A Level H1 Physics Practice Paper 2

TuitionGoWhere Practice Paper — Physics H1 A-Level

Answer Key — Version 2 of 5

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Section A: Multiple Choice [10 marks]

1. Answer: B

At the highest point of a vertical trajectory, the ball momentarily stops (velocity = 0), but gravity still acts on it, giving a downward acceleration of $9.8 \text{ m s}^{-2}$. Acceleration due to gravity is constant throughout the motion (ignoring air resistance).

Common mistake: Choosing A — students often assume that if velocity is zero, acceleration must also be zero. Acceleration is the rate of change of velocity, not dependent on the instantaneous velocity itself.

[1 mark]

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2. Answer: B

Using $s = \frac{1}{2}(u + v)t$: $s = \frac{1}{2}(0 + 24)(8.0) = \frac{1}{2}(24)(8.0) = 96 \text{ m}$

Alternatively, using $a = \frac{v - u}{t} = \frac{24}{8.0} = 3.0 \text{ m s}^{-2}$, then $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(64) = 96 \text{ m}$.

[1 mark]

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3. Answer: D

Momentum ($p = mv$) 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 only.

[1 mark]

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4. Answer: B

Using conservation of linear momentum (perfectly inelastic collision): $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(2.0)(3.0) + (4.0)(0) = (2.0 + 4.0)v$ $6.0 = 6.0v$ $v = 1.0 \text{ m s}^{-1}$

[1 mark]

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5. Answer: C

Work done: $W = F \times d = 12 \times 5.0 = 60 \text{ J}$

Work done is the product of force and displacement in the direction of the force.

[1 mark]

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6. Answer: C

In projectile motion (ignoring air resistance), there is no horizontal force, so the horizontal velocity remains constant. The vertical acceleration is always $g = 9.8 \text{ m s}^{-2}$ downwards and does not change.

[1 mark]

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7. Answer: C

Total downward force = weight of beam + load = $40 + 20 = 60 \text{ N}$

By symmetry (uniform beam, load at centre), each support carries half: $\text{Reaction at each support} = \frac{60}{2} = 30 \text{ N}$

[1 mark]

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8. Answer: B

First, find the deceleration using $v^2 = u^2 + 2as$: $0 = (20)^2 + 2a(50)$ $0 = 400 + 100a$ $a = -4.0 \text{ m s}^{-2}$

Then, using $F = ma$: $F = 1000 \times 4.0 = 4000 \text{ N}$

[1 mark]

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9. Answer: C

Gain in gravitational potential energy: $\Delta E_p = mgh = 5.0 \times 9.8 \times 10 = 490 \text{ J}$

[1 mark]

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10. Answer: C

Taking the initial direction as positive: $\text{Initial momentum} = 0.50 \times 6.0 = 3.0 \text{ kg m s}^{-1}$ $\text{Final momentum} = 0.50 \times (-4.0) = -2.0 \text{ kg m s}^{-1}$

Change in momentum: $\Delta p = p_{\text{final}} - p_{\text{initial}} = (-2.0) - (3.0) = -5.0 \text{ kg m s}^{-1}$

Magnitude of change in momentum = $5.0 \text{ kg m s}^{-1}$

Common mistake: Students may subtract speeds directly ($6.0 - 4.0 = 2.0$) and multiply by mass to get $1.0$, which is incorrect. Momentum is a vector; direction matters.

[1 mark]

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Section B: Structured Questions [30 marks]

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11. (a) [B1] Newton's first law: An object remains at rest or continues to move at constant velocity [B1] unless acted upon by a resultant (net) external force.

Teaching note: This is also called the law of inertia. The key idea is that a resultant force is needed to change motion, not to maintain it.

[2 marks]

(b) Free-body diagram should show:

• Weight ($W$) acting vertically downward from the centre of the book, labelled $W = mg$ or "weight"
• Normal reaction force ($N$ or $R$) acting vertically upward from the bottom surface of the book, labelled "normal reaction" or "$N$"

[B1] for correct downward force (weight), correctly labelled [B1] for correct upward force (normal reaction), correctly labelled, approximately equal in length to weight arrow

Note: The two arrows should be approximately equal in length since the book is in equilibrium.

[2 marks]

(c) The book remains at rest because the resultant force on it is zero — the weight is balanced by the normal reaction. By Newton's first law, since there is no resultant force, the book remains in its state of rest.

[1 mark]

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12. (a) Using $v = u + at$: $v = 0 + (2.5)(10) = 25 \text{ m s}^{-1}$

[B1] for correct equation or method [B1] for correct answer with unit: $25 \text{ m s}^{-1}$

[2 marks]

(b) Using $s = ut + \frac{1}{2}at^2$: $s = 0 + \frac{1}{2}(2.5)(10)^2 = \frac{1}{2}(2.5)(100) = 125 \text{ m}$

Alternatively, $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 25)(10) = 125 \text{ m}$

[B1] for correct equation or method [B1] for correct answer with unit: $125 \text{ m}$

[2 marks]

(c) Distance during constant velocity phase: $s_2 = v \times t = 25 \times 20 = 500 \text{ m}$

Total distance: $s_{\text{total}} = 125 + 500 = 625 \text{ m}$

[B1] for calculating distance during constant velocity phase [B1] for correct total distance with unit: $625 \text{ m}$

[2 marks]

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13. (a) [B1] In a closed/isolated system, the total momentum remains constant [B1] provided no external forces act on the system.

Teaching note: A "closed system" means no mass enters or leaves, and "no external forces" means the net external force is zero. Both conditions are needed for momentum conservation.

[2 marks]

(b) Taking the initial direction of the ball as positive:

(i) Initial momentum: $p_i = 0.060 \times 25 = 1.50 \text{ kg m s}^{-1}$

Final momentum: $p_f = 0.060 \times (-30) = -1.80 \text{ kg m s}^{-1}$

Change in momentum: $\Delta p = p_f - p_i = (-1.80) - (1.50) = -3.30 \text{ kg m s}^{-1}$

Magnitude = $3.30 \text{ kg m s}^{-1}$

[B1] for correct substitution into momentum change formula [B1] for correct answer: $3.30 \text{ kg m s}^{-1}$ (or $3.3 \text{ kg m s}^{-1}$)

[2 marks]

(ii) Using the impulse-momentum theorem: $F \Delta t = \Delta p$ $F = \frac{|\Delta p|}{\Delta t} = \frac{3.30}{0.0050} = 660 \text{ N}$

[B1] for correct use of $F = \frac{\Delta p}{\Delta t}$ [B1] for correct answer with unit: $660 \text{ N}$

[2 marks]

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14. (a) Horizontal component: $v_x = v \cos\theta = 20 \cos 35° = 20 \times 0.8192 = 16.4 \text{ m s}^{-1} \text{ (to 3 s.f.)}$

[B1] for correct working shown

[1 mark]

(b) Vertical component: $v_y = v \sin\theta = 20 \sin 35° = 20 \times 0.5736 = 11.5 \text{ m s}^{-1} \text{ (to 3 s.f.)}$

[B1] for correct answer: $11.5 \text{ m s}^{-1}$

[1 mark]

(c) At maximum height, vertical velocity $v_y = 0$. Using $v^2 = u^2 + 2as$ vertically: $0 = (11.5)^2 + 2(-9.8)h$ $0 = 132.25 - 19.6h$ $h = \frac{132.25}{19.6} = 6.75 \text{ m}$

[B1] for correct equation [B1] for correct substitution [B1] for correct answer: $6.75 \text{ m}$ (or $6.7 \text{ m}$ to 2 s.f.)

[3 marks]

(d) Time of flight: Using $v = u + at$ for the upward phase: $0 = 11.5 - 9.8t_{\text{up}}$ $t_{\text{up}} = \frac{11.5}{9.8} = 1.173 \text{ s}$

Total time of flight: $T = 2 \times t_{\text{up}} = 2 \times 1.173 = 2.347 \text{ s}$

Horizontal range: $R = v_x \times T = 16.4 \times 2.347 = 38.5 \text{ m}$

Alternatively, using the range formula: $R = \frac{v^2 \sin 2\theta}{g} = \frac{(20)^2 \sin 70°}{9.8} = \frac{400 \times 0.9397}{9.8} = 38.4 \text{ m}$

[B1] for finding time of flight (or using range formula) [B1] for correct substitution [B1] for correct answer: $38.4 \text{ m}$ or $38.5 \text{ m}$ (accept $38 \text{ m}$ to 2 s.f.)

[3 marks]

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15. (a) Weight: $W = mg = 500 \times 9.8 = 4900 \text{ N}$

[B1] for correct answer with unit: $4900 \text{ N}$

[1 mark]

(b) Since the load moves at constant speed, the tension in the cable equals the weight. Work done: $W = F \times d = 4900 \times 12 = 58\,800 \text{ J} = 58.8 \text{ kJ}$

[B1] for correct force (equal to weight since constant speed) [B1] for correct answer: $58\,800 \text{ J}$ or $58.8 \text{ kJ}$

[2 marks]

(c) Power: $P = \frac{W}{t} = \frac{58\,800}{15} = 3920 \text{ W} = 3.92 \text{ kW}$

[B1] for correct formula [B1] for correct answer: $3920 \text{ W}$ or $3.92 \text{ kW}$

[2 marks]

(d) Gain in gravitational potential energy: $\Delta E_p = mgh = 500 \times 9.8 \times 12 = 58\,800 \text{ J}$

This is equal to the work done by the crane because the load moves at constant speed (no change in kinetic energy). By the work-energy principle, all the work done by the crane goes into increasing the gravitational potential energy of the load.

[B1] for correct value of GPE gain: $58\,800 \text{ J}$ [B1] for correct explanation linking constant speed (no $\Delta KE$) to work done = GPE gain

[2 marks]

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Section C: Data Interpretation & Extended Response [20 marks]

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16. (a) The car accelerates uniformly from rest, reaching a velocity of $20 \text{ m s}^{-1}$ in $10 \text{ s}$.

[B1] for stating uniform/increasing velocity/acceleration from rest

[1 mark]

(b) Acceleration = gradient of velocity–time graph: $a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{10 - 0} = \frac{20}{10} = 2.0 \text{ m s}^{-2}$

[B1] for correct method (gradient or $a = \frac{\Delta v}{\Delta t}$) [B1] for correct answer with unit: $2.0 \text{ m s}^{-2}$

[2 marks]

(c) Total distance = area under the velocity–time graph.

The graph forms a trapezium (or can be split into a triangle + rectangle + triangle):

• Area of triangle (0–10 s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
• Area of rectangle (10–20 s): $10 \times 20 = 200 \text{ m}$
• Area of triangle (20–30 s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$

Total distance: $100 + 200 + 100 = 400 \text{ m}$

Alternatively, using the trapezium formula: $s = \frac{1}{2}(20 + 30) \times 20 = \frac{1}{2}(50)(20) = 500 \text{ m}$

Wait — let me recalculate. The total time is 30 s. The shape is a trapezium with parallel sides of 20 s (at $v = 20$) and 0 s... Actually, it's simpler to use the three regions:

• Triangle: $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
• Rectangle: $10 \times 20 = 200 \text{ m}$
• Triangle: $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$

Total: $100 + 200 + 100 = 400 \text{ m}$

[B1] for identifying area under graph as distance [B1] for correct calculation of at least two areas [B1] for correct total: $400 \text{ m}$

[3 marks]

(d) Acceleration–time graph:

• $0$ to $10 \text{ s}$: $a = +2.0 \text{ m s}^{-2}$ (uniform acceleration)
• $10$ to $20 \text{ s}$: $a = 0$ (constant velocity)
• $20$ to $30 \text{ s}$: $a = \frac{0 - 20}{10} = -2.0 \text{ m s}^{-2}$ (uniform deceleration)

[B1] for correct shape (three horizontal segments) [B1] for correct values ($+2.0$, $0$, $-2.0 \text{ m s}^{-2}$) [B1] for correct time intervals ($0$–$10$, $10$–$20$, $20$–$30$ s)

[3 marks]

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17. (a) The data suggests uniform acceleration (constant acceleration). Justification: the velocity increases by $0.40 \text{ m s}^{-1}$ in every $0.5 \text{ s}$ interval, i.e., the rate of change of velocity is constant. Alternatively, velocity is directly proportional to time ($v \propto t$), which is characteristic of uniform acceleration from rest.

[B1] for identifying uniform/constant acceleration [B1] for valid justification (equal velocity increments in equal time intervals, or $v \propto t$)

[2 marks]

(b) Graph requirements:

• Axes correctly labelled with units: "Time / s" (horizontal), "Velocity / m s⁻¹" (vertical)
• Suitable scales used (e.g., 1 cm = 0.5 s horizontally, 1 cm = 0.4 m s⁻¹ vertically)
• All 7 points plotted correctly to within ±½ small square
• Straight best-fit line drawn through the origin

[B1] for correct axes labels and units with suitable scales [B1] for correct plotting of points (allow ±½ small square) [B1] for correct best-fit straight line through origin

[3 marks]

(c) Acceleration = gradient of $v$–$t$ graph.

Using two points on the best-fit line, e.g., $(0, 0)$ and $(3.0, 2.40)$: $a = \frac{\Delta v}{\Delta t} = \frac{2.40 - 0}{3.0 - 0} = \frac{2.40}{3.0} = 0.80 \text{ m s}^{-2}$

[B1] for correct method (gradient calculation shown) [B1] for correct answer: $0.80 \text{ m s}^{-2}$

[2 marks]

(d) The acceleration will not change. On a friction-compensated slope, the component of gravitational force along the slope is $mg\sin\theta$. By Newton's second law, $ma = mg\sin\theta$, so $a = g\sin\theta$. The acceleration is independent of mass — it depends only on $g$ and the slope angle $\theta$. Since neither changes, the acceleration remains the same.

[B1] for stating that acceleration does not change [B1] for correct explanation ($a = g\sin\theta$, independent of mass)

[2 marks]

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18. (a) Work done by a force is defined as the product of the force and the displacement of the object in the direction of the force. Equivalently, work done = force × displacement × $\cos\theta$, where $\theta$ is the angle between the force and displacement.

Accept: $W = Fd\cos\theta$ with explanation of symbols.

[1 mark]

(b) (i) Horizontal component of applied force: $F_x = 50 \cos 25° = 50 \times 0.9063 = 45.3 \text{ N}$

Work done by applied force: $W = F_x \times d = 45.3 \times 6.0 = 272 \text{ J}$

Or directly: $W = Fd\cos\theta = 50 \times 6.0 \times \cos 25° = 300 \times 0.9063 = 272 \text{ J}$

[B1] for correct method ($W = Fd\cos\theta$ or resolving force) [B1] for correct answer: $272 \text{ J}$ (or $270 \text{ J}$ to 2 s.f.)

[2 marks]

(ii) Work done against friction: $W_f = f \times d = 18 \times 6.0 = 108 \text{ J}$

[B1] for correct answer: $108 \text{ J}$

[1 mark]

(iii) By the work-energy principle, the net work done on the box equals the change in kinetic energy:

$W_{\text{net}} = \Delta KE$ $W_{\text{applied}} - W_{\text{friction}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Since the box starts from rest, $u = 0$: $272 - 108 = \frac{1}{2}(8.0)v^2$ $164 = 4.0v^2$ $v^2 = 41$ $v = \sqrt{41} = 6.4 \text{ m s}^{-1}$

[B1] for correct application of work-energy principle [B1] for correct substitution [B1] for correct answer: $6.4 \text{ m s}^{-1}$

[3 marks]

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END OF ANSWER KEY

Total: 60 marks