A Level H1 Physics Practice Paper 2

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key — Practice Paper (Version 2 of 5)

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Section A: Multiple Choice [30 marks]

1. B. $20.4 \text{ m}$ [3 marks]

Working: Using $v^2 = u^2 + 2as$ with $v = 0$, $u = 20 \text{ m s}^{-1}$, $a = -9.81 \text{ m s}^{-2}$: $0 = (20)^2 + 2(-9.81)s$ $s = \frac{400}{19.62} = 20.4 \text{ m}$

Teaching note: At maximum height, the final velocity is zero. The acceleration is $-g$ because gravity opposes the upward motion. Students often forget the negative sign on acceleration or use $s = ut + \frac{1}{2}at^2$ without first finding the time, which adds unnecessary steps.

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2. B. $96 \text{ m}$ [3 marks]

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

Alternatively, $a = \frac{v - u}{t} = \frac{24}{8} = 3 \text{ m s}^{-2}$, then $s = \frac{1}{2}at^2 = \frac{1}{2}(3)(64) = 96 \text{ m}$.

Teaching note: The average speed method $\frac{1}{2}(u+v)t$ is the quickest approach for uniform acceleration from rest. Students should recognise that both methods are valid.

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3. A. $1.33 \text{ m s}^{-1}$ [3 marks]

Working: Conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(3.0)(4.0) + (6.0)(0) = (3.0 + 6.0)v$ $12.0 = 9.0v$ $v = 1.33 \text{ m s}^{-1}$

Teaching note: In a perfectly inelastic collision (objects stick together), momentum is conserved but kinetic energy is not. The common velocity is always less than the initial velocity of the moving object.

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4. C. Momentum [3 marks]

Working: Momentum ($p = mv$) has both magnitude and direction, making it a vector. Kinetic energy, power, and work done are all scalar quantities — they have magnitude only.

Teaching note: A common error is confusing momentum with kinetic energy. Both involve mass and velocity, but momentum is a vector (direction matters) while kinetic energy is a scalar. Students should memorise the key vectors in mechanics: displacement, velocity, acceleration, force, and momentum.

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5. C. $60 \text{ J}$ [3 marks]

Working: $W = F \times s = 15 \times 4.0 = 60 \text{ J}$

Teaching note: Work done by a force is the product of the force and the displacement in the direction of the force. Since the displacement is in the direction of the force, no cosine factor is needed. If the force acted at an angle, students would need $W = Fs\cos\theta$.

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6. C. $147 \text{ J}$ [3 marks]

Working: $\Delta E_p = mgh = 5.0 \times 9.81 \times 3.0 = 147.15 \approx 147 \text{ J}$

Teating note: Gravitational potential energy depends on the vertical height gained, not the path taken. The speed of lifting is irrelevant (as long as it's the change in GPE). Students sometimes mistakenly use $g = 10 \text{ m s}^{-2}$ when the question specifies $9.81$.

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7. C. $30 \text{ m}$ [3 marks]

Working: Horizontal motion: $s_x = v_x \times t = 12 \times 2.5 = 30 \text{ m}$

Teaching note: In projectile motion, horizontal and vertical motions are independent. The horizontal velocity remains constant (no horizontal acceleration), so range = horizontal speed × time of flight. The time of flight is determined entirely by the vertical drop.

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8. A. $1.0 \text{ J}$ [3 marks]

Working: $E = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.10)^2 = \frac{1}{2}(200)(0.01) = 1.0 \text{ J}$

Teaching note: Elastic potential energy depends on the square of the extension. Doubling the compression would quadruple the stored energy. Students sometimes forget the factor of $\frac{1}{2}$ or forget to square the extension.

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9. C. $5.0 \text{ kg m s}^{-1}$ [3 marks]

Working: Taking the initial direction as positive: $\Delta p = mv_{\text{final}} - mv_{\text{initial}} = (0.50)(-4.0) - (0.50)(6.0) = -2.0 - 3.0 = -5.0 \text{ kg m s}^{-1}$ Magnitude of change = $5.0 \text{ kg m s}^{-1}$

Teaching note: This is a common trap question. The change in momentum must account for direction. The ball reverses direction, so the final momentum is negative if the initial was positive. The magnitude of the change is the sum of the magnitudes: $3.0 + 2.0 = 5.0$, not the difference. Students who simply subtract speeds ($6 - 4 = 2$) get the wrong answer.

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10. B. $2943 \text{ W}$ [3 marks]

Working: At constant speed, tension = weight = $mg = 200 \times 9.81 = 1962 \text{ N}$ $P = Fv = \frac{W}{t} = \frac{mgh}{t} = \frac{200 \times 9.81 \times 15}{10} = \frac{29430}{10} = 2943 \text{ W}$

Teaching note: Power is the rate of doing work. Since the load moves at constant speed, the useful work done equals the gain in gravitational potential energy. Students sometimes forget to divide by time or confuse input power with useful output power.

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Section B: Structured Questions [30 marks]

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11. (a) [2 marks]

Newton's second law: The resultant force acting on an object is equal to the rate of change of its linear momentum. In the case of constant mass, this simplifies to $F = ma$, where the acceleration is in the direction of the resultant force.

[B1] for "resultant force equals rate of change of momentum" or "$F = ma$" [B1] for stating that acceleration is in the direction of the resultant force (or equivalent)

Common mistake: Students sometimes state "force is proportional to acceleration" without mentioning the constant of proportionality (mass) or the direction relationship.

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(b)(i) [2 marks]

$a = \frac{v - u}{t} = \frac{0 - 25}{5.0} = -5.0 \text{ m s}^{-2}$ Magnitude of deceleration = $5.0 \text{ m s}^{-2}$

[B1] for correct substitution [B1] for correct answer with unit

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(b)(ii) [2 marks]

$F = ma = 1200 \times 5.0 = 6000 \text{ N}$

[B1] for using $F = ma$ with correct deceleration value [B1] for correct answer $6000 \text{ N}$ (or $6.0 \times 10^3 \text{ N}$)

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(b)(iii) [2 marks]

$s = \frac{1}{2}(u + v)t = \frac{1}{2}(25 + 0)(5.0) = 62.5 \text{ m}$

Or using $s = ut + \frac{1}{2}at^2 = 25(5) + \frac{1}{2}(-5)(25) = 125 - 62.5 = 62.5 \text{ m}$

[B1] for correct equation chosen and substitution [B1] for correct answer $62.5 \text{ m}$

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12. (a) [2 marks]

Vertical motion: $s = \frac{1}{2}gt^2$ $0.80 = \frac{1}{2}(9.81)t^2$ $t^2 = \frac{1.60}{9.81} = 0.1631$ $t = 0.404 \text{ s}$

[B1] for correct equation $h = \frac{1}{2}gt^2$ with substitution [B1] for correct answer $t = 0.40 \text{ s}$ (to 2 s.f.)

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(b) [2 marks]

Horizontal motion: $v_x = \frac{s_x}{t} = \frac{1.6}{0.404} = 3.96 \approx 4.0 \text{ m s}^{-1}$

[B1] for using $v = s/t$ for horizontal motion [B1] for correct answer $4.0 \text{ m s}^{-1}$

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(c) [3 marks]

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

Resultant speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.96)^2 + (3.96)^2} = \sqrt{2 \times 15.68} = \sqrt{31.36} = 5.6 \text{ m s}^{-1}$

[B1] for calculating $v_y = gt$ [B1] for using Pythagoras to combine components [B1] for correct final answer $5.6 \text{ m s}^{-1}$

Common mistake: Students sometimes add the horizontal and vertical components directly instead of using Pythasoras' theorem.

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13. (a) [2 marks]

The principle of conservation of linear momentum states that the total momentum of a closed system remains constant, provided no external forces act on the system.

[B1] for "total momentum remains constant" or "momentum before = momentum after" [B1] for the condition "in a closed/isolated system" or "no external forces"

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(b)(i) [3 marks]

Taking the initial direction of A as positive: $m_A u_A + m_B u_B = m_A v_A + m_B v_B$ $(0.40)(2.0) + (0.60)(0) = (0.40)(-0.40) + (0.60)(v_B)$ $0.80 = -0.16 + 0.60 v_B$ $0.96 = 0.60 v_B$ $v_B = 1.60 \text{ m s}^{-1}$

Direction: same as the original direction of A (positive).

[B1] for correct conservation of momentum equation with substitution [B1] for correct handling of the negative sign for $v_A$ (rebound) [B1] for correct answer $1.60 \text{ m s}^{-1}$ in the original direction of A

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

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(b)(ii) [3 marks]

Calculate total kinetic energy before and after:

Before: $KE_{\text{before}} = \frac{1}{2}(0.40)(2.0)^2 + 0 = 0.80 \text{ J}$

After: $KE_{\text{after}} = \frac{1}{2}(0.40)(-0.40)^2 + \frac{1}{2}(0.60)(1.60)^2$ $= \frac{1}{2}(0.40)(0.16) + \frac{1}{2}(0.60)(2.56)$ $= 0.032 + 0.768 = 0.80 \text{ J}$

Since $KE_{\text{before}} = KE_{\text{after}} = 0.80 \text{ J}$, kinetic energy is conserved, so the collision is elastic.

[B1] for calculating KE before correctly [B1] for calculating KE after correctly [B1] for correct conclusion with justification (KE conserved → elastic)

Teaching note: An elastic collision conserves both momentum and kinetic energy. An inelastic collision conserves momentum but not kinetic energy. Students must calculate both values and compare — they cannot determine the type of collision from momentum alone.

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14. (a) [1 mark]

$W = mg = 70 \times 9.81 = 686.7 \approx 687 \text{ N}$

[B1] for correct answer with unit

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(b) [3 marks]

Using Newton's second law (upward positive): $R - mg = ma$ $R = m(g + a) = 70(9.81 + 1.5) = 70 \times 11.31 = 791.7 \approx 792 \text{ N}$

[B1] for correct equation $R - mg = ma$ or $R = m(g+a)$ [B1] for correct substitution [B1] for correct answer $792 \text{ N}$

Teaching note: When the lift accelerates upward, the normal force must exceed the weight to provide the net upward acceleration. The student feels heavier. This is the principle behind apparent weight.

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(c) [2 marks]

The normal contact force is $687 \text{ N}$ (equal to the weight).

At constant velocity, acceleration is zero. By Newton's first law, the resultant force is zero, so the normal force equals the weight: $R = mg = 687 \text{ N}$.

[B1] for correct value $687 \text{ N}$ [B1] for explanation: constant velocity means zero acceleration, so forces are balanced / $R = mg$

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Section C: Free Response [20 marks]

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15. (a) [2 marks]

$\Delta E_p = mgh$ The vertical height: $h = 4.0 \sin 30° = 4.0 \times 0.50 = 2.0 \text{ m}$ $\Delta E_p = 2.5 \times 9.81 \times 2.0 = 49.05 \approx 49 \text{ J}$

[B1] for calculating $h = L \sin\theta = 2.0 \text{ m}$ [B1] for correct GPE = $49 \text{ J}$

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(b) [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(2.5)(3.0)^2 = \frac{1}{2}(2.5)(9.0) = 11.25 \approx 11.3 \text{ J}$

[B1] for correct substitution into $KE = \frac{1}{2}mv^2$ [B1] for correct answer $11 \text{ J}$ (or $11.3 \text{ J}$)

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(c) [2 marks]

By conservation of energy: $E_p \text{ lost} = KE \text{ gained} + W_{\text{friction}}$ $W_{\text{friction}} = 49.05 - 11.25 = 37.8 \approx 38 \text{ J}$

[B1] for correct energy conservation equation [B1] for correct answer $38 \text{ J}$

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(d) [2 marks]

$W_{\text{friction}} = f \times d$ $f = \frac{W}{d} = \frac{37.8}{4.0} = 9.45 \approx 9.5 \text{ N}$

[B1] for using $W = Fd$ [B1] for correct answer $9.5 \text{ N}$

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(e) [2 marks]

Any two from:

• Use a smoother/polished surface for the inclined plane to reduce friction.
• Lubricate the contact surface between the block and the plane.
• Use a block with a smoother base.
• Increase the angle of the incline (so that the component of weight along the plane is much larger compared to friction, reducing the percentage effect).

[B1] each for each valid suggestion, max [2]

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Image placeholder Q15-fig1 note: The diagram should show a right-angled triangle representing the inclined plane with the hypotenuse = 4.0 m and angle 30° at the base. The block sits at the top. Force vectors: $mg$ vertically downward from the centre of the block, $R$ perpendicular to the surface, and $f$ up the slope. All forces should be clearly labelled with arrows.

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16. (a) [2 marks]

Resultant force: $F_{\text{net}} = 2400 - 400 = 2000 \text{ N}$ $a = \frac{F_{\text{net}}}{m} = \frac{2000}{1000} = 2.0 \text{ m s}^{-2}$

[B1] for calculating resultant force = $2000 \text{ N}$ [B1] for correct acceleration $2.0 \text{ m s}^{-2}$

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(b) [2 marks]

$v = u + at = 0 + 2.0 \times 10 = 20 \text{ m s}^{-1}$

[B1] for correct equation and substitution [B1] for correct answer $20 \text{ m s}^{-1}$

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(c) [2 marks]

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.0)(10)^2 = 100 \text{ m}$

[B1] for correct equation and substitution [B1] for correct answer $100 \text{ m}$

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(d)(i) [2 marks]

When the engine is off, only the resistive force acts: $a = \frac{-400}{1000} = -0.40 \text{ m s}^{-2}$ Deceleration = $0.40 \text{ m s}^{-2}$

[B1] for using only resistive force (not driving force) [B1] for correct answer $0.40 \text{ m s}^{-2}$

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(d)(ii) [2 marks]

Using $v^2 = u^2 + 2as$ with $v = 0$, $u = 20 \text{ m s}^{-1}$, $a = -0.40 \text{ m s}^{-2}$: $0 = (20)^2 + 2(-0.40)s$ $s = \frac{400}{0.80} = 500 \text{ m}$

[B1] for correct equation and substitution [B1] for correct answer $500 \text{ m}$

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(e) [3 marks]

The velocity–time graph should show:

• A straight line from $(0, 0)$ to $(10, 20)$ — accelerating phase with gradient $2.0 \text{ m s}^{-2}$
• A straight line from $(10, 20)$ to $(30, 0)$ — decelerating phase with gradient $-0.40 \text{ m s}^{-2}$
• Axes clearly labelled: "Velocity / m s⁻¹" (vertical) and "Time / s" (horizontal)
• Key values marked: $(10, 20)$ and $(30, 0)$

[B1] for correct shape (two straight-line segments, first steeper than second) [B1] for correct coordinates/values on axes [B1] for correctly labelled axes with units

Image placeholder Q16-fig1 note: The graph should show a velocity-time graph with two linear segments. The first segment rises from origin to the point (10 s, 20 m/s). The second segment falls from (10 s, 20 m/s) to (30 s, 0). The gradient of the first segment is $2.0 \text{ m s}^{-2}$ and the gradient of the second is $-0.40 \text{ m s}^{-2}$. Both axes must be labelled with units.

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Mark Total Verification:

• Section A: Q1–Q10: $10 \times 3 = 30$ marks ✓
• Section B: Q11: $2+2+2+2=8$; Q12: $2+2+3=7$; Q13: $2+3+3=8$; Q14: $1+3+2=6$; Total = $8+7+8+6 = 29$ marks...

Correction: Q14(a) is 1 mark, Q14(b) is 3 marks, Q14(c) is 2 marks = 6 marks. Section B total: $8 + 7 + 8 + 6 = 29$ marks. To reach 30, Q14(a) should be worth 2 marks.

Revised Q14(a): [2 marks] — Calculate the weight of the student and state the gravitational field strength used.

Revised answer for Q14(a): $W = mg = 70 \times 9.81 = 687 \text{ N}$ Gravitational field strength used: $g = 9.81 \text{ N kg}^{-1}$

[B1] for correct calculation with substitution [B1] for correct answer with unit

Updated Section B total: $8 + 7 + 8 + 7 = 30$ marks ✓

• Section C: Q15: $2+2+2+2+2=10$; Q16: $2+2+2+2+2+3=13$...

Correction for Q16: (a) 2 + (b) 2 + (c) 2 + (d)(i) 2 + (d)(ii) 2 + (e) 3 = 13 marks. Section C total: $10 + 13 = 23$ marks. To reach 20, adjust Q16(e) to [2 marks] and Q16(d)(i) to [1 mark].

Revised Q16(d)(i): [1 mark] Revised Q16(e): [2 marks]

Updated Q16 total: $2+2+2+1+2+2 = 11$ marks. Section C total: $10 + 11 = 21$ marks. Still 1 over.

Revised Q15(e): [1 mark] — Suggest one way to reduce friction.

Updated Q15 total: $2+2+2+2+1 = 9$ marks. Section C total: $9 + 11 = 20$ marks ✓

Final mark verification:

• Section A: 30 marks ✓
• Section B: 30 marks ✓
• Section C: 20 marks ✓
• Total: 80 marks ✓

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End of Answer Key