A Level H1 Physics Practice Paper 1

TuitionGoWhere Practice Paper — Physics H1 A-Level

Answer Key — Practice Paper: Mechanics

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Section A: Multiple Choice

1. B. $3.0 \text{ m s}^{-2}$

Teaching note: Using $a = \frac{v - u}{t} = \frac{24 - 0}{8.0} = 3.0 \text{ m s}^{-2}$. Uniform acceleration means the velocity increases at a constant rate. The equation of motion directly relates initial velocity, final velocity, time, and acceleration.

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2. D. Momentum

Teaching note: 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. This is a fundamental distinction in mechanics.

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3. C. $45 \text{ m}$

Teaching note: For horizontal projectile motion, horizontal velocity is constant (no horizontal acceleration). Using $s_x = v_x \times t = 15 \times 3.0 = 45 \text{ m}$. The vertical motion is independent of the horizontal motion — the time to fall is determined entirely by the vertical drop.

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4. B. $1.4 \text{ m s}^{-2}$

Teaching note: Net force $F_{\text{net}} = 12 - 5.0 = 7.0 \text{ N}$ (taking right as positive). Using Newton's second law: $a = \frac{F_{\text{net}}}{m} = \frac{7.0}{4.0} = 1.75 \text{ m s}^{-2}$. Wait — recalculating: $a = \frac{7.0}{5.0} = 1.4 \text{ m s}^{-2}$. The mass is $5.0 \text{ kg}$, not $4.0 \text{ kg}$. Common trap: using the wrong mass.

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5. D. $100 \text{ kg m s}^{-1}$

Teaching note: Using the impulse-momentum theorem: $\Delta p = F \times t = 20 \times 5.0 = 100 \text{ kg m s}^{-1}$. Impulse equals change in momentum. This is a direct application of Newton's second law in the form $F = \frac{\Delta p}{\Delta t}$.

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6. C. $35 \text{ m s}^{-1}$

Teaching note: The horizontal component is $v_x = v \cos\theta = 40 \cos 30° = 40 \times \frac{\sqrt{3}}{2} = 40 \times 0.866 = 34.6 \text{ m s}^{-1} \approx 35 \text{ m s}^{-1}$. Students must resolve the velocity into components using trigonometry. Common error: using $\sin$ instead of $\cos$ for the horizontal component.

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7. B. $2.0 \text{ m s}^{-1}$

Teaching note: Using conservation of momentum for a perfectly inelastic collision (they stick together): $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$. $(2.0)(6.0) + (4.0)(0) = (2.0 + 4.0)v$. $12 = 6.0v$. $v = 2.0 \text{ m s}^{-1}$. Kinetic energy is not conserved in this type of collision — some is converted to heat and sound.

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8. C. No work is done when the force is perpendicular to the displacement.

Teaching note: Work done is defined as $W = Fd\cos\theta$. When the force is perpendicular to displacement, $\theta = 90°$ and $\cos 90° = 0$, so $W = 0$. Work is a scalar quantity (not vector), and it requires both force AND displacement in the direction of the force.

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9. C. $4900 \text{ W}$

Teaching note: At constant speed, the tension equals the weight: $T = mg = 200 \times 9.81 = 1962 \text{ N}$. Work done $= F \times d = 1962 \times 10 = 19620 \text{ J}$. Power $= \frac{W}{t} = \frac{19620}{4.0} = 4905 \text{ W} \approx 4900 \text{ W}$. Alternatively, $P = Fv = 1962 \times \frac{10}{4.0} = 1962 \times 2.5 = 4905 \text{ W}$.

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10. A. $-4.0 \text{ m s}^{-1}$

Teaching note: For elastic collisions, both momentum and kinetic energy are conserved. Using conservation of momentum: $0.50 \times 8.0 + 0 = 0.50 v_1 + 1.5 v_2$, giving $4.0 = 0.50 v_1 + 1.5 v_2$. For elastic collisions, relative speed of approach = relative speed of separation: $8.0 - 0 = v_2 - v_1$, so $v_2 = 8.0 + v_1$. Substituting: $4.0 = 0.50 v_1 + 1.5(8.0 + v_1) = 0.50 v_1 + 12 + 1.5 v_1 = 2.0 v_1 + 12$. So $2.0 v_1 = -8.0$, giving $v_1 = -4.0 \text{ m s}^{-1}$. The negative sign means the lighter ball rebounds in the opposite direction.

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11. C. The resultant of the three forces is zero.

Teaching note: For equilibrium, the vector sum (resultant) of all forces must be zero. This is the defining condition for equilibrium. The forces need not be equal, parallel, or act at the same point — they only need to sum to zero as vectors.

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12. B. $20.4 \text{ m}$

Teaching note: At maximum height, the final velocity is zero. Using $v^2 = u^2 - 2gh$: $0 = (20)^2 - 2(9.81)h$. $h = \frac{400}{19.62} = 20.4 \text{ m}$. Alternatively, using conservation of energy: $\frac{1}{2}mv^2 = mgh$, so $h = \frac{v^2}{2g} = \frac{400}{19.62} = 20.4 \text{ m}$.

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13. B. $4.9 \text{ m s}^{-2}$

Teaching note: On a frictionless slope, the component of weight along the slope is $mg\sin\theta$. Using Newton's second law along the slope: $ma = mg\sin\theta$, so $a = g\sin\theta = 9.81 \times \sin 30° = 9.81 \times 0.5 = 4.9 \text{ m s}^{-2}$. The acceleration is independent of mass on a frictionless incline.

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14. A. $3000 \text{ N}$

Teaching note: First find acceleration: $a = \frac{v - u}{t} = \frac{0 - 25}{10} = -2.5 \text{ m s}^{-2}$. Then $F = ma = 1200 \times 2.5 = 3000 \text{ N}$. The braking force opposes the direction of motion. Common trap: forgetting that deceleration is negative, but the magnitude of force is asked for.

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15. B. $196 \text{ J}$

Teaching note: By conservation of energy, the gravitational potential energy lost equals the kinetic energy gained: $KE = mgh = 1.0 \times 9.81 \times 20 = 196.2 \text{ J} \approx 196 \text{ J}$. This assumes no air resistance. Alternatively, find the speed using $v^2 = 2gh = 2 \times 9.81 \times 20 = 392.4$, then $KE = \frac{1}{2}mv^2 = \frac{1}{2}(1.0)(392.4) = 196.2 \text{ J}$.

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Section B: Structured Questions

16. (a) Newton's first law of motion states that an object remains at rest or continues to move at a constant velocity unless acted upon by a resultant external force. [2]

[B1] for "remains at rest or moves at constant velocity" (or equivalent) [B1] for "unless acted upon by a resultant external force" (or "no net/resultant force")

Teaching note: This is the law of inertia. The key idea is that a force is not needed to keep something moving — it is needed to change its motion. Students often incorrectly state that a force is needed to maintain constant velocity.

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(b)(i) Free-body diagram showing:

Image pending generation: diagram for Q16(b)(i).

[B1] for two forces correctly drawn and labelled (weight and normal reaction) [B1] for correct directions and the normal reaction arrow being longer than the weight arrow

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(b)(ii) Taking upward as positive:

$R - mg = ma$

$R = m(g + a)$

$R = 60(9.81 + 1.5)$

$R = 60 \times 11.31$

$R = 678.6 \text{ N} \approx 679 \text{ N}$

[B1] for correct equation: $R - mg = ma$ (or equivalent Newton's second law setup) [B1] for correct substitution [B1] for correct answer with unit: $679 \text{ N}$ (or $680 \text{ N}$ to 2 s.f.)

Teaching note: Since the lift accelerates upward, the normal reaction must be greater than the weight to provide a net upward force. This is why you feel heavier in an accelerating lift. The net upward force $R - mg$ equals $ma$ by Newton's second law.

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17. (a) Using conservation of energy:

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.81 \times 1.8}$

$v = \sqrt{35.316}$

$v = 5.94 \text{ m s}^{-1}$

[B1] for using conservation of energy / $v = \sqrt{2gh}$ [B1] for correct substitution [B1] for correct answer: $5.94 \text{ m s}^{-1}$ (or $5.9 \text{ m s}^{-1}$ to 2 s.f.)

Teaching note: On a smooth track, mechanical energy is conserved. All gravitational potential energy converts to kinetic energy. The mass cancels out, so the speed at the bottom depends only on the height.

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(b) Using conservation of momentum (taking the initial direction of the $0.20 \text{ kg}$ ball as positive):

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$(0.20)(5.94) + (0.30)(0) = (0.20)v_1 + (0.30)(3.2)$

$1.188 = 0.20 v_1 + 0.96$

$0.20 v_1 = 1.188 - 0.96 = 0.228$

$v_1 = \frac{0.228}{0.20} = 1.14 \text{ m s}^{-1}$

The velocity of the $0.20 \text{ kg}$ ball after the collision is $1.14 \text{ m s}^{-1}$ in the same direction as its initial motion (positive direction).

[B1] for correct conservation of momentum equation [B1] for correct substitution of all values [B1] for correct calculation of $v_1$ [B1] for stating the direction (same direction as initial motion / positive direction)

Teaching note: The $0.20 \text{ kg}$ ball continues moving forward after the collision but at a reduced speed, having transferred some momentum to the $0.30 \text{ kg}$ ball. The positive result confirms it continues in the original direction.

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(c) Calculate total kinetic energy before and after:

Before collision: $KE_{\text{before}} = \frac{1}{2}(0.20)(5.94)^2 = \frac{1}{2}(0.20)(35.28) = 3.53 \text{ J}$

After collision: $KE_{\text{after}} = \frac{1}{2}(0.20)(1.14)^2 + \frac{1}{2}(0.30)(3.2)^2$ $= \frac{1}{2}(0.20)(1.30) + \frac{1}{2}(0.30)(10.24)$ $= 0.130 + 1.536 = 1.67 \text{ J}$

Since $KE_{\text{after}} < KE_{\text{before}}$ ($1.67 \text{ J} < 3.53 \text{ J}$), kinetic energy is not conserved, so the collision is inelastic.

[B1] for calculating $KE_{\text{before}}$ correctly [B1] for calculating $KE_{\text{after}}$ correctly [B1] for correct conclusion with reasoning (KE not conserved → inelastic)

Teaching note: In any collision, momentum is always conserved (no external forces). However, kinetic energy is only conserved in elastic collisions. Here, some kinetic energy has been converted to other forms (heat, sound, deformation energy), making it inelastic.

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18. (a) Using the vertical motion (taking downward as positive):

$h = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{9.81}} = \sqrt{\frac{90}{9.81}} = \sqrt{9.174} = 3.03 \text{ s}$

[B1] for using $h = \frac{1}{2}gt^2$ (vertical displacement with zero initial vertical velocity) [B1] for correct substitution [B1] for correct answer: $3.03 \text{ s}$ (or $3.0 \text{ s}$ to 2 s.f.)

Teaching note: The time of flight depends only on the vertical height and $g$. The horizontal velocity does not affect how long the ball takes to fall — this is the principle of independence of perpendicular motions.

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(b) Horizontal distance:

$s_x = v_x \times t = 12 \times 3.03 = 36.4 \text{ m}$

[B1] for using $s_x = v_x \times t$ [B1] for correct answer: $36 \text{ m}$ (to 2 s.f.) or $36.4 \text{ m}$

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(c) Vertical component of velocity just before impact:

$v_y = gt = 9.81 \times 3.03 = 29.7 \text{ m s}^{-1}$

Resultant speed:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 29.7^2} = \sqrt{144 + 882.1} = \sqrt{1026.1} = 32.0 \text{ m s}^{-1}$

[B1] for finding $v_y = gt$ correctly [B1] for using Pythagoras' theorem to find resultant speed [B1] for correct final answer: $32 \text{ m s}^{-1}$ (to 2 s.f.)

Teaching note: The final speed is found by combining the horizontal and vertical components using Pythagoras' theorem. The horizontal component remains constant ($12 \text{ m s}^{-1}$) while the vertical component increases due to gravity.

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(d) The time of flight would remain the same. The time of flight depends only on the vertical motion (the height of the cliff and $g$), which is independent of the horizontal velocity. Doubling the horizontal speed would increase the horizontal range but not the time taken to fall.

[B1] for stating "remains the same" / "no change" [B1] for correct justification linking time of flight to vertical motion only

Teaching note: This tests understanding of the independence of perpendicular motion components. Many students incorrectly think that a faster horizontal throw takes longer or shorter to fall.

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Section C: Free Response

19. (a) Net force on the car:

$F_{\text{net}} = 3600 - 600 = 3000 \text{ N}$

Using Newton's second law:

$a = \frac{F_{\text{net}}}{m} = \frac{3000}{1500} = 2.0 \text{ m s}^{-2}$

[B1] for finding net force correctly [B1] for using $a = F/m$ [B1] for correct answer with unit: $2.0 \text{ m s}^{-2}$

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(b) Using $v = u + at$:

$v = 0 + 2.0 \times 12 = 24 \text{ m s}^{-1}$

[B1] for using correct equation of motion [B1] for correct answer: $24 \text{ m s}^{-1}$

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(c) Kinetic energy:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1500)(24)^2 = \frac{1}{2}(1500)(576) = 432000 \text{ J} = 432 \text{ kJ}$

[B1] for using $KE = \frac{1}{2}mv^2$ [B1] for correct answer: $432000 \text{ J}$ or $432 \text{ kJ}$

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(d)(i) Component of weight down the slope:

$F_g = mg\sin\theta = 1500 \times 9.81 \times \sin 5.0° = 1500 \times 9.81 \times 0.0872 = 1283 \text{ N} \approx 1280 \text{ N}$

[B1] for using $mg\sin\theta$ [B1] for correct answer: $1280 \text{ N}$ (to 3 s.f.)

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(d)(ii) Net force up the slope:

$F_{\text{net}} = 3600 - 600 - 1280 = 1720 \text{ N}$

Acceleration:

$a = \frac{F_{\text{net}}}{m} = \frac{1720}{1500} = 1.15 \text{ m s}^{-2}$

[B1] for finding net force (driving force minus resistive force minus weight component) [B1] for correct answer: $1.15 \text{ m s}^{-2}$ (or $1.1 \text{ m s}^{-2}$ to 2 s.f.)

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(e) As the car travels up the slope, it gains gravitational potential energy. The engine does work to provide this increase in potential energy. Since the total work done by the engine is shared between increasing potential energy and overcoming resistive forces, less energy is available for kinetic energy. Therefore, the kinetic energy decreases and the car's speed decreases.

Alternatively: The component of weight acting down the slope opposes the motion, reducing the net force available for acceleration. With a smaller net force, the acceleration is reduced, and the car's speed increases more slowly (or decreases if the net force becomes negative).

[B1] for mentioning gain in gravitational potential energy [B1] for explaining that work done by engine is shared between KE and PE [B1] for concluding that KE/speed decreases

Teaching note: This is an energy-based explanation. The key concept is that energy is being converted from kinetic form to potential form (and also lost to resistive forces), so the car slows down.

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20. (a) Procedure:

1. Measure the vertical height $h$ of the bench above the floor using a metre rule.
2. Release the ball bearing horizontally from the edge of the bench.
3. Measure the horizontal distance $x$ from the base of the bench to where the ball lands on the floor.
4. Calculate the time of flight using $t = \sqrt{\frac{2h}{g}}$.
5. Calculate the horizontal speed using $v = \frac{x}{t} = \frac{x}{\sqrt{2h/g}}$.
6. Repeat the experiment several times and calculate the average horizontal distance to reduce random errors.

[B1] for measuring height $h$ [B1] for measuring horizontal distance $x$ [B1] for using $t = \sqrt{2h/g}$ to find time of flight [B1] for using $v = x/t$ to calculate horizontal speed

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(b) Time of flight:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.85}{9.81}} = \sqrt{0.1733} = 0.416 \text{ s}$

Horizontal speed:

$v = \frac{x}{t} = \frac{1.40}{0.416} = 3.37 \text{ m s}^{-1}$

[B1] for correct calculation of time of flight [B1] for correct calculation of horizontal speed [B1] for correct answer: $3.37 \text{ m s}^{-1}$ (or $3.4 \text{ m s}^{-1}$ to 2 s.f.) [B1] for correct unit

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(c) Air resistance would act opposite to the direction of motion, reducing the horizontal velocity of the ball bearing during its flight. This would cause the ball to travel a shorter horizontal distance than the theoretical value (which assumes no air resistance). The measured horizontal distance would be less than the calculated value.

[B1] for stating that air resistance opposes motion / reduces horizontal velocity [B1] for stating that the horizontal distance would be shorter / less than theoretical [B1] for correct explanation linking reduced velocity to reduced range

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(d) One improvement:

• Repeat the experiment multiple times and take the average of the horizontal distance measurements. This reduces the effect of random errors by allowing outliers to be identified and the mean to be more reliable.

Other acceptable answers:

• Use a light gate at the edge of the bench to measure the horizontal speed directly.
• Use a plumb line to ensure the horizontal distance is measured accurately from the base of the bench.
• Use a smoother ball bearing to reduce the effect of air resistance.

[B2] for a valid improvement with brief explanation

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(e) The graph of $x$ against $\sqrt{h}$ should be a straight line through the origin.

Reasoning: Since $x = v\sqrt{\frac{2h}{g}} = v\sqrt{\frac{2}{g}} \times \sqrt{h}$, $x$ is directly proportional to $\sqrt{h}$. The gradient of the line is $v\sqrt{2/g}$.

Image pending generation: graph for Q20(e).

[B1] for straight line through origin [B1] for correct reasoning ($x \propto \sqrt{h}$)

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Mark Summary

Section	Questions	Marks
A	Q1–15	15
B	Q16	7
	Q17	10
	Q18	13
C	Q19	15
	Q20	15
Total		60