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A Level H2 Physics Practice Paper 5

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI)

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

Instructions

Answer all questions.
The number of marks for each question or part-question is shown in brackets [ ].
You are advised to show all working clearly, including formulas and substitutions.
Numerical answers should be given to an appropriate number of significant figures.
Assume $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.
A list of formulae and constants is provided at the end of this paper.

Section A: Multiple Choice [10 marks]

Questions 1–10: Each question carries 1 mark. Choose the single best answer.

1. A ball is thrown vertically upward with an initial speed of $20 \text{ m s}^{-1}$ . Ignoring air resistance, what is the acceleration of the ball at the highest point of its trajectory?

A. $0 \text{ m s}^{-2}$ B. $9.81 \text{ m s}^{-2}$ downward C. $9.81 \text{ m s}^{-2}$ upward D. $20 \text{ m s}^{-2}$ downward

2. Which of the following is a vector quantity?

A. Kinetic energy B. Power C. Momentum D. Work

3. A car of mass $1000 \text{ kg}$ accelerates uniformly from rest to $25 \text{ m s}^{-1}$ in $10 \text{ s}$ . What is the net force acting on the car?

A. $250 \text{ N}$ B. $2500 \text{ N}$ C. $4000 \text{ N}$ D. $25000 \text{ N}$

4. An object moves in a horizontal circle of radius $2.0 \text{ m}$ at a constant speed of $4.0 \text{ m s}^{-1}$ . What is the magnitude of the centripetal acceleration?

A. $2.0 \text{ m s}^{-2}$ B. $4.0 \text{ m s}^{-2}$ C. $8.0 \text{ m s}^{-2}$ D. $16.0 \text{ m s}^{-2}$

5. A force of $12 \text{ N}$ acts on an object for $0.50 \text{ s}$ . What is the impulse delivered to the object?

A. $6.0 \text{ N s}$ B. $12 \text{ N s}$ C. $24 \text{ N s}$ D. $48 \text{ N s}$

6. A stone is released from rest and falls freely under gravity. What is the ratio of the distance fallen in the first $2.0 \text{ s}$ to the distance fallen in the next $2.0 \text{ s}$ ?

A. $1:1$ B. $1:2$ C. $1:3$ D. $1:4$

7. A uniform rod of length $3.0 \text{ m}$ and weight $60 \text{ N}$ is pivoted at one end. What is the minimum vertical force required at the other end to hold the rod horizontal?

A. $20 \text{ N}$ B. $30 \text{ N}$ C. $60 \text{ N}$ D. $120 \text{ N}$

8. A satellite orbits Earth at a height where the gravitational field strength is $2.45 \text{ N kg}^{-1}$ . If the radius of Earth is $6.37 \times 10^6 \text{ m}$ , what is the orbital radius of the satellite? (Mass of Earth $= 5.97 \times 10^{24} \text{ kg}$ )

A. $1.27 \times 10^7 \text{ m}$ B. $1.59 \times 10^7 \text{ m}$ C. $2.55 \times 10^7 \text{ m}$ D. $3.18 \times 10^7 \text{ m}$

9. A $0.50 \text{ kg}$ object attached to a spring oscillates with a period of $1.2 \text{ s}$ . What is the spring constant?

A. $3.4 \text{ N m}^{-1}$ B. $6.8 \text{ N m}^{-1}$ C. $13.7 \text{ N m}^{-1}$ D. $27.4 \text{ N m}^{-1}$

10. Two objects collide and stick together. Which of the following is conserved in this collision?

A. Kinetic energy only B. Momentum only C. Both kinetic energy and momentum D. Neither kinetic energy nor momentum

Section B: Structured Questions [30 marks]

Answer all questions. Show all working.

11. (6 marks)

A student investigates projectile motion by launching a small ball horizontally from a table of height $0.80 \text{ m}$ with an initial horizontal speed of $3.0 \text{ m s}^{-1}$ .

(a) Calculate the time taken for the ball to reach the floor. [2]

(b) Determine the horizontal distance from the base of the table where the ball lands. [2]

12. (8 marks)

A $1500 \text{ kg}$ car travelling at $20 \text{ m s}^{-1}$ collides head-on with a stationary $2500 \text{ kg}$ van. After the collision, the two vehicles move together.

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

(b) Calculate the common velocity of the car and van immediately after the collision. [3]

(d) Explain whether the collision is elastic or inelastic, justifying your answer. [2]

13. (8 marks)

A small object of mass $0.20 \text{ kg}$ is attached to one end of a light inextensible string of length $1.5 \text{ m}$ . The other end of the string is fixed. The object is made to move in a horizontal circle with the string at an angle of $30°$ to the vertical, as shown in the diagram below.

<image_placeholder> id: Q13-fig1 type: diagram linked_question: Q13 description: A conical pendulum diagram showing a mass of 0.20 kg attached to a string of length 1.5 m, making an angle of 30° with the vertical, moving in a horizontal circle. The string traces out a cone shape. labels: string length L = 1.5 m, angle θ = 30°, mass m = 0.20 kg, radius of circular path r, tension T in string, weight mg acting downward values: L = 1.5 m, θ = 30°, m = 0.20 kg, g = 9.81 m s⁻² must_show: The string at 30° to vertical, the mass at the bottom, the circular path (dashed horizontal circle), the radius r labelled, the vertical height component, tension T along the string, weight mg vertically downward, and the angle 30° clearly marked between string and vertical.

</image_placeholder>

(a) Calculate the radius of the circular path. [2]

(b) Draw a free-body diagram showing all forces acting on the object. [1]

(d) Determine the speed of the object. [3]

14. (8 marks)

A $2.0 \text{ kg}$ block is initially at rest on a smooth horizontal surface. A constant horizontal force of $8.0 \text{ N}$ is applied to the block for $4.0 \text{ s}$ , after which the force is removed. The block then travels up a smooth inclined plane at $25°$ to the horizontal.

(a) Calculate the acceleration of the block on the horizontal surface. [2]

(b) Determine the speed of the block at the instant the force is removed. [1]

(d) The inclined plane is now rough. Without further calculation, state and explain how the distance travelled up the plane would change. [2]

Section C: Long Structured Questions [20 marks]

Answer all questions. Show all working.

15. (10 marks)

A ball of mass $0.40 \text{ kg}$ is dropped from a height of $5.0 \text{ m}$ above the ground. After hitting the ground, it rebounds to a height of $3.2 \text{ m}$ . Assume air resistance is negligible.

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

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

(d) The ball is in contact with the ground for $0.020 \text{ s}$ . Calculate the average force exerted by the ground on the ball. [3]

16. (10 marks)

A geostationary satellite orbits the Earth such that it remains above the same point on the equator.

(a) State two features of a geostationary orbit. [2]

(b) The mass of the Earth is $5.97 \times 10^{24} \text{ kg}$ and the gravitational constant $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ .

(i) Show that the orbital radius $r$ of the satellite is related to its period $T$ by the expression:

$r^3 = \frac{GMT^2}{4\pi^2}$

[3]

(ii) The period of a geostationary orbit is $24 \text{ hours}$ . Calculate the orbital radius. [3]

(iii) Hence determine the height of the satellite above the Earth's surface. (Radius of Earth $= 6.37 \times 10^6 \text{ m}$ ) [2]

Formulae and Constants

Quantity	Formula
Kinematics ( $a$ constant)	$v = u + at$ , $s = ut + \frac{1}{2}at^2$ , $v^2 = u^2 + 2as$
Newton's second law	$F = ma$
Momentum	$p = mv$
Impulse	$J = F\Delta t = \Delta p$
Work done	$W = Fs\cos\theta$
Kinetic energy	$KE = \frac{1}{2}mv^2$
Gravitational potential energy	$PE = mgh$
Power	$P = Fv = \frac{W}{t}$
Centripetal acceleration	$a = \frac{v^2}{r} = r\omega^2$
Centripetal force	$F = \frac{mv^2}{mr} = mr\omega^2$
Gravitational force	$F = \frac{GMm}{r^2}$
Gravitational field strength	$g = \frac{GM}{r^2}$
Gravitational potential energy	$U = -\frac{GMm}{r}$
SHM acceleration	$a = -\omega^2 x$
SHM period (mass-spring)	$T = 2\pi\sqrt{\frac{m}{k}}$
SHM period (pendulum)	$T = 2\pi\sqrt{\frac{L}{g}}$

Constants:

$g = 9.81 \text{ m s}^{-2}$
$G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
Mass of Earth $= 5.97 \times 10^{24} \text{ kg}$
Radius of Earth $= 6.37 \times 10^6 \text{ m}$

End of Paper

Mark Summary:

Section	Marks
A: Multiple Choice (Q1–10)	10
B: Structured Questions (Q11–14)	30
C: Long Structured Questions (Q15–16)	20
Total	60

Answers

TuitionGoWhere Practice Paper — Physics H2 A-Level

Answer Key — Mechanics Focus (Version 5 of 5)

Section A: Multiple Choice

1. B — $9.81 \text{ m s}^{-2}$ downward [1]

Teaching note: At the highest point, the ball's velocity is momentarily zero, but the acceleration is still $g = 9.81 \text{ m s}^{-2}$ downward. The only force acting on the ball throughout its flight is gravity (ignoring air resistance), so the acceleration is constant at $g$ downward at every point in the trajectory. A common mistake is choosing A, assuming zero velocity means zero acceleration.

2. C — Momentum [1]

Teaching note: Momentum ( $p = mv$ ) is a vector quantity because it has both magnitude and direction (the direction of velocity). Kinetic energy, power, and work are all scalar quantities — they have magnitude but no direction.

3. B — $2500 \text{ N}$ [1]

Working: $a = \frac{v - u}{t} = \frac{25 - 0}{10} = 2.5 \text{ m s}^{-2}$ $F = ma = 1000 \times 2.5 = 2500 \text{ N}$

4. C — $8.0 \text{ m s}^{-2}$ [1]

Working: $a_c = \frac{v^2}{r} = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8.0 \text{ m s}^{-2}$

5. A — $6.0 \text{ N s}$ [1]

Working: Impulse $= F \times \Delta t = 12 \times 0.50 = 6.0 \text{ N s}$

6. C — $1:3$ [1]

Working: Distance fallen in first $2.0 \text{ s}$ : $s_1 = \frac{1}{2}(9.81)(2.0)^2 = 19.62 \text{ m}$ Distance fallen in first $4.0 \text{ s}$ : $s_{total} = \frac{1}{2}(9.81)(4.0)^2 = 78.48 \text{ m}$ Distance fallen in next $2.0 \text{ s}$ : $s_2 = 78.48 - 19.62 = 58.86 \text{ m}$ Ratio $s_1 : s_2 = 19.62 : 58.86 = 1 : 3$

Teaching note: For an object starting from rest under uniform acceleration, distances travelled in successive equal time intervals are in the ratio $1:3:5:7:\ldots$ . This is a standard result worth remembering.

7. B — $30 \text{ N}$ [1]

Working: Taking moments about the pivot (one end): The weight of the rod ( $60 \text{ N}$ ) acts at the centre, $1.5 \text{ m}$ from the pivot. Let $F$ be the upward force at the other end, $3.0 \text{ m}$ from the pivot. For equilibrium: $F \times 3.0 = 60 \times 1.5$ $F = \frac{90}{3.0} = 30 \text{ N}$

8. B — $1.59 \times 10^7 \text{ m}$ [1]

Working: $g = \frac{GM}{r^2}$ $r^2 = \frac{GM}{g} = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{2.45} = \frac{3.983 \times 10^{14}}{2.45} = 1.626 \times 10^{14}$ $r = \sqrt{1.626 \times 10^{14}} = 1.275 \times 10^7 \text{ m}$

Wait — let me recalculate: $r^2 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{2.45} = \frac{3.982 \times 10^{14}}{2.45} = 1.625 \times 10^{14}$ $r = \sqrt{1.625 \times 10^{14}} = 1.275 \times 10^7 \text{ m}$

Hmm, this gives $1.27 \times 10^7$ m which is option A. Let me re-examine the question values.

Actually, with $g = 2.45$ N kg⁻²: $r = \sqrt{\frac{GM}{g}} = \sqrt{\frac{3.982 \times 10^{14}}{2.45}} = \sqrt{1.625 \times 10^{14}} = 1.275 \times 10^7$ m

This rounds to $1.27 \times 10^7$ m → A.

Correction: The answer is A — $1.27 \times 10^7 \text{ m}$

9. C — $13.7 \text{ N m}^{-1}$ [1]

Working: $T = 2\pi\sqrt{\frac{m}{k}}$ $k = \frac{4\pi^2 m}{T^2} = \frac{4\pi^2(0.50)}{(1.2)^2} = \frac{19.74}{1.44} = 13.7 \text{ N m}^{-1}$

10. B — Momentum only [1]

Teaching note: In any collision (elastic or inelastic), the total momentum of the system is conserved provided no external forces act. When objects stick together, the collision is perfectly inelastic, and kinetic energy is not conserved — some is converted to other forms such as heat and sound. The key distinction: momentum is always conserved in collisions; kinetic energy is only conserved in elastic collisions.

Section B: Structured Questions

11. (6 marks)

(a) [2]

Using $s = \frac{1}{2}gt^2$ for the vertical motion: $0.80 = \frac{1}{2}(9.81)t^2$ $t^2 = \frac{2 \times 0.80}{9.81} = \frac{1.60}{9.81} = 0.1631$ $t = 0.404 \text{ s}$

Marking:

[1] for correct formula/substitution
[1] for correct answer $t = 0.40 \text{ s}$ (2 s.f.)

(b) [2]

Horizontal distance: $x = v_x \times t = 3.0 \times 0.404 = 1.21 \text{ m}$

Marking:

[1] for using horizontal velocity × time
[1] for correct answer $x = 1.2 \text{ m}$ (2 s.f.)

(c) [2]

Vertical component of velocity just before hitting the floor: $v_y = gt = 9.81 \times 0.404 = 3.96 \text{ m s}^{-1}$

Speed $= \sqrt{v_x^2 + v_y^2} = \sqrt{(3.0)^2 + (3.96)^2} = \sqrt{9.00 + 15.68} = \sqrt{24.68} = 4.97 \text{ m s}^{-1}$

Marking:

[1] for finding vertical component of velocity
[1] for combining components to get speed $= 5.0 \text{ m s}^{-1}$ (2 s.f.)

12. (8 marks)

(a) [1]

The total momentum of a system remains constant (is conserved) provided no external resultant force acts on the system.

Marking:

[1] for complete statement including the condition (no external force / closed system)

Common trap: Simply stating "momentum is conserved" without mentioning the condition of no external forces will not earn the mark.

(b) [3]

Using conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(1500)(20) + (2500)(0) = (1500 + 2500)v$ $30000 = 4000v$ $v = 7.5 \text{ m s}^{-1}$

Marking:

[1] for correct formula/principle stated
[1] for correct substitution
[1] for correct answer $v = 7.5 \text{ m s}^{-1}$

(c) [2]

Initial kinetic energy: $KE_i = \frac{1}{2}(1500)(20)^2 = \frac{1}{2}(1500)(400) = 300000 \text{ J}$

Final kinetic energy: $KE_f = \frac{1}{2}(4000)(7.5)^2 = \frac{1}{2}(4000)(56.25) = 112500 \text{ J}$

Kinetic energy lost: $\Delta KE = 300000 - 112500 = 187500 \text{ J} = 1.88 \times 10^5 \text{ J}$

Marking:

[1] for calculating initial and final KE correctly
[1] for correct energy lost $= 1.88 \times 10^5 \text{ J}$

(d) [2]

The collision is inelastic because kinetic energy is not conserved — there is a loss of $1.88 \times 10^5 \text{ J}$ of kinetic energy. In an elastic collision, kinetic energy would be conserved. Since the two vehicles stick together after the collision, this is a perfectly inelastic collision, which is the type that results in the maximum loss of kinetic energy for a given situation.

Marking:

[1] for stating "inelastic" with correct justification (KE not conserved)
[1] for noting that the objects stick together (perfectly inelastic)

13. (8 marks)

(a) [2]

The radius of the circular path is: $r = L\sin\theta = 1.5 \times \sin 30° = 1.5 \times 0.50 = 0.75 \text{ m}$

Marking:

[1] for correct formula $r = L\sin\theta$
[1] for correct answer $r = 0.75 \text{ m}$

(b) [1]

Free-body diagram should show:

Weight $mg$ acting vertically downward
Tension $T$ acting along the string toward the pivot

Marking:

[1] for both forces correctly drawn and labelled

Expected diagram features: The diagram should show the mass with two forces: tension T directed along the string (at 30° to vertical, toward the pivot point above) and weight mg directed vertically downward from the mass. Both arrows should originate from the centre of the mass.

(c) [2]

Resolving vertically (no vertical acceleration): $T\cos\theta = mg$ $T = \frac{mg}{\cos\theta} = \frac{0.20 \times 9.81}{\cos 30°} = \frac{1.962}{0.866} = 2.27 \text{ N}$

Marking:

[1] for correct vertical equilibrium equation
[1] for correct answer $T = 2.3 \text{ N}$ (2 s.f.)

(d) [3]

Resolving horizontally (provides centripetal force): $T\sin\theta = \frac{mv^2}{r}$

From part (c): $T = 2.265$ N, and $r = 0.75$ m

$2.265 \times \sin 30° = \frac{0.20 \times v^2}{0.75}$ $2.265 \times 0.50 = \frac{0.20 v^2}{0.75}$ $1.1325 = 0.2667 v^2$ $v^2 = \frac{1.1325}{0.2667} = 4.247$ $v = 2.06 \text{ m s}^{-1}$

Marking:

[1] for correct horizontal force equation ( $T\sin\theta = mv^2/r$ )
[1] for correct substitution
[1] for correct answer $v = 2.1 \text{ m s}^{-1}$ (2 s.f.)

14. (8 marks)

(a) [2]

Using Newton's second law: $F = ma$ $a = \frac{F}{m} = \frac{8.0}{2.0} = 4.0 \text{ m s}^{-2}$

Marking:

[1] for correct formula
[1] for correct answer $a = 4.0 \text{ m s}^{-2}$

(b) [1]

$v = u + at = 0 + 4.0 \times 4.0 = 16 \text{ m s}^{-1}$

Marking:

[1] for correct answer $v = 16 \text{ m s}^{-1}$

(c) [3]

On the smooth inclined plane, the only force decelerating the block is the component of weight along the plane: $ma' = -mg\sin 25°$ $a' = -g\sin 25° = -9.81 \times 0.4226 = -4.15 \text{ m s}^{-2}$

Using $v^2 = u^2 + 2as$ with final velocity $= 0$ : $0 = (16)^2 + 2(-4.15)s$ $s = \frac{256}{8.30} = 30.8 \text{ m}$

Marking:

[1] for correct deceleration along the plane
[1] for correct kinematic equation and substitution
[1] for correct answer $s = 31 \text{ m}$ (2 s.f.)

(d) [2]

The distance travelled up the plane would decrease. With a rough inclined plane, friction acts down the plane (opposing motion), providing an additional decelerating force. This means the total deceleration is greater than $g\sin 25°$ , so the block comes to rest over a shorter distance.

Marking:

[1] for stating the distance decreases
[1] for correct explanation involving friction adding to the decelerating force

Section C: Long Structured Questions

15. (10 marks)

(a) [2]

Using conservation of energy (or kinematics): $v^2 = u^2 + 2gh = 0 + 2(9.81)(5.0) = 98.1$ $v = \sqrt{98.1} = 9.90 \text{ m s}^{-1}$

Marking:

[1] for correct formula/substitution
[1] for correct answer $v = 9.9 \text{ m s}^{-1}$ (2 s.f.) directed downward

(b) [2]

Using conservation of energy for the rebound: $v'^2 = 2gh' = 2(9.81)(3.2) = 62.784$ $v' = \sqrt{62.784} = 7.92 \text{ m s}^{-1}$

Marking:

[1] for correct formula/substitution
[1] for correct answer $v' = 7.9 \text{ m s}^{-1}$ (2 s.f.) directed upward

(c) [3]

Taking upward as positive:

Momentum just before impact: $p_i = 0.40 \times (-9.90) = -3.96 \text{ kg m s}^{-1}$

Momentum just after impact: $p_f = 0.40 \times (+7.92) = +3.168 \text{ kg m s}^{-1}$

Change in momentum: $\Delta p = p_f - p_i = 3.168 - (-3.96) = 3.168 + 3.96 = 7.13 \text{ kg m s}^{-1}$

Marking:

[1] for correct sign convention
[1] for correct values of momentum before and after
[1] for correct change in momentum $\Delta p = 7.1 \text{ kg m s}^{-1}$ (upward)

Common trap: Students often subtract magnitudes without considering direction. The change in momentum must account for the reversal of direction, so the magnitudes add.

(d) [3]

Using the impulse-momentum theorem: $F_{avg} \times \Delta t = \Delta p$ $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{7.128}{0.020} = 356.4 \text{ N}$

However, this is the net force. The average force exerted by the ground must also support the weight: $F_{ground} = F_{net} + mg = 356.4 + (0.40 \times 9.81) = 356.4 + 3.92 = 360.3 \text{ N}$

Marking:

[1] for using impulse-momentum theorem correctly
[1] for correct net force calculation
[1] for adding weight to get the force by the ground $= 360 \text{ N}$ (2 s.f.)

Common trap: Forgetting to add the weight of the ball. The ground must both provide the impulse to reverse the ball's momentum AND support the ball's weight.

16. (10 marks)

(a) [2]

Two features of a geostationary orbit:

The orbital period is 24 hours (equal to the Earth's rotational period).
The satellite orbits in the equatorial plane (above the equator), moving in the same direction as Earth's rotation.

Marking:

[1] for each correct feature (2 marks total)

(b)(i) [3]

The gravitational force provides the centripetal force: $\frac{GMm}{r^2} = mr\omega^2$

Substituting $\omega = \frac{2\pi}{T}$ : $\frac{GMm}{r^2} = mr\left(\frac{2\pi}{T}\right)^2 = \frac{4\pi^2 mr}{T^2}$

Dividing both sides by $m$ : $\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}$

Rearranging: $GM T^2 = 4\pi^2 r^3$

$\boxed{r^3 = \frac{GMT^2}{4\pi^2}}$

Marking:

[1] for equating gravitational force to centripetal force
[1] for substituting $\omega = 2\pi/T$
[1] for correct algebraic manipulation to obtain the expression

(b)(ii) [3]

$T = 24 \text{ hours} = 24 \times 3600 = 86400 \text{ s}$

$r^3 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(86400)^2}{4\pi^2}$

$(86400)^2 = 7.465 \times 10^9$

$r^3 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(7.465 \times 10^9)}{39.48}$

Numerator: $(6.67 \times 10^{-11})(5.97 \times 10^{24}) = 3.982 \times 10^{14}$ $(3.982 \times 10^{14})(7.465 \times 10^9) = 2.973 \times 10^{24}$

$r^3 = \frac{2.973 \times 10^{24}}{39.48} = 7.530 \times 10^{22}$

$r = (7.530 \times 10^{22})^{1/3} = 4.225 \times 10^7 \text{ m}$

Marking:

[1] for correct conversion of period to seconds
[1] for correct substitution into the formula
[1] for correct answer $r = 4.23 \times 10^7 \text{ m}$ (3 s.f.)

(b)(iii) [2]

Height above Earth's surface: $h = r - R_E = 4.225 \times 10^7 - 6.37 \times 10^6 = 3.588 \times 10^7 \text{ m}$

$h \approx 3.59 \times 10^7 \text{ m} \approx 35900 \text{ km}$

Marking:

[1] for correct method (subtracting Earth's radius)
[1] for correct answer $h = 3.59 \times 10^7 \text{ m}$

End of Answer Key

Mark Summary:

Section	Marks
A: Multiple Choice (Q1–10)	10
B: Structured Questions (Q11–14)	30
C: Long Structured Questions (Q15–16)	20
Total	60