A Level H2 Physics Practice Paper 4

TuitionGoWhere Practice Paper — Physics H2 A-Level

Answer Key: Practice Paper — Mechanics (Version 4)

Section A: Multiple Choice [10 marks]

1. B — $20.4 \text{ m}$ [1 mark]

Teaching note: At maximum height, the final velocity is zero. Using $v^2 = u^2 - 2gh$ with $v = 0$: $0 = (20)^2 - 2(9.81)h \implies h = \frac{400}{19.62} = 20.4 \text{ m}$ This uses the kinematic equation for constant acceleration (deceleration due to gravity). The key insight is that at the highest point, the ball momentarily stops before falling back down.

2. B — $90 \text{ m}$ [1 mark]

Teaching note: For uniform acceleration from rest, $s = \frac{1}{2}(u+v)t = \frac{1}{2}(0+30)(6.0) = 90 \text{ m}$. Alternatively, find acceleration first: $a = \frac{v-u}{t} = \frac{30}{6} = 5 \text{ m s}^{-2}$, then $s = \frac{1}{2}at^2 = \frac{1}{2}(5)(36) = 90 \text{ m}$.

3. D — Momentum [1 mark]

Teaching note: Momentum $\mathbf{p} = m\mathbf{v}$ is a vector quantity because it has both magnitude and direction (the direction of velocity). Energy, power, and speed are all scalar quantities — they have magnitude but no direction. This is a fundamental distinction in mechanics.

4. B — $4.0 \text{ m s}^{-2}$ [1 mark]

Teaching note: From Newton's second law, $F = ma$, so $a = \frac{F}{m} = \frac{12}{3.0} = 4.0 \text{ m s}^{-2}$. This is a direct application of $F = ma$ on a frictionless surface where the net force equals the applied force.

5. C — $8.0 \text{ m s}^{-2}$ [1 mark]

Teaching note: Centripetal acceleration is given by $a_c = \frac{v^2}{r} = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8.0 \text{ m s}^{-2}$. This acceleration is always directed toward the center of the circular path. Even though the speed is constant, the changing direction of velocity means the object is accelerating.

6. C — $34.6 \text{ m s}^{-1}$ [1 mark]

Teaching note: The horizontal component is $u_x = u\cos\theta = 40\cos 30° = 40 \times 0.866 = 34.6 \text{ m s}^{-1}$. Students often confuse $\cos$ and $\sin$ — remember that the horizontal component uses $\cos$ when the angle is measured from the horizontal.

7. B — $2.0 \text{ m s}^{-1}$ [1 mark]

Teaching note: This is a perfectly inelastic collision (objects stick together). Using conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(2.0)(5.0) + (3.0)(0) = (2.0 + 3.0)v$ $10 = 5.0v \implies v = 2.0 \text{ m s}^{-1}$ Note that kinetic energy is NOT conserved in this type of collision — some energy is converted to heat, sound, and deformation.

8. C — $122.6 \text{ N}$ [1 mark]

Teaching note: Taking moments about the left support (so the left reaction has zero moment): $\text{Anticlockwise} = \text{Clockwise}$ $R_R \times 4.0 = (10 \times 9.81) \times 2.0 + (20 \times 9.81) \times 1.0$ $R_R \times 4.0 = 196.2 + 196.2 = 392.4$ $R_R = 98.1 \text{ N}$

Wait — let me recalculate. The plank's weight acts at its center (2.0 m from left), and the child sits 1.0 m from the left: $R_R \times 4.0 = (10 \times 9.81) \times 2.0 + (20 \times 9.81) \times 1.0$ $R_R \times 4.0 = 196.2 + 196.2 = 392.4$ $R_R = 98.1 \text{ N}$

Hmm, that gives 98.1 N which is option B. Let me recheck: $10 \times 9.81 = 98.1$, times 2.0 = 196.2. $20 \times 9.81 = 196.2$, times 1.0 = 196.2. Sum = 392.4. Divided by 4.0 = 98.1 N.

Corrected answer: B — $98.1 \text{ N}$ [1 mark]

Teaching note: Taking moments about the left support: The plank's weight (98.1 N) acts at the center (2.0 m from left), and the child's weight (196.2 N) acts 1.0 m from the left. Setting anticlockwise moments = clockwise moments: $R_R \times 4.0 = 98.1 \times 2.0 + 196.2 \times 1.0 = 196.2 + 196.2 = 392.4$, giving $R_R = 98.1 \text{ N}$.

9. A — It becomes $\frac{1}{\sqrt{2}}$ times the original. [1 mark]

Teaching note: For a circular orbit, $\frac{GMm}{r^2} = \frac{mv^2}{r}$, so $v = \sqrt{\frac{GM}{r}}$. If $r$ doubles, $v$ becomes $\sqrt{\frac{GM}{2r}} = \frac{1}{\sqrt{2}}\sqrt{\frac{GM}{r}}$. So the orbital speed decreases by a factor of $\frac{1}{\sqrt{2}}$. This is an important result in orbital mechanics — higher orbits are slower.

10. B — $1.0 \text{ J}$ [1 mark]

Teaching note: Elastic potential energy is given by $E = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2 = \frac{1}{2}(200)(0.01) = 1.0 \text{ J}$. This energy is stored in the spring and can be released when the spring returns to its natural length.

Section B: Structured Questions [30 marks]

11. [6 marks]

(a) Newton's second law states that the net force acting on an object is equal to the rate of change of its momentum. For constant mass, this simplifies to $F = ma$, where the acceleration is in the direction of the net force. [1 mark]

Marking: Award 1 mark for stating that net force equals rate of change of momentum, OR $F = ma$ with correct description of direction.

(b)(i) The normal contact force is found by resolving forces vertically. The applied force has an upward vertical component $F\sin\theta$ that partially supports the weight:

$N + F\sin\theta = mg$ $N = mg - F\sin\theta$ $N = (5.0)(9.81) - (30)\sin 25°$ $N = 49.05 - 30 \times 0.4226$ $N = 49.05 - 12.68$ $\boxed{N = 36.4 \text{ N}}$

[2 marks] — 1 mark for correct equation, 1 mark for correct numerical answer.

Common mistake: Students often forget that the upward component of the applied force reduces the normal force. If the force were applied downward at an angle, it would increase the normal force.

(b)(ii) The friction force is: $f = \mu_k N = 0.35 \times 36.4 = 12.7 \text{ N}$

The net horizontal force is: $F_{\text{net}} = F\cos\theta - f = 30\cos 25° - 12.7$ $F_{\text{net}} = 30 \times 0.9063 - 12.7 = 27.19 - 12.7 = 14.5 \text{ N}$

The acceleration is: $a = \frac{F_{\text{net}}}{m} = \frac{14.5}{5.0}$ $\boxed{a = 2.90 \text{ m s}^{-2}}$

[3 marks] — 1 mark for friction force, 1 mark for net force, 1 mark for acceleration.

Teaching note: This is a multi-step problem requiring careful resolution of forces. The key steps are: (1) find the normal force by vertical equilibrium, (2) calculate friction, (3) find net horizontal force, (4) apply Newton's second law.

12. [7 marks]

(a) The principle of conservation of linear momentum states that the total momentum of a closed system remains constant (is conserved) provided that no external resultant force acts on the system. [1 mark]

Marking: Award 1 mark for a complete statement including both the constancy of total momentum AND the condition of no external force / closed system.

(b)(i) Taking the initial direction of trolley A as positive:

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$ $(0.80)(2.5) + (1.2)(0) = (0.80)(-0.50) + (1.2)v_B$ $2.0 = -0.40 + 1.2v_B$ $1.2v_B = 2.4$ $\boxed{v_B = 2.0 \text{ m s}^{-1}}$

The positive sign means trolley B moves in the original direction of trolley A.

[3 marks] — 1 mark for correct equation with signs, 1 mark for correct substitution, 1 mark for correct answer with unit.

Common mistake: Forgetting that trolley A rebounds, so $v_A$ must be negative. This is the most common sign error in collision problems.

(b)(ii) Calculate total kinetic energy before and after:

Before: $KE_{\text{before}} = \frac{1}{2}(0.80)(2.5)^2 + 0 = \frac{1}{2}(0.80)(6.25) = 2.50 \text{ J}$

After: $KE_{\text{after}} = \frac{1}{2}(0.80)(-0.50)^2 + \frac{1}{2}(1.2)(2.0)^2 = \frac{1}{2}(0.80)(0.25) + \frac{1}{2}(1.2)(4.0)$ $= 0.10 + 2.40 = 2.50 \text{ J}$

Since $KE_{\text{before}} = KE_{\text{after}} = 2.50 \text{ J}$, kinetic energy is conserved, so the collision is elastic. [3 marks]

Marking: 1 mark for calculating KE before, 1 mark for calculating KE after, 1 mark for correct conclusion.

Teaching note: An elastic collision is one where both momentum AND kinetic energy are conserved. An inelastic collision conserves momentum but not kinetic energy. In a perfectly inelastic collision, the objects stick together and the maximum kinetic energy is lost.

13. [6 marks]

(a) Horizontal component: $u_x = u\cos\theta = 20\cos 37° = 20 \times 0.7986 = 15.97 \approx 16.0 \text{ m s}^{-1}$

Vertical component (upward positive): $u_y = u\sin\theta = 20\sin 37° = 20 \times 0.6018 = 12.04 \approx 12.0 \text{ m s}^{-1}$

[2 marks] — 1 mark each for horizontal and vertical components.

Note: Using $\cos 37° \approx 0.80$ and $\sin 37° \approx 0.60$ gives $u_x = 16.0 \text{ m s}^{-1}$ and $u_y = 12.0 \text{ m s}^{-1}$.

(b) Taking upward as positive, the vertical displacement is $-45 \text{ m}$ (the ball ends up 45 m below the launch point):

$s_y = u_y t + \frac{1}{2}a_y t^2$ $-45 = 12.0t + \frac{1}{2}(-9.81)t^2$ $-45 = 12.0t - 4.905t^2$ $4.905t^2 - 12.0t - 45 = 0$

Using the quadratic formula: $t = \frac{12.0 \pm \sqrt{(-12.0)^2 + 4(4.905)(45)}}{2(4.905)}$ $t = \frac{12.0 \pm \sqrt{144 + 882.9}}{9.81}$ $t = \frac{12.0 \pm \sqrt{1026.9}}{9.81}$ $t = \frac{12.0 \pm 32.05}{9.81}$

Taking the positive root: $t = \frac{12.0 + 32.05}{9.81} = \frac{44.05}{9.81}$ $\boxed{t = 4.49 \text{ s}}$

[2 marks] — 1 mark for correct equation, 1 mark for correct answer.

(c) The horizontal distance from the base of the cliff: $R = u_x \times t = 16.0 \times 4.49$ $\boxed{R = 71.8 \text{ m}}$

[2 marks] — 1 mark for using horizontal velocity × time, 1 mark for correct answer. Allow ecf from (a) and (b).

Teaching note: Projectile motion problems are solved by treating horizontal and vertical motions independently. Horizontal motion has zero acceleration (constant velocity), while vertical motion has constant acceleration $g$ downward.

14. [5 marks]

(a) Centripetal force is the net force directed toward the center of a circular path that keeps an object moving in a circle. It is not a new type of force but is provided by forces such as tension, gravity, friction, or a combination. [1 mark]

Marking: Must mention "toward the center" and "circular path" for the mark.

(b)(i) $F_c = \frac{mv^2}{r} = \frac{900 \times (15)^2}{50} = \frac{900 \times 225}{50} = \frac{202500}{50}$ $\boxed{F_c = 4050 \text{ N}}$

[2 marks] — 1 mark for correct formula, 1 mark for correct answer.

(b)(ii) For a banked curve with no friction, resolving forces:

Vertically: $N\cos\theta = mg$ ... (i) Horizontally: $N\sin\theta = \frac{mv^2}{r}$ ... (ii)

Dividing (ii) by (i): $\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}$ $\tan\theta = \frac{v^2}{rg}$ $\theta = \tan^{-1}\left(\frac{v^2}{rg}\right)$

Substituting values: $\theta = \tan^{-1}\left(\frac{(15)^2}{50 \times 9.81}\right) = \tan^{-1}\left(\frac{225}{490.5}\right) = \tan^{-1}(0.4587)$ $\boxed{\theta = 24.6°}$

[2 marks] — 1 mark for derivation, 1 mark for correct angle.

Teaching note: On a banked curve, the horizontal component of the normal force provides the centripetal force. The faster the car or the tighter the bend, the greater the banking angle needed.

15. [6 marks]

(a) The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy: $W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ [1 mark]

(b)(i) Using conservation of energy (or $v^2 = u^2 + 2as$): $v^2 = 0 + 2(9.81)(10.0) = 196.2$ $\boxed{v = 14.0 \text{ m s}^{-1}}$

[2 marks] — 1 mark for correct method, 1 mark for correct answer.

(b)(ii) Using $v^2 = u^2 + 2gh$ for the rebound (final speed at max height = 0): $0 = v_{\text{after}}^2 - 2(9.81)(6.4)$ $v_{\text{after}}^2 = 2 \times 9.81 \times 6.4 = 125.57$ $\boxed{v_{\text{after}} = 11.2 \text{ m s}^{-1}}$

[1 mark] for correct answer.

(b)(iii) Energy lost = $KE_{\text{before}} - KE_{\text{after}}$ $= \frac{1}{2}(0.50)(14.0)^2 - \frac{1}{2}(0.50)(11.2)^2$ $= \frac{1}{2}(0.50)(196) - \frac{1}{2}(0.50)(125.44)$ $= 49.0 - 31.36$ $\boxed{= 17.6 \text{ J}}$

[2 marks] — 1 mark for correct KE values, 1 mark for correct energy lost.

Teaching note: The energy lost during the collision is converted to other forms — sound, heat, and permanent deformation of the ball or ground. The coefficient of restitution can also be found: $e = \frac{11.2}{14.0} = 0.80$.

Section C: Long Structured Questions [20 marks]

16. [10 marks]

(a) The component of gravitational force down the slope: $F_{\text{parallel}} = mg\sin\theta = 1.5 \times 9.81 \times \sin 15°$ $= 1.5 \times 9.81 \times 0.2588$ $\boxed{F_{\text{parallel}} = 3.81 \text{ N}}$

[2 marks] — 1 mark for correct formula, 1 mark for correct answer.

(b) By Newton's second law along the slope (frictionless): $a = \frac{F_{\text{parallel}}}{m} = g\sin\theta = 9.81 \times \sin 15°$ $\boxed{a = 2.54 \text{ m s}^{-2}}$

[2 marks] — 1 mark for correct method, 1 mark for correct answer.

(c) Using $v^2 = u^2 + 2as$ with $u = 0$: $v^2 = 0 + 2(2.54)(2.0) = 10.16$ $\boxed{v = 3.19 \text{ m s}^{-1}}$

[2 marks] — 1 mark for correct equation, 1 mark for correct answer.

(d)(i) The measured speed is less than theoretical because of friction between the trolley and the slope (or air resistance). [1 mark]

(d)(ii) Using the work-energy theorem. The net work done equals the change in kinetic energy:

Theoretical KE at bottom (frictionless): $\frac{1}{2}mv_{\text{theoretical}}^2 = \frac{1}{2}(1.5)(3.19)^2 = 7.63 \text{ J}$

Actual KE at bottom: $\frac{1}{2}mv_{\text{measured}}^2 = \frac{1}{2}(1.5)(1.8)^2 = \frac{1}{2}(1.5)(3.24) = 2.43 \text{ J}$

Energy lost to friction = $7.63 - 2.43 = 5.20 \text{ J}$

This energy loss equals the work done by friction: $f \times d = 5.20$ $f \times 2.0 = 5.20$ $\boxed{f = 2.60 \text{ N}}$

Alternatively, using forces: Net force with friction: $mg\sin\theta - f = ma_{\text{measured}}$ From $v^2 = 2as$: $a_{\text{measured}} = \frac{(1.8)^2}{2(2.0)} = \frac{3.24}{4.0} = 0.81 \text{ m s}^{-2}$ $f = mg\sin\theta - ma_{\text{measured}} = 3.81 - 1.5 \times 0.81 = 3.81 - 1.215 = 2.60 \text{ N}$

[3 marks] — 1 mark for calculating measured acceleration or KE, 1 mark for correct method, 1 mark for correct frictional force.

Teaching note: This question tests the work-energy theorem in a practical context. The difference between theoretical and measured values is a common theme in experimental physics and is how we quantify real-world effects like friction.

17. [10 marks]

(a) Newton's law of universal gravitation states that every particle attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres:

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

where:

• $F$ = gravitational force (N)
• $G$ = gravitational constant ($6.674 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$)
• $M$ = mass of the Earth (kg)
• $m$ = mass of the spacecraft (kg)
• $r$ = distance between centres (m)

[2 marks] — 1 mark for equation, 1 mark for defining symbols.

(b) Orbital radius: $r = R_E + h = 6.37 \times 10^6 + 400 \times 10^3 = 6.77 \times 10^6 \text{ m}$

$F = \frac{GMm}{r^2} = \frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})(2000)}{(6.77 \times 10^6)^2}$ $= \frac{(6.674 \times 10^{-11})(1.194 \times 10^{28})}{4.583 \times 10^{13}}$ $= \frac{7.969 \times 10^{17}}{4.583 \times 10^{13}}$ $\boxed{F = 1.74 \times 10^4 \text{ N} \approx 17400 \text{ N}}$

[3 marks] — 1 mark for correct orbital radius, 1 mark for correct substitution, 1 mark for correct answer.

(c) For a circular orbit, the gravitational force provides the centripetal force:

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

Cancel $m$ from both sides: $\frac{GM}{r^2} = \frac{v^2}{r}$

Multiply both sides by $r$: $\frac{GM}{r} = v^2$

Therefore: $\boxed{v = \sqrt{\frac{GM}{r}}}$

[2 marks] — 1 mark for equating gravitational force to centripetal force, 1 mark for correct derivation to final expression.

(d) $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6}}$ $= \sqrt{\frac{3.984 \times 10^{14}}{6.77 \times 10^6}}$ $= \sqrt{5.885 \times 10^7}$ $\boxed{v = 7671 \text{ m s}^{-1} \approx 7.67 \times 10^3 \text{ m s}^{-1}}$

[2 marks] — 1 mark for correct substitution, 1 mark for correct answer.

(e) The orbital period: $T = \frac{2\pi r}{v} = \frac{2\pi(6.77 \times 10^6)}{7671} = \frac{4.254 \times 10^7}{7671}$ $\boxed{T = 5546 \text{ s} \approx 92.4 \text{ minutes}}$

[1 mark] for correct answer.

Teaching note: Orbital mechanics connects Newton's law of gravitation with circular motion. The key insight is that gravity provides the centripetal force. Notice that the orbital speed and period are independent of the mass of the orbiting object — a heavier satellite orbits at the same speed as a lighter one at the same altitude.

Total: 60 marks

Mark Summary: