Secondary 4 Pure Physics Mechanics Quiz

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Secondary 4 Pure Physics Quiz - Mechanics

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: ______ / 40

Duration: 45 minutes
Total Marks: 40

Instructions:

• Answer all questions.
• Write your answers in the spaces provided.
• Show all working for calculation questions.
• Use $g = 10 \text{ N/kg}$ unless otherwise stated.
• The number of marks is given in brackets [ ] at the end of each question or part question.

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Section A: Multiple Choice Questions (10 marks)

Answer all questions. Choose the correct option and write the letter (A, B, C, or D) in the box provided.

1. A car of mass 1200 kg accelerates uniformly from rest to 25 m/s in 10 s. What is the average resultant force acting on the car? [1]

A. 300 N
B. 3000 N
C. 30 000 N
D. 300 000 N

☐

2. A ball is thrown vertically upwards. At the highest point of its motion, which statement is correct? [1]

A. Velocity is zero and acceleration is zero.
B. Velocity is zero and acceleration is $10 \text{ m/s}^2$ downwards.
C. Velocity is maximum and acceleration is zero.
D. Velocity is maximum and acceleration is $10 \text{ m/s}^2$ downwards.

☐

3. A block of weight 50 N rests on a rough horizontal surface. A horizontal force of 30 N is applied to the block, but the block does not move. What is the magnitude of the frictional force acting on the block? [1]

A. 0 N
B. 20 N
C. 30 N
D. 50 N

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4. The diagram shows a uniform metre rule pivoted at the 40 cm mark. A weight of 2 N is suspended at the 10 cm mark. What weight must be suspended at the 90 cm mark to balance the rule? [1]

<image_placeholder> id: Q4-fig1 type: diagram linked_question: Q4 description: Uniform metre rule pivoted at 40 cm mark. Weight of 2 N suspended at 10 cm mark. Unknown weight W suspended at 90 cm mark. Rule is horizontal and in equilibrium. labels: Pivot at 40 cm, 2 N at 10 cm, W at 90 cm, weight of rule acting at 50 cm mark values: Length = 100 cm, pivot at 40 cm, 2 N at 10 cm, W at 90 cm must_show: Horizontal rule, pivot position, all forces with arrows, distances from pivot </image_placeholder>

A. 1.0 N
B. 1.5 N
C. 2.0 N
D. 3.0 N

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5. A skydiver of mass 80 kg reaches terminal velocity. What is the magnitude of the air resistance acting on the skydiver at terminal velocity? [1]

A. 0 N
B. 80 N
C. 800 N
D. 8000 N

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6. A force of 20 N acts on an object of mass 4 kg for 3 s. The object starts from rest. What is the final velocity of the object? [1]

A. 5 m/s
B. 15 m/s
C. 20 m/s
D. 60 m/s

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7. Which of the following is a vector quantity? [1]

A. Distance
B. Speed
C. Work done
D. Momentum

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8. A car travels at a constant speed around a circular track. Which statement about the car is correct? [1]

A. Its velocity is constant.
B. Its acceleration is zero.
C. It experiences a resultant force directed towards the centre of the circle.
D. It experiences a resultant force directed away from the centre of the circle.

☐

9. A 2 kg object moving at 5 m/s collides head-on with a stationary 3 kg object. After the collision, the 2 kg object moves at 1 m/s in the same direction. What is the velocity of the 3 kg object after the collision? [1]

A. 2.0 m/s
B. 2.7 m/s
C. 3.0 m/s
D. 3.3 m/s

☐

10. The graph shows how the velocity of a 5 kg object varies with time.

<image_placeholder> id: Q10-fig1 type: graph linked_question: Q10 description: Velocity-time graph for a 5 kg object. Graph shows: from t=0 to t=4 s, velocity increases linearly from 0 to 8 m/s; from t=4 to t=8 s, velocity remains constant at 8 m/s; from t=8 to t=12 s, velocity decreases linearly to 0 m/s. labels: Time (s) on x-axis from 0 to 12, Velocity (m/s) on y-axis from 0 to 8 values: (0,0), (4,8), (8,8), (12,0) must_show: Three distinct segments: acceleration, constant velocity, deceleration </image_placeholder>

What is the resultant force acting on the object during the first 4 seconds? [1]

A. 5 N
B. 10 N
C. 20 N
D. 40 N

☐

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Section B: Short Answer and Structured Questions (18 marks)

Answer all questions in the spaces provided.

11. A student investigates the motion of a toy car on a horizontal track. The car starts from rest and accelerates uniformly. The student measures the distance travelled and the time taken.

(a) State the equation that relates average speed, distance, and time. [1]

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(b) In one trial, the car travels 2.0 m in 1.6 s. Calculate the average speed of the car. [1]

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(c) The car has a mass of 0.5 kg and its final velocity is 2.5 m/s. Calculate the kinetic energy of the car at this velocity. [2]

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(d) The student suggests that the work done by the resultant force equals the gain in kinetic energy. State the principle that supports this suggestion. [1]

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12. A box of mass 15 kg is pulled along a rough horizontal floor by a horizontal force of 80 N. The box accelerates at $2.0 \text{ m/s}^2$.

(a) Calculate the resultant force acting on the box. [1]

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(b) Determine the magnitude of the frictional force acting on the box. [2]

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(c) The box is pulled a distance of 5.0 m. Calculate the work done by the applied force of 80 N. [1]

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(d) Calculate the work done against friction. [1]

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13. A uniform beam AB of length 4.0 m and weight 200 N is hinged at A to a vertical wall. The beam is held horizontal by a cable attached at B, making an angle of 30° with the beam. A load of 300 N is suspended from the beam at a point 1.0 m from A.

<image_placeholder> id: Q13-fig1 type: diagram linked_question: Q13 description: Uniform beam AB, 4.0 m long, weight 200 N acting at centre (2.0 m from A). Hinged at A to vertical wall. Cable attached at B at 30° to beam. Load of 300 N suspended 1.0 m from A. Beam is horizontal. labels: Hinge at A, cable at B at 30°, weight of beam 200 N at 2.0 m from A, load 300 N at 1.0 m from A, tension T in cable values: Length = 4.0 m, beam weight = 200 N at midpoint, load = 300 N at 1.0 m from A, cable angle = 30° must_show: Horizontal beam, hinge at A, cable at B with 30° angle, all forces with arrows and labels, perpendicular distances from A </image_placeholder>

(a) State the principle of moments. [1]

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(b) By taking moments about A, calculate the tension in the cable. [3]

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(c) The hinge at A exerts a force on the beam. State the direction of the vertical component of this force. [1]

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14. A ball of mass 0.2 kg is dropped from a height of 1.8 m onto a hard floor. It rebounds to a height of 1.2 m. The ball is in contact with the floor for 0.05 s. Take $g = 10 \text{ N/kg}$.

(a) Calculate the speed of the ball just before it hits the floor. [2]

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(b) Calculate the speed of the ball just after it leaves the floor. [2]

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(c) Calculate the change in momentum of the ball during the collision. [2]

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(d) Calculate the average force exerted by the floor on the ball during the collision. [2]

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15. A rocket of initial mass 500 kg (including fuel) is launched vertically upwards from rest. The rocket engine provides a constant thrust of 12 000 N. The mass of the rocket decreases at a constant rate of 2 kg/s as fuel is burnt. Assume $g = 10 \text{ N/kg}$ and ignore air resistance.

(a) Calculate the initial acceleration of the rocket. [2]

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(b) Explain why the acceleration of the rocket increases during the first few seconds of flight, even though the thrust remains constant. [2]

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(c) After 10 s, the mass of the rocket is 480 kg. Calculate the acceleration of the rocket at this instant. [2]

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(d) The rocket engine stops after 20 s. Describe the motion of the rocket after the engine stops until it reaches its maximum height. [2]

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16. A block of mass 2.0 kg slides down a rough inclined plane of length 5.0 m, inclined at 30° to the horizontal. The block starts from rest at the top and reaches the bottom with a speed of 4.0 m/s. Take $g = 10 \text{ N/kg}$.

<image_placeholder> id: Q16-fig1 type: diagram linked_question: Q16 description: Rough inclined plane, length 5.0 m, angle 30° to horizontal. Block of mass 2.0 kg at top, slides down from rest. Reaches bottom with speed 4.0 m/s. Forces shown: weight mg vertically down, normal reaction perpendicular to plane, friction up the plane. labels: Mass = 2.0 kg, length = 5.0 m, angle = 30°, initial velocity = 0, final velocity = 4.0 m/s, weight = 20 N, components of weight parallel and perpendicular to plane values: m = 2.0 kg, L = 5.0 m, θ = 30°, u = 0, v = 4.0 m/s, g = 10 N/kg must_show: Inclined plane at 30°, block on plane, all forces labelled with arrows, components of weight </image_placeholder>

(a) Calculate the loss in gravitational potential energy of the block. [2]

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(b) Calculate the gain in kinetic energy of the block. [1]

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(c) Determine the work done against friction. [1]

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(d) Calculate the magnitude of the frictional force acting on the block. [2]

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(e) Calculate the coefficient of friction between the block and the plane. [2]

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17. Two ice skaters, A and B, stand at rest on a frictionless horizontal ice rink. Skater A has mass 60 kg and skater B has mass 40 kg. Skater A pushes skater B, causing skater B to move away with a velocity of 3.0 m/s.

(a) State the principle of conservation of momentum. [1]

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(b) Calculate the velocity of skater A after the push. [2]

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(c) Calculate the total kinetic energy of the two skaters after the push. [2]

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(d) Explain where this kinetic energy came from. [1]

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18. A car of mass 1000 kg travels at a constant speed of 20 m/s around a banked circular track of radius 80 m. The track is banked at an angle of 15° to the horizontal.

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: Banked circular track, radius 80 m, banking angle 15°. Car of mass 1000 kg moving at 20 m/s. Forces: weight mg down, normal reaction N perpendicular to track surface, friction f (if needed) parallel to track surface. labels: Mass = 1000 kg, speed = 20 m/s, radius = 80 m, banking angle = 15°, weight = 10 000 N values: m = 1000 kg, v = 20 m/s, r = 80 m, θ = 15°, g = 10 N/kg must_show: Banked track cross-section, car on track, all forces with arrows, banking angle labelled </image_placeholder>

(a) Calculate the centripetal force required to keep the car moving in a circle. [2]

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(b) Assuming no friction is required, calculate the normal reaction force exerted by the track on the car. [2]

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(c) Determine whether friction is actually required for the car to travel at this speed on this banked track. If so, state the direction of the frictional force. [3]

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19. A 0.5 kg mass is attached to a spring of spring constant 200 N/m on a smooth horizontal surface. The mass is pulled 0.1 m from its equilibrium position and released from rest.

(a) Calculate the maximum acceleration of the mass during its motion. [2]

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(b) Calculate the period of oscillation. [2]

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(c) Calculate the maximum speed of the mass. [2]

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(d) On the axes below, sketch a graph of acceleration against displacement for one complete oscillation. Label the maximum acceleration and maximum displacement. [2]

<image_placeholder> id: Q19-fig1 type: graph linked_question: Q19 description: Axes for acceleration vs displacement graph. x-axis: displacement (m) from -0.1 to 0.1. y-axis: acceleration (m/s²) from -40 to 40. Graph should be a straight line through origin with negative gradient. labels: Displacement (m) on x-axis, Acceleration (m/s²) on y-axis values: Max displacement = 0.1 m, Max acceleration = 40 m/s² must_show: Straight line through origin with negative slope, intercepts at (±0.1, ∓40) </image_placeholder>

20. A satellite of mass 500 kg orbits the Earth in a circular orbit of radius 7.0 × 10⁶ m. The gravitational field strength at this orbit is 8.0 N/kg.

(a) Calculate the gravitational force acting on the satellite. [1]

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(b) Calculate the orbital speed of the satellite. [2]

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(c) Calculate the period of the orbit. [2]

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(d) The satellite loses energy due to atmospheric drag and moves to a lower orbit of radius 6.5 × 10⁶ m. State and explain what happens to its orbital speed. [2]

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Secondary 4 Pure Physics Quiz - Mechanics (Answer Key)

Total Marks: 40

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Section A: Multiple Choice Questions (10 marks)

1. Answer: B [1]

Working:
Acceleration $a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m/s}^2$
Resultant force $F = ma = 1200 \times 2.5 = 3000 \text{ N}$

Teaching note: Use $F = ma$ with acceleration calculated from change in velocity over time. Common mistake: forgetting to calculate acceleration first and using velocity directly.

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2. Answer: B [1]

Explanation:
At the highest point, the ball momentarily stops (velocity = 0) but still experiences gravitational force, so acceleration = $g = 10 \text{ m/s}^2$ downwards.

Teaching note: Velocity is zero at the top, but acceleration is not zero because gravity still acts. This is a common misconception.

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3. Answer: C [1]

Explanation:
Since the block does not move, it is in equilibrium. The frictional force equals the applied force (30 N) but acts in the opposite direction. Static friction matches the applied force up to its maximum value.

Teaching note: Static friction adjusts to match the applied force until the limiting friction is reached. Here, 30 N is less than the maximum static friction.

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4. Answer: A [1]

Working:
For a uniform metre rule, weight ≈ 1 N acting at the 50 cm mark (10 cm from pivot).
Taking moments about the pivot (40 cm mark):
Clockwise moments = Anticlockwise moments
$(1 \times 10) + (W \times 50) = 2 \times 30$
$10 + 50W = 60$
$50W = 50$
$W = 1.0 \text{ N}$

Teaching note: The weight of a uniform metre rule is typically 1 N. If the rule's weight were negligible, the answer would be 1.2 N (not an option), so the rule's weight must be considered.

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5. Answer: C [1]

Working:
At terminal velocity, air resistance = weight = $mg = 80 \times 10 = 800 \text{ N}$.

Teaching note: Terminal velocity occurs when resultant force is zero, so air resistance equals weight.

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6. Answer: B [1]

Working:
Acceleration $a = \frac{F}{m} = \frac{20}{4} = 5 \text{ m/s}^2$
Final velocity $v = u + at = 0 + 5 \times 3 = 15 \text{ m/s}$

Teaching note: Use $F=ma$ to find acceleration, then $v=u+at$.

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7. Answer: D [1]

Explanation:
Momentum = mass × velocity. Since velocity is a vector, momentum is a vector. Distance, speed, and work done are scalars.

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8. Answer: C [1]

Explanation:
In uniform circular motion, speed is constant but velocity changes direction, so there is centripetal acceleration towards the centre, requiring a centripetal force towards the centre.

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9. Answer: B [1]

Working:
Conservation of momentum:
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$2 \times 5 + 3 \times 0 = 2 \times 1 + 3 \times v_2$
$10 = 2 + 3v_2$
$3v_2 = 8$
$v_2 = \frac{8}{3} = 2.67 \text{ m/s} \approx 2.7 \text{ m/s}$

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10. Answer: B [1]

Working:
From graph, gradient in first 4 s = $\frac{8 - 0}{4 - 0} = 2 \text{ m/s}^2$
Resultant force $F = ma = 5 \times 2 = 10 \text{ N}$

Teaching note: Gradient of velocity-time graph = acceleration.

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Section B: Short Answer and Structured Questions (18 marks)

11. (a) Average speed = $\frac{\text{total distance}}{\text{total time}}$ [1]

(b) Average speed = $\frac{2.0}{1.6} = 1.25 \text{ m/s}$ [1]

(c) Kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times (2.5)^2 = 0.25 \times 6.25 = 1.5625 \text{ J} \approx 1.56 \text{ J}$ [2]
Mark breakdown: 1 mark for formula/substitution, 1 mark for correct answer with unit.

(d) Work-energy theorem (or principle of conservation of energy applied to work and kinetic energy) [1]

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12. (a) Resultant force = $ma = 15 \times 2.0 = 30 \text{ N}$ [1]

(b) Applied force - Friction = Resultant force
$80 - F_{\text{friction}} = 30$
$F_{\text{friction}} = 50 \text{ N}$ [2]
Mark breakdown: 1 mark for equation/principle, 1 mark for correct answer.

(c) Work done = Force × distance = $80 \times 5.0 = 400 \text{ J}$ [1]

(d) Work done against friction = Friction × distance = $50 \times 5.0 = 250 \text{ J}$ [1]

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13. (a) Principle of moments: For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point. [1]

(b) Taking moments about A (hinge):
Clockwise moments = Anticlockwise moments
$(200 \times 2.0) + (300 \times 1.0) = T \sin 30^\circ \times 4.0$
$400 + 300 = T \times 0.5 \times 4.0$
$700 = 2T$
$T = 350 \text{ N}$ [3]
Mark breakdown: 1 mark for correct moment arms identified, 1 mark for correct equation, 1 mark for correct answer with unit.

Teaching note: The vertical component of tension ($T \sin 30^\circ$) provides the anticlockwise moment. The perpendicular distance from A to the line of action of this component is 4.0 m.

(c) Upwards [1]
Explanation: The hinge must provide an upward force to help support the weight of the beam and the load, since the vertical component of tension (175 N) is less than the total downward force (500 N).

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14. (a) Using $v^2 = u^2 + 2as$:

$v^2 = 0 + 2 \times 10 \times 1.8 = 36$
$v = 6 \text{ m/s}$ (downwards) [2]
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(b) Using $v^2 = u^2 + 2as$ for upward motion after bounce:
$0 = u^2 - 2 \times 10 \times 1.2$
$u^2 = 24$
$u = \sqrt{24} = 4.9 \text{ m/s}$ (upwards) [2]
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(c) Change in momentum = $m(v - u)$
Taking downwards as positive:
$u = +6 \text{ m/s}$, $v = -4.9 \text{ m/s}$
$\Delta p = 0.2 \times (-4.9 - 6) = 0.2 \times (-10.9) = -2.18 \text{ kg m/s}$
Magnitude = $2.18 \text{ kg m/s}$ [2]
Mark breakdown: 1 mark for correct signs/substitution, 1 mark for correct magnitude.

(d) Average force = $\frac{\text{change in momentum}}{\text{time}} = \frac{2.18}{0.05} = 43.6 \text{ N}$ [2]
Mark breakdown: 1 mark for formula, 1 mark for correct answer with unit.

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15. (a) Initial weight = $500 \times 10 = 5000 \text{ N}$

Resultant force = Thrust - Weight = $12000 - 5000 = 7000 \text{ N}$
Initial acceleration $a = \frac{F}{m} = \frac{7000}{500} = 14 \text{ m/s}^2$ [2]
Mark breakdown: 1 mark for resultant force, 1 mark for acceleration.

(b) As fuel burns, the mass of the rocket decreases while thrust remains constant. Since $a = \frac{F_{\text{net}}}{m}$ and $F_{\text{net}} = \text{Thrust} - mg$, as $m$ decreases, both the denominator decreases and the weight $mg$ decreases, so the resultant force increases. Both effects cause acceleration to increase. [2]
Mark breakdown: 1 mark for mass decreases, 1 mark for resultant force increases/acceleration increases.

(c) Mass at 10 s = 480 kg
Weight = $480 \times 10 = 4800 \text{ N}$
Resultant force = $12000 - 4800 = 7200 \text{ N}$
Acceleration $a = \frac{7200}{480} = 15 \text{ m/s}^2$ [2]
Mark breakdown: 1 mark for resultant force, 1 mark for acceleration.

(d) After engine stops, the only force acting is weight (downwards). The rocket decelerates at $10 \text{ m/s}^2$ until its velocity becomes zero at maximum height, then accelerates downwards at $10 \text{ m/s}^2$. [2]
Mark breakdown: 1 mark for deceleration upward, 1 mark for acceleration downward/maximum height.

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16. (a) Vertical height lost = $5.0 \times \sin 30^\circ = 5.0 \times 0.5 = 2.5 \text{ m}$

Loss in GPE = $mgh = 2.0 \times 10 \times 2.5 = 50 \text{ J}$ [2]
Mark breakdown: 1 mark for height calculation, 1 mark for GPE loss.

(b) Gain in KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \times (4.0)^2 = 16 \text{ J}$ [1]

(c) Work done against friction = Loss in GPE - Gain in KE = $50 - 16 = 34 \text{ J}$ [1]

(d) Work done against friction = Friction × distance
$34 = F_{\text{friction}} \times 5.0$
$F_{\text{friction}} = 6.8 \text{ N}$ [2]
Mark breakdown: 1 mark for formula, 1 mark for correct answer.

(e) Normal reaction $R = mg \cos 30^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ N}$
$F_{\text{friction}} = \mu R$
$\mu = \frac{6.8}{17.3} = 0.39$ [2]
Mark breakdown: 1 mark for normal reaction, 1 mark for coefficient.

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17. (a) Principle of conservation of momentum: The total momentum of a closed system remains constant if no external resultant force acts on the system. [1]

(b) Initial momentum = 0
Final momentum: $60 v_A + 40 \times 3.0 = 0$
$60 v_A = -120$
$v_A = -2.0 \text{ m/s}$ (opposite direction to B) [2]
Mark breakdown: 1 mark for conservation equation, 1 mark for correct answer with sign/direction.

(c) KE of A = $\frac{1}{2} \times 60 \times (2.0)^2 = 120 \text{ J}$
KE of B = $\frac{1}{2} \times 40 \times (3.0)^2 = 180 \text{ J}$
Total KE = $120 + 180 = 300 \text{ J}$ [2]
Mark breakdown: 1 mark for each KE calculation, 1 mark for sum (or 1 mark for method, 1 mark for answer).

(d) The kinetic energy came from the chemical potential energy stored in the skater's muscles, converted to kinetic energy during the push. [1]

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18. (a) Centripetal force $F_c = \frac{mv^2}{r} = \frac{1000 \times (20)^2}{80} = \frac{400000}{80} = 5000 \text{ N}$ [2]

Mark breakdown: 1 mark for formula/substitution, 1 mark for correct answer.

(b) Resolving forces vertically (no vertical acceleration):
$N \cos 15^\circ = mg = 10000$
$N = \frac{10000}{\cos 15^\circ} = \frac{10000}{0.9659} = 10353 \text{ N} \approx 10400 \text{ N}$ [2]
Mark breakdown: 1 mark for vertical equilibrium equation, 1 mark for correct answer.

(c) Horizontal component of normal reaction = $N \sin 15^\circ = 10353 \times 0.2588 = 2680 \text{ N}$
This is less than the required centripetal force (5000 N).
Therefore, friction is required. The friction must act up the slope (towards the centre of the circle) to provide the additional centripetal force needed. [3]
Mark breakdown: 1 mark for horizontal component calculation, 1 mark for comparison/conclusion, 1 mark for direction.

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19. (a) Maximum acceleration $a_{\text{max}} = \omega^2 x_0$

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \text{ rad/s}$
$a_{\text{max}} = (20)^2 \times 0.1 = 400 \times 0.1 = 40 \text{ m/s}^2$ [2]
Mark breakdown: 1 mark for $\omega$ or $F_{\text{max}} = kx_0$, 1 mark for $a_{\text{max}}$.

(b) Period $T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = \frac{\pi}{10} = 0.314 \text{ s}$ [2]
Mark breakdown: 1 mark for formula, 1 mark for correct answer.

(c) Maximum speed $v_{\text{max}} = \omega x_0 = 20 \times 0.1 = 2.0 \text{ m/s}$ [2]
Mark breakdown: 1 mark for formula, 1 mark for correct answer.

(d) Graph: Straight line through origin with negative gradient.
Intercepts: at $x = +0.1 \text{ m}$, $a = -40 \text{ m/s}^2$; at $x = -0.1 \text{ m}$, $a = +40 \text{ m/s}^2$.
Equation: $a = -\omega^2 x = -400x$ [2]
Mark breakdown: 1 mark for straight line through origin with negative slope, 1 mark for correct intercepts labelled.

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20. (a) Gravitational force $F = mg = 500 \times 8.0 = 4000 \text{ N}$ [1]

(b) Gravitational force provides centripetal force:
$F = \frac{mv^2}{r}$
$4000 = \frac{500 \times v^2}{7.0 \times 10^6}$
$v^2 = \frac{4000 \times 7.0 \times 10^6}{500} = 5.6 \times 10^7$
$v = \sqrt{5.6 \times 10^7} = 7483 \text{ m/s} \approx 7500 \text{ m/s}$ [2]
Mark breakdown: 1 mark for equating forces/formula, 1 mark for correct answer.

(c) Period $T = \frac{2\pi r}{v} = \frac{2\pi \times 7.0 \times 10^6}{7483} = 5880 \text{ s} \approx 5900 \text{ s}$ (or 1.63 hours) [2]
Mark breakdown: 1 mark for formula, 1 mark for correct answer.

(d) Orbital speed increases.
Explanation: In a lower orbit, the gravitational field strength is stronger, so the gravitational force is larger. This provides a larger centripetal force, requiring a higher orbital speed for circular motion ($v = \sqrt{gr}$). As $r$ decreases, $g$ increases more than $r$ decreases, so $v$ increases. [2]
Mark breakdown: 1 mark for "increases", 1 mark for explanation.