A Level H1 Physics Practice Paper 4

A-Level Physics H1 Quiz - Mechanics

Answer Key

Section A: Multiple Choice [15 marks]

1. D. Displacement
Displacement is a vector quantity because it has both magnitude and direction. Speed, distance, and energy are scalar quantities — they have magnitude only.
Common mistake: Students often confuse speed (scalar) with velocity (vector), or distance (scalar) with displacement (vector).

2. B. $-9.81 \text{ m s}^{-2}$
At the highest point, the ball's velocity is momentarily zero, but its acceleration is still due to gravity throughout the flight. Taking upward as positive, gravitational acceleration is $-9.81 \text{ m s}^{-2}$.
Common mistake: Students incorrectly assume acceleration is zero when velocity is zero.

3. B. $90 \text{ m}$
Using $v = u + at$: $30 = 0 + a(6.0)$, so $a = 5.0 \text{ m s}^{-2}$.
Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(5.0)(6.0)^2 = 90 \text{ m}$.
Alternatively, average speed = $\frac{0+30}{2} = 15 \text{ m s}^{-1}$, distance = $15 \times 6.0 = 90 \text{ m}$.

4. B. $1.6 \text{ m s}^{-1}$
By conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$
$(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v$
$8.0 = 5.0v$
$v = 1.6 \text{ m s}^{-1}$
This is a perfectly inelastic collision — the objects stick together.

5. B. $3.0 \text{ m s}^{-2}$
Using Newton's Second Law: $F = ma$
$12 = 4.0 \times a$
$a = 3.0 \text{ m s}^{-2}$

6. B. Action and reaction forces are equal in magnitude and opposite in direction.
Newton's Third Law states that for every action, there is an equal and opposite reaction. These forces act on different bodies, so they do not cancel each other out. They can be contact or non-contact (e.g., gravitational) forces.
Common mistake: Students think action-reaction pairs act on the same body or cancel out.

7. C. $49 \text{ N}$
The normal contact force balances the weight of the block: $R = mg = 5.0 \times 9.81 = 49.05 \text{ N} \approx 49 \text{ N}$.
Common mistake: Students forget to multiply by $g$ and simply write the mass value.

8. B. $20\sqrt{3} \text{ m s}^{-1}$ (or $34.6 \text{ m s}^{-1}$)
Horizontal component: $v_x = v \cos\theta = 40 \cos 30^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.6 \text{ m s}^{-1}$.
Note: Both B and C are numerically equivalent, but B is the exact form. In an exam, either would be accepted.

9. C. $15.0 \text{ kg m s}^{-1}$
Taking the initial direction as positive:
Initial momentum = $1.5 \times 6.0 = 9.0 \text{ kg m s}^{-1}$
Final momentum = $1.5 \times (-4.0) = -6.0 \text{ kg m s}^{-1}$
Change in momentum = final − initial = $-6.0 - 9.0 = -15.0 \text{ kg m s}^{-1}$
Magnitude = $15.0 \text{ kg m s}^{-1}$
Common mistake: Students subtract speeds instead of momenta, or forget to account for the direction change.

10. C. Energy and work
Both energy and work have the unit joule (J). Work is the transfer of energy.

• Work and force: work is in joules (J), force is in newtons (N) — different.
• Power and momentum: power is in watts (W), momentum is in kg m s⁻¹ — different.
• Impulse and energy: impulse is in N s (or kg m s⁻¹), energy is in joules (J) — different.

11. B. $9600 \text{ N}$
Centripetal force: $F = \frac{mv^2}{r} = \frac{1200 \times (20)^2}{50} = \frac{1200 \times 400}{50} = \frac{480000}{50} = 9600 \text{ N}$

12. D. $400 \text{ N m}^{-1}$
Elastic potential energy: $E = \frac{1}{2}kx^2$
$2.0 = \frac{1}{2} \times k \times (0.10)^2$
$2.0 = \frac{1}{2} \times k \times 0.01$
$k = \frac{2.0 \times 2}{0.01} = 400 \text{ N m}^{-1}$

13. B. $98 \text{ J}$
By conservation of energy: loss in gravitational potential energy = gain in kinetic energy
$KE = mgh = 0.50 \times 9.81 \times 20 = 98.1 \text{ J} \approx 98 \text{ J}$
Common mistake: Students use $v^2 = 2gh$ and then $\frac{1}{2}mv^2$ but make arithmetic errors.

14. A. The net force on the object is zero.
By Newton's First Law, an object moving with constant velocity (which includes being at rest) has zero net force acting on it. This does not mean no forces act — it means the forces are balanced.
Common mistake: Students confuse "no net force" with "no forces at all."

15. C. $147 \text{ J}$
Work done against gravity: $W = mgh = 3.0 \times 9.81 \times 5.0 = 147.15 \text{ J} \approx 147 \text{ J}$
Common mistake: Students forget to include $g$ in the calculation.

Section B: Structured Questions [35 marks]

16. (a) [2 marks]
The principle of conservation of linear momentum states that the total momentum of a closed/isolated system remains constant, provided that no external forces act on the system.
Equivalently: the total momentum before an interaction equals the total momentum after the interaction.
Marking: [B1] for "total momentum is constant/unchanged"; [B1] for "no external forces" or "closed/isolated system" qualifier.

(b) (i) [2 marks]
Taking the initial direction of motion as positive:
Initial momentum = $0.060 \times 25 = 1.5 \text{ kg m s}^{-1}$
Final momentum = $0.060 \times (-18) = -1.08 \text{ kg m s}^{-1}$
Change in momentum = final − initial = $-1.08 - 1.5 = -2.58 \text{ kg m s}^{-1}$
Magnitude of change in momentum = $2.58 \text{ kg m s}^{-1}$ (or $2.6 \text{ kg m s}^{-1}$ to 2 s.f.)
Marking: [M1] for correct substitution into momentum change formula; [A1] for correct answer with unit.

(b) (ii) [2 marks]
Using the impulse-momentum theorem: $F \Delta t = \Delta p$
$F = \frac{\Delta p}{\Delta t} = \frac{2.58}{0.020} = 129 \text{ N} \approx 130 \text{ N}$
Marking: [M1] for using $F = \Delta p / \Delta t$; [A1] for correct answer.

(b) (iii) [1 mark]
The force exerted by the wall on the ball is in the opposite direction to the initial motion of the ball (i.e., away from the wall).
This is because the wall pushes the ball back, causing it to rebound.
Marking: [B1] for correct direction stated.

17. (a) [2 marks]
Using $v^2 = u^2 + 2as$:
$(30)^2 = 0 + 2 \times a \times 200$
$900 = 400a$
$a = 2.25 \text{ m s}^{-2}$
Marking: [M1] for correct substitution; [A1] for correct answer with unit.

(b) [2 marks]
Using $v = u + at$:
$30 = 0 + 2.25 \times t$
$t = \frac{30}{2.25} = 13.3 \text{ s}$
Marking: [M1] for correct method; [A1] for correct answer.

(c) [2 marks]
Work done = force × distance = $ma \times s$
$F = 1500 \times 2.25 = 3375 \text{ N}$
$W = 3375 \times 200 = 675000 \text{ J} = 675 \text{ kJ}$
Alternatively, work done = change in kinetic energy = $\frac{1}{2}mv^2 - 0 = \frac{1}{2}(1500)(30)^2 = 675000 \text{ J}$
Marking: [M1] for correct method; [A1] for correct answer.

(d) [2 marks]
Average power = $\frac{\text{Work done}}{\text{time}} = \frac{675000}{13.3} = 50752 \text{ W} \approx 50.8 \text{ kW}$
Alternatively, $P = Fv_{\text{average}} = 3375 \times 15 = 50625 \text{ W} \approx 50.6 \text{ kW}$
Marking: [M1] for correct method; [A1] for correct answer.

18. (a) [1 mark]
$v_x = v \cos\theta = 28 \cos 45^\circ = 28 \times \frac{\sqrt{2}}{2} = 19.8 \text{ m s}^{-1}$
Marking: [B1] for correct answer.

(b) [1 mark]
$v_y = v \sin\theta = 28 \sin 45^\circ = 28 \times \frac{\sqrt{2}}{2} = 19.8 \text{ m s}^{-1}$
Marking: [B1] for correct answer.

(c) [2 marks]
At maximum height, vertical velocity = 0.
Using $v_y^2 = u_y^2 - 2gh$:
$0 = (19.8)^2 - 2(9.81)h$
$h = \frac{(19.8)^2}{2 \times 9.81} = \frac{392.04}{19.62} = 19.98 \text{ m} \approx 20.0 \text{ m}$
Marking: [M1] for correct method; [A1] for correct answer.

(d) [2 marks]
Time of flight: $t = \frac{2u_y}{g} = \frac{2 \times 19.8}{9.81} = 4.04 \text{ s}$
Range: $R = v_x \times t = 19.8 \times 4.04 = 79.9 \text{ m} \approx 80.0 \text{ m}$
Alternatively, using $R = \frac{v^2 \sin 2\theta}{g} = \frac{(28)^2 \sin 90^\circ}{9.81} = \frac{784}{9.81} = 79.9 \text{ m}$
Marking: [M1] for correct method; [A1] for correct answer.

(e) [2 marks]
Air resistance would decrease the range.
Reason: Air resistance opposes the motion of the ball, reducing both the horizontal and vertical components of velocity throughout the flight. This reduces the horizontal speed (shortening the range) and also reduces the maximum height and time of flight.
Marking: [B1] for "decrease"; [B1] for valid reason involving opposition to motion/energy loss.

19. (a) [2 marks]
The free-body diagram should show:

• Weight ($mg$) acting vertically downward from the centre of the block
• Normal reaction ($R$) acting perpendicular to the surface, away from the surface
• Frictional force ($f$) acting up the slope (opposing the direction of motion/tendency to slide)
  Marking: [B1] for weight and normal reaction correctly drawn; [B1] for friction correctly drawn in the correct direction.

(b) [1 mark]
Component of weight down the slope = $mg\sin\theta = 4.0 \times 9.81 \times \sin 30^\circ = 4.0 \times 9.81 \times 0.5 = 19.62 \text{ N} \approx 19.6 \text{ N}$
Marking: [B1] for correct answer.

(c) [1 mark]
Normal reaction = $mg\cos\theta = 4.0 \times 9.81 \times \cos 30^\circ = 4.0 \times 9.81 \times 0.866 = 33.98 \text{ N} \approx 34.0 \text{ N}$
Marking: [B1] for correct answer.

(d) [1 mark]
Frictional force = $\mu R = 0.35 \times 33.98 = 11.89 \text{ N} \approx 11.9 \text{ N}$
Marking: [B1] for correct answer.

(e) [2 marks]
The component of weight down the slope ($19.6 \text{ N}$) is greater than the maximum frictional force ($11.9 \text{ N}$).
Therefore, there is a net force down the slope, and the block will slide down.
Net force down slope = $19.6 - 11.9 = 7.7 \text{ N}$
Marking: [M1] for comparing the two forces; [A1] for correct conclusion that the block will slide.

Section C: Free Response [30 marks]

20. (a) [2 marks]
Centripetal acceleration is the acceleration directed towards the centre of the circular path. It is responsible for changing the direction of the velocity (not the speed) of an object moving in a circle. Its magnitude is given by $a = \frac{v^2}{r}$ or $a = \omega^2 r$.
Marking: [B1] for "towards the centre"; [B1] for "changes direction of velocity" or equivalent.

(b) (i) [2 marks]
$a_c = \frac{v^2}{r} = \frac{(6.0)^2}{1.5} = \frac{36}{1.5} = 24 \text{ m s}^{-2}$
Marking: [M1] for correct substitution; [A1] for correct answer with unit.

(b) (ii) [2 marks]
At the lowest point, the tension acts upward and weight acts downward. The net force towards the centre (upward) provides the centripetal force:
$T - mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r} = (0.20 \times 9.81) + \frac{0.20 \times (6.0)^2}{1.5}$
$T = 1.962 + 4.8 = 6.762 \text{ N} \approx 6.8 \text{ N}$
Marking: [M1] for correct equation $T - mg = mv^2/r$; [A1] for correct answer.

(c) (i) [2 marks]
At the highest point, for the string to remain taut with minimum tension (T = 0), the weight alone provides the centripetal force:
$mg = \frac{mv_{\text{top}}^2}{r}$
$v_{\text{top}}^2 = gr = 9.81 \times 1.5 = 14.715$
$v_{\text{top}} = \sqrt{14.715} = 3.84 \text{ m s}^{-1}$
Marking: [M1] for setting $mg = mv^2/r$; [A1] for correct answer.

(c) (ii) [3 marks]
Using conservation of energy between the lowest and highest points:
The height difference between lowest and highest point = $2r = 2 \times 1.5 = 3.0 \text{ m}$
$\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)$
$\frac{1}{2}v_{\text{bottom}}^2 = \frac{1}{2}v_{\text{top}}^2 + g(2r)$
$\frac{1}{2}v_{\text{bottom}}^2 = \frac{1}{2}(14.715) + 9.81 \times 3.0$
$\frac{1}{2}v_{\text{bottom}}^2 = 7.3575 + 29.43 = 36.7875$
$v_{\text{bottom}}^2 = 73.575$
$v_{\text{bottom}} = \sqrt{73.575} = 8.58 \text{ m s}^{-1}$
Marking: [M1] for correct energy conservation equation; [M1] for correct substitution of values; [A1] for correct final answer.

(d) [2 marks]
The tension varies because the weight of the sphere has a component that either adds to or subtracts from the tension depending on the position in the circle:

• At the lowest point: tension must support the weight AND provide the centripetal force, so $T = mg + \frac{mv^2}{r}$ (maximum tension).
• At the highest point: weight acts towards the centre, so $T = \frac{mv^2}{r} - mg$ (minimum tension).
• At intermediate positions: the radial component of weight varies as $\cos\theta$, causing the tension to vary continuously.
  Additionally, the speed varies with height due to energy conservation, which also affects the tension.
  Marking: [B1] for explaining that weight's radial component varies with position; [B1] for explaining the difference between top and bottom positions.

End of Answer Key

Total: 80 marks