5 s. [2 marks]

Answer: As the velocity increases, the air resistance (drag force) increases. When the air resistance becomes equal to the weight of the mass, the net force becomes zero. According to Newton's first law, when net force is zero, the object moves with constant velocity (terminal velocity).

Marking:

1 mark: Air resistance increases with velocity and eventually equals weight
1 mark: Net force zero → constant velocity (terminal velocity)

(b) Calculate the acceleration of the mass during the first 1.5 s. [2 marks]

Answer: $a = \frac{\Delta v}{\Delta t} = \frac{8.0 - 0}{1.5} = 5.33$ m s⁻²

Marking:

1 mark: Correct formula
1 mark: Correct answer 5.33 m s⁻²

Answer: At terminal velocity, net force = 0 Weight = Air resistance $R = mg = 0.50 \times 9.81 = 4.91$ N

Marking:

1 mark: Recognition that air resistance equals weight at terminal velocity
1 mark: Correct answer 4.91 N (or 4.9 N)

17. (a) Show that the gradient of the T² against L graph is equal to $\frac{4\pi^2}{g}$ . [2 marks]

Answer: $T = 2\pi\sqrt{\frac{L}{g}}$ $T^2 = 4\pi^2\frac{L}{g}$ $T^2 = \frac{4\pi^2}{g}L$

This is of the form $y = mx$ , where $y = T^2$ , $x = L$ , and gradient $m = \frac{4\pi^2}{g}$ .

Marking:

1 mark: Correct squaring to obtain $T^2 = 4\pi^2 L/g$
1 mark: Identification of gradient as $4\pi^2/g$

(b) Use the gradient to calculate a value for g. [2 marks]

Answer: Gradient = $\frac{4\pi^2}{g} = 4.05$ $g = \frac{4\pi^2}{4.05} = \frac{39.48}{4.05} = 9.75$ m s⁻²

Marking:

1 mark: Correct rearrangement
1 mark: Correct answer 9.75 m s⁻²

Answer: Percentage difference = $\frac{|9.75 - 9.81|}{9.81} \times 100\% = \frac{0.06}{9.81} \times 100\% = 0.61\%$

Marking:

1 mark: Correct answer 0.61% (accept 0.6%)

(d) Suggest one reason for the difference, and state how the experiment could be improved. [2 marks]

Answer (any valid pair):

Reason: Reaction time error in starting/stopping the stopwatch. Improvement: Use a light gate and electronic timer for more accurate period measurement.
Reason: The amplitude of oscillation was too large, so the motion was not simple harmonic. Improvement: Ensure the angular displacement is small (less than about 10°).
Reason: Parallax error in measuring the length of the pendulum. Improvement: Use a metre rule with a vernier scale, and ensure eye level is perpendicular to the scale.

Marking:

1 mark: Valid reason for discrepancy
1 mark: Corresponding valid improvement

18. Use dimensional analysis to determine the values of α, β, and γ. [4 marks]

Answer: $F = km^\alpha v^\beta r^\gamma$

Dimensions: $[F] = \text{MLT}^{-2}$ $[m] = \text{M}$ $[v] = \text{LT}^{-1}$ $[r] = \text{L}$

Substituting: $\text{MLT}^{-2} = \text{M}^\alpha (\text{LT}^{-1})^\beta \text{L}^\gamma$ $\text{MLT}^{-2} = \text{M}^\alpha \text{L}^\beta \text{T}^{-\beta} \text{L}^\gamma$ $\text{MLT}^{-2} = \text{M}^\alpha \text{L}^{\beta+\gamma} \text{T}^{-\beta}$

Equating powers: M: $1 = \alpha$ → $\alpha = 1$ L: $1 = \beta + \gamma$ T: $-2 = -\beta$ → $\beta = 2$

From L: $1 = 2 + \gamma$ → $\gamma = -1$

Therefore: $\alpha = 1$ , $\beta = 2$ , $\gamma = -1$

This gives $F = km^1 v^2 r^{-1} = k\frac{mv^2}{r}$ , which matches the known formula.

Marking:

1 mark: Correct dimensions for F, m, v, r
1 mark: Correct dimensional equation set up
1 mark: Correct values for α and β
1 mark: Correct value for γ

19. (a) Explain the shape of the graph. [3 marks]

Answer: Initially, as the applied force F increases, the frictional force f increases equally to match it. This is static friction, where the block does not move. The frictional force adjusts to exactly balance the applied force up to a maximum value (5.9 N).

When F exceeds the maximum static friction (5.9 N), the block begins to move. Once moving, the frictional force becomes kinetic friction, which is approximately constant at 5.9 N (slightly less than maximum static friction in reality, but shown as equal here). The kinetic friction does not increase with further increases in applied force.

Marking:

1 mark: Linear increase region explained as static friction matching applied force
1 mark: Maximum point identified as limiting static friction
1 mark: Constant region explained as kinetic friction after motion begins

(b) Determine the coefficient of static friction between the block and the surface. [2 marks]

Answer: Maximum static friction: $f_s^{\max} = \mu_s N = \mu_s mg$ $5.9 = \mu_s \times 2.0 \times 9.81$ $\mu_s = \frac{5.9}{19.62} = 0.30$

Marking:

1 mark: Correct equation $f_s^{\max} = \mu_s mg$
1 mark: Correct answer 0.30

Answer: Kinetic friction = 5.9 N Net force = Applied force - Friction = $8.0 - 5.9 = 2.1$ N Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{2.1}{2.0} = 1.05$ m s⁻²

Marking:

1 mark: Correct net force (2.1 N)
1 mark: Correct answer 1.05 m s⁻²

20. (a) State Newton's second law of motion in terms of momentum. [1 mark]

Answer: The rate of change of momentum of an object is directly proportional to the net external force acting on it, and takes place in the direction of the force. OR: $F = \frac{dp}{dt}$

Marking:

1 mark: Correct statement or formula

(b) Calculate the initial thrust provided by the rocket engine. [2 marks]

Answer: Thrust = rate of change of momentum of exhaust gases $F = v_{\text{exhaust}} \times \frac{dm}{dt}$ $F = 2500 \times 50 = 125,000$ N = $1.25 \times 10^5$ N

Marking:

1 mark: Correct formula $F = v(dm/dt)$
1 mark: Correct answer 1.25 × 10⁵ N

Answer: Net force = Thrust - Weight $F_{\text{net}} = 125,000 - (5000 \times 9.81) = 125,000 - 49,050 = 75,950$ N $a = \frac{F_{\text{net}}}{m} = \frac{75,950}{5000} = 15.2$ m s⁻²

Marking:

1 mark: Correct net force calculation
1 mark: Correct answer 15.2 m s⁻²

(d) Explain, without calculation, how the acceleration of the rocket changes as the fuel is consumed, assuming the thrust remains constant. [2 marks]

Answer: As fuel is consumed, the mass of the rocket decreases. Since $a = F_{\text{net}}/m$ , and the net force increases (weight decreases while thrust remains constant), the acceleration increases. The rocket accelerates at an increasing rate as more fuel is burned.

Marking:

1 mark: Recognition that mass decreases
1 mark: Explanation that acceleration increases because $a = F/m$ (or net force increases as weight decreases)

END OF ANSWER KEY