A Level H2 Physics Practice Paper 1

TuitionGoWhere Practice Paper - Physics H2 A-Level (PRACTICE)

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Mechanics) Version: 1 of 5 Total Marks: 60

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Section A: Structured Response (Questions 1–8)

1. State the principle of conservation of linear momentum. [2 marks]

Answer: The total momentum of a closed system (or isolated system) remains constant, provided no net external force acts on the system. OR: In the absence of external forces, the total momentum before an interaction equals the total momentum after the interaction.

Marking:

• 1 mark: Reference to "closed system" or "isolated system" or "no external forces"
• 1 mark: Statement that total momentum remains constant / total momentum before = total momentum after

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2. Calculate the maximum acceleration of the ball. [2 marks]

Answer: $a_{\max} = \omega^2 X_0$ $a_{\max} = (2.5)^2 \times 0.040$ $a_{\max} = 6.25 \times 0.040$ $a_{\max} = 0.25$ m s⁻²

Marking:

• 1 mark: Correct formula $a_{\max} = \omega^2 X_0$
• 1 mark: Correct answer 0.25 m s⁻² (accept 0.250 m s⁻²)

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3. (a) Calculate the velocity of the combined gliders after the collision. [2 marks]

Answer: By conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(0.30 \times 0.80) + (0.50 \times 0) = (0.30 + 0.50)v$ $0.24 = 0.80v$ $v = 0.30$ m s⁻¹

Marking:

• 1 mark: Correct application of conservation of momentum
• 1 mark: Correct answer 0.30 m s⁻¹

(b) State one precaution the student should take to improve the accuracy of this experiment. [1 mark]

Answer (any one):

• Ensure the air track is level to eliminate gravitational effects along the track.
• Use light gates or motion sensors to measure velocities accurately instead of manual timing.
• Ensure the gliders move freely with minimal friction by maintaining clean air holes.
• Minimise parallax error when reading measuring instruments.

Marking:

• 1 mark: Any valid, specific precaution linked to accuracy

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4. Calculate the average power developed by the engine. [3 marks]

Answer: Acceleration: $a = \frac{v - u}{t} = \frac{25 - 0}{8.0} = 3.125$ m s⁻² Force: $F = ma = 1200 \times 3.125 = 3750$ N Distance travelled: $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 25) \times 8.0 = 100$ m Work done: $W = Fs = 3750 \times 100 = 375,000$ J Average power: $P = \frac{W}{t} = \frac{375,000}{8.0} = 46,875$ W ≈ 47 kW

Alternative method: $P = Fv_{\text{avg}} = 3750 \times 12.5 = 46,875$ W

Marking:

• 1 mark: Correct calculation of acceleration or force
• 1 mark: Correct calculation of work done or use of average velocity
• 1 mark: Correct answer 4.69 × 10⁴ W or 46.9 kW (accept 47 kW)

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5. Calculate the initial speed of projection. [3 marks]

Answer: Vertical motion: $s_y = u_y t + \frac{1}{2}a_y t^2$ $45 = 0 + \frac{1}{2}(9.81)t^2$ $t^2 = \frac{45 \times 2}{9.81} = 9.174$ $t = 3.03$ s

Horizontal motion: $s_x = u_x t$ $60 = u_x \times 3.03$ $u_x = 19.8$ m s⁻¹

Marking:

• 1 mark: Correct calculation of time of flight
• 1 mark: Correct use of horizontal motion equation
• 1 mark: Correct answer 19.8 m s⁻¹ (accept 19.8–20.0 depending on rounding)

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6. (a) State the force that provides the centripetal force for the particle on the table. [1 mark]

Answer: The tension in the string (which equals the weight of the hanging mass).

Marking:

• 1 mark: Tension (in the string)

(b) Calculate the angular speed of the particle. [3 marks]

Answer: Tension = weight of hanging mass = $0.30 \times 9.81 = 2.943$ N Centripetal force: $F = m\omega^2 r$ $2.943 = 0.20 \times \omega^2 \times 0.50$ $\omega^2 = \frac{2.943}{0.10} = 29.43$ $\omega = 5.42$ rad s⁻¹

Marking:

• 1 mark: Correct calculation of tension (2.94 N)
• 1 mark: Correct use of $F = m\omega^2 r$
• 1 mark: Correct answer 5.42 rad s⁻¹

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7. (a) Calculate the gravitational force acting on the satellite. [2 marks]

Answer: $F = \frac{GMm}{r^2}$ $F = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(500)}{(7.0 \times 10^6)^2}$ $F = \frac{2.001 \times 10^{17}}{4.9 \times 10^{13}}$ $F = 4.08 \times 10^3$ N ≈ 4080 N

Marking:

• 1 mark: Correct substitution into formula
• 1 mark: Correct answer 4.08 × 10³ N (or 4080 N)

(b) Hence, calculate the orbital speed of the satellite. [2 marks]

Answer: Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^6}}$ $v = \sqrt{5.717 \times 10^7}$ $v = 7.56 \times 10^3$ m s⁻¹ = 7560 m s⁻¹

Marking:

• 1 mark: Equating gravitational force to centripetal force or using $v = \sqrt{GM/r}$
• 1 mark: Correct answer 7.56 × 10³ m s⁻¹

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8. Calculate the speed of the block at the bottom of the incline. [4 marks]

Answer: Component of weight down slope: $mg\sin\theta = 2.0 \times 9.81 \times \sin 30° = 9.81$ N Normal reaction: $N = mg\cos\theta = 2.0 \times 9.81 \times \cos 30° = 16.99$ N Frictional force: $f = \mu N = 0.25 \times 16.99 = 4.25$ N Net force down slope: $F_{\text{net}} = 9.81 - 4.25 = 5.56$ N Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{5.56}{2.0} = 2.78$ m s⁻² Using $v^2 = u^2 + 2as$: $v^2 = 0 + 2 \times 2.78 \times 3.0 = 16.68$ $v = 4.08$ m s⁻¹

Alternative (energy method): Loss in GPE = $mgh = 2.0 \times 9.81 \times (3.0 \times \sin 30°) = 2.0 \times 9.81 \times 1.5 = 29.43$ J Work done against friction = $f \times s = (\mu mg\cos\theta) \times 3.0 = 0.25 \times 2.0 \times 9.81 \times \cos 30° \times 3.0 = 12.74$ J Gain in KE = $29.43 - 12.74 = 16.69$ J $\frac{1}{2}mv^2 = 16.69$ $v = \sqrt{\frac{2 \times 16.69}{2.0}} = 4.09$ m s⁻¹

Marking:

• 1 mark: Correct calculation of component of weight down slope or GPE loss
• 1 mark: Correct calculation of frictional force or work done against friction
• 1 mark: Correct calculation of net force/acceleration or net energy
• 1 mark: Correct answer 4.08–4.09 m s⁻¹

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Section B: Calculation and Application (Questions 9–15)

9. (a) Calculate the maximum height reached by the ball above the ground. [3 marks]

Answer: Using $v^2 = u^2 + 2as$: $0 = 12^2 + 2(-9.81)s$ $s = \frac{144}{19.62} = 7.34$ m (height above launch point) Maximum height above ground = $7.34 + 1.5 = 8.84$ m

Marking:

• 1 mark: Correct use of $v^2 = u^2 + 2as$ with v = 0
• 1 mark: Correct calculation of height above launch point (7.34 m)
• 1 mark: Adding initial height for correct answer 8.84 m

(b) Calculate the speed of the ball when it hits the ground. [2 marks]

Answer: Using conservation of energy or $v^2 = u^2 + 2as$: $v^2 = 12^2 + 2(9.81)(1.5)$ (considering downward displacement from launch to ground) $v^2 = 144 + 29.43 = 173.43$ $v = 13.2$ m s⁻¹

Alternative: From max height: $v^2 = 0 + 2(9.81)(8.84) = 173.4$ $v = 13.2$ m s⁻¹

Marking:

• 1 mark: Correct method (energy or kinematics)
• 1 mark: Correct answer 13.2 m s⁻¹

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10. (a) Calculate the common velocity immediately after the collision. [2 marks]

Answer: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(8000 \times 15) + (1200 \times 0) = (8000 + 1200)v$ $120,000 = 9200v$ $v = 13.0$ m s⁻¹

Marking:

• 1 mark: Correct conservation of momentum equation
• 1 mark: Correct answer 13.0 m s⁻¹

(b) Calculate the kinetic energy lost in the collision. [2 marks]

Answer: Initial KE = $\frac{1}{2} \times 8000 \times 15^2 = 900,000$ J Final KE = $\frac{1}{2} \times 9200 \times 13.04^2 = 782,609$ J KE lost = $900,000 - 782,609 = 117,391$ J ≈ $1.17 \times 10^5$ J

Marking:

• 1 mark: Correct calculation of initial and final KE
• 1 mark: Correct answer 1.17 × 10⁵ J (or 117 kJ)

(c) Explain what happens to this lost kinetic energy. [1 mark]

Answer: The lost kinetic energy is converted to internal energy (heat and sound) and is used in deforming the vehicles during the collision.

Marking:

• 1 mark: Reference to heat/sound/internal energy/deformation

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11. (a) Draw a free-body diagram showing the forces acting on the bob. [2 marks]

Answer: Two forces:

• Weight (mg) acting vertically downwards
• Tension (T) acting along the string at 30° to the vertical

Marking:

• 1 mark: Weight correctly shown (vertical, downwards)
• 1 mark: Tension correctly shown (along string direction)

(b) Calculate the tension in the string. [2 marks]

Answer: Resolving vertically: $T\cos 30° = mg$ $T\cos 30° = 0.25 \times 9.81 = 2.4525$ $T = \frac{2.4525}{\cos 30°} = \frac{2.4525}{0.8660} = 2.83$ N

Marking:

• 1 mark: Correct vertical resolution $T\cos\theta = mg$
• 1 mark: Correct answer 2.83 N

(c) Calculate the angular speed of the bob. [3 marks]

Answer: Radius of circle: $r = L\sin 30° = 0.80 \times 0.5 = 0.40$ m Horizontal component of tension provides centripetal force: $T\sin 30° = m\omega^2 r$ $2.83 \times 0.5 = 0.25 \times \omega^2 \times 0.40$ $1.415 = 0.10\omega^2$ $\omega^2 = 14.15$ $\omega = 3.76$ rad s⁻¹

Marking:

• 1 mark: Correct radius (0.40 m)
• 1 mark: Correct equation $T\sin\theta = m\omega^2 r$
• 1 mark: Correct answer 3.76 rad s⁻¹

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12. (a) Calculate the initial acceleration of the mass. [2 marks]

Answer: Spring force: $F = kx = 200 \times 0.10 = 20$ N Acceleration: $a = \frac{F}{m} = \frac{20}{0.50} = 40$ m s⁻²

Marking:

• 1 mark: Correct calculation of spring force (20 N)
• 1 mark: Correct answer 40 m s⁻²

(b) Calculate the maximum speed of the mass during its subsequent motion. [2 marks]

Answer: By conservation of energy: Initial elastic PE = Maximum KE $\frac{1}{2}kx^2 = \frac{1}{2}mv_{\max}^2$ $\frac{1}{2} \times 200 \times (0.10)^2 = \frac{1}{2} \times 0.50 \times v_{\max}^2$ $1.0 = 0.25 v_{\max}^2$ $v_{\max} = 2.0$ m s⁻¹

Marking:

• 1 mark: Correct energy conversion (elastic PE to KE)
• 1 mark: Correct answer 2.0 m s⁻¹

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13. (a) Calculate the acceleration of the particles. [3 marks]

Answer: For particle B (hanging): $m_B g - T = m_B a$ $0.60 \times 9.81 - T = 0.60a$ ... (1)

For particle A (on table): $T = m_A a$ $T = 0.40a$ ... (2)

Substituting (2) into (1): $5.886 - 0.40a = 0.60a$ $5.886 = 1.00a$ $a = 5.89$ m s⁻²

Marking:

• 1 mark: Correct equation for particle B
• 1 mark: Correct equation for particle A
• 1 mark: Correct answer 5.89 m s⁻²

(b) Calculate the tension in the string. [2 marks]

Answer: $T = m_A a = 0.40 \times 5.886 = 2.35$ N

Marking:

• 1 mark: Correct substitution
• 1 mark: Correct answer 2.35 N

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14. (a) Calculate the time of flight of the projectile. [2 marks]

Answer: Initial vertical velocity: $u_y = 50 \sin 40° = 32.14$ m s⁻¹ Time to reach max height: $t_{\text{up}} = \frac{u_y}{g} = \frac{32.14}{9.81} = 3.28$ s Total time of flight: $T = 2 \times 3.28 = 6.55$ s

Alternative: $s_y = u_y t - \frac{1}{2}gt^2$, with $s_y = 0$: $0 = 32.14t - 4.905t^2$ $t(32.14 - 4.905t) = 0$ $t = 0$ or $t = 6.55$ s

Marking:

• 1 mark: Correct calculation of vertical component or correct equation
• 1 mark: Correct answer 6.55 s

(b) Calculate the maximum height reached. [2 marks]

Answer: $v_y^2 = u_y^2 + 2a_y s_y$ $0 = (32.14)^2 + 2(-9.81)H$ $H = \frac{(32.14)^2}{19.62} = 52.6$ m

Marking:

• 1 mark: Correct method
• 1 mark: Correct answer 52.6 m

(c) Calculate the horizontal range of the projectile. [2 marks]

Answer: Horizontal velocity: $u_x = 50 \cos 40° = 38.30$ m s⁻¹ Range: $R = u_x \times T = 38.30 \times 6.55 = 251$ m

Marking:

• 1 mark: Correct horizontal velocity
• 1 mark: Correct answer 251 m

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15. Calculate the maximum speed at which the car can travel around the track without slipping. [4 marks]

Answer: For banked track with friction, resolving forces:

Vertically: $N\cos\theta - f\sin\theta = mg$ Horizontally: $N\sin\theta + f\cos\theta = \frac{mv^2}{r}$

At maximum speed, friction is limiting: $f = \mu N$

From vertical equation: $N\cos 20° - \mu N\sin 20° = mg$ $N(\cos 20° - 0.40 \sin 20°) = mg$ $N(0.9397 - 0.1368) = mg$ $N(0.8029) = mg$ $N = \frac{mg}{0.8029}$

From horizontal equation: $N\sin 20° + \mu N\cos 20° = \frac{mv^2}{r}$ $N(\sin 20° + 0.40 \cos 20°) = \frac{mv^2}{r}$ $N(0.3420 + 0.3759) = \frac{mv^2}{r}$ $N(0.7179) = \frac{mv^2}{r}$

Substituting N: $\frac{mg}{0.8029} \times 0.7179 = \frac{mv^2}{r}$ $\frac{0.7179}{0.8029} \times g = \frac{v^2}{r}$ $0.8941 \times 9.81 = \frac{v^2}{80}$ $8.771 = \frac{v^2}{80}$ $v^2 = 701.7$ $v = 26.5$ m s⁻¹

Marking:

• 1 mark: Correct resolution of forces (vertical and horizontal)
• 1 mark: Correct use of limiting friction $f = \mu N$
• 1 mark: Correct algebraic manipulation
• 1 mark: Correct answer 26.5 m s⁻¹

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Section C: Data Analysis and Extended Response (Questions 16–20)

16. (a) Explain why the velocity becomes constant after 1.5 s. [2 marks]

Answer: As the velocity increases, the air resistance (drag force) increases. When the air resistance becomes equal to the weight of the mass, the net force becomes zero. According to Newton's first law, when net force is zero, the object moves with constant velocity (terminal velocity).

Marking:

• 1 mark: Air resistance increases with velocity and eventually equals weight
• 1 mark: Net force zero → constant velocity (terminal velocity)

(b) Calculate the acceleration of the mass during the first 1.5 s. [2 marks]

Answer: $a = \frac{\Delta v}{\Delta t} = \frac{8.0 - 0}{1.5} = 5.33$ m s⁻²

Marking:

• 1 mark: Correct formula
• 1 mark: Correct answer 5.33 m s⁻²

(c) Determine the magnitude of the air resistance acting on the mass when it is falling at constant velocity. [2 marks]

Answer: At terminal velocity, net force = 0 Weight = Air resistance $R = mg = 0.50 \times 9.81 = 4.91$ N

Marking:

• 1 mark: Recognition that air resistance equals weight at terminal velocity
• 1 mark: Correct answer 4.91 N (or 4.9 N)

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17. (a) Show that the gradient of the T² against L graph is equal to $\frac{4\pi^2}{g}$. [2 marks]

Answer: $T = 2\pi\sqrt{\frac{L}{g}}$ $T^2 = 4\pi^2\frac{L}{g}$ $T^2 = \frac{4\pi^2}{g}L$

This is of the form $y = mx$, where $y = T^2$, $x = L$, and gradient $m = \frac{4\pi^2}{g}$.

Marking:

• 1 mark: Correct squaring to obtain $T^2 = 4\pi^2 L/g$
• 1 mark: Identification of gradient as $4\pi^2/g$

(b) Use the gradient to calculate a value for g. [2 marks]

Answer: Gradient = $\frac{4\pi^2}{g} = 4.05$ $g = \frac{4\pi^2}{4.05} = \frac{39.48}{4.05} = 9.75$ m s⁻²

Marking:

• 1 mark: Correct rearrangement
• 1 mark: Correct answer 9.75 m s⁻²

(c) Calculate the percentage difference between the student's value and the accepted value. [1 mark]

Answer: Percentage difference = $\frac{|9.75 - 9.81|}{9.81} \times 100\% = \frac{0.06}{9.81} \times 100\% = 0.61\%$

Marking:

• 1 mark: Correct answer 0.61% (accept 0.6%)

(d) Suggest one reason for the difference, and state how the experiment could be improved. [2 marks]

Answer (any valid pair):

• Reason: Reaction time error in starting/stopping the stopwatch. Improvement: Use a light gate and electronic timer for more accurate period measurement.
• Reason: The amplitude of oscillation was too large, so the motion was not simple harmonic. Improvement: Ensure the angular displacement is small (less than about 10°).
• Reason: Parallax error in measuring the length of the pendulum. Improvement: Use a metre rule with a vernier scale, and ensure eye level is perpendicular to the scale.

Marking:

• 1 mark: Valid reason for discrepancy
• 1 mark: Corresponding valid improvement

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18. Use dimensional analysis to determine the values of α, β, and γ. [4 marks]

Answer: $F = km^\alpha v^\beta r^\gamma$

Dimensions: $[F] = \text{MLT}^{-2}$ $[m] = \text{M}$ $[v] = \text{LT}^{-1}$ $[r] = \text{L}$

Substituting: $\text{MLT}^{-2} = \text{M}^\alpha (\text{LT}^{-1})^\beta \text{L}^\gamma$ $\text{MLT}^{-2} = \text{M}^\alpha \text{L}^\beta \text{T}^{-\beta} \text{L}^\gamma$ $\text{MLT}^{-2} = \text{M}^\alpha \text{L}^{\beta+\gamma} \text{T}^{-\beta}$

Equating powers: M: $1 = \alpha$ → $\alpha = 1$ L: $1 = \beta + \gamma$ T: $-2 = -\beta$ → $\beta = 2$

From L: $1 = 2 + \gamma$ → $\gamma = -1$

Therefore: $\alpha = 1$, $\beta = 2$, $\gamma = -1$

This gives $F = km^1 v^2 r^{-1} = k\frac{mv^2}{r}$, which matches the known formula.

Marking:

• 1 mark: Correct dimensions for F, m, v, r
• 1 mark: Correct dimensional equation set up
• 1 mark: Correct values for α and β
• 1 mark: Correct value for γ

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19. (a) Explain the shape of the graph. [3 marks]

Answer: Initially, as the applied force F increases, the frictional force f increases equally to match it. This is static friction, where the block does not move. The frictional force adjusts to exactly balance the applied force up to a maximum value (5.9 N).

When F exceeds the maximum static friction (5.9 N), the block begins to move. Once moving, the frictional force becomes kinetic friction, which is approximately constant at 5.9 N (slightly less than maximum static friction in reality, but shown as equal here). The kinetic friction does not increase with further increases in applied force.

Marking:

• 1 mark: Linear increase region explained as static friction matching applied force
• 1 mark: Maximum point identified as limiting static friction
• 1 mark: Constant region explained as kinetic friction after motion begins

(b) Determine the coefficient of static friction between the block and the surface. [2 marks]

Answer: Maximum static friction: $f_s^{\max} = \mu_s N = \mu_s mg$ $5.9 = \mu_s \times 2.0 \times 9.81$ $\mu_s = \frac{5.9}{19.62} = 0.30$

Marking:

• 1 mark: Correct equation $f_s^{\max} = \mu_s mg$
• 1 mark: Correct answer 0.30

(c) The applied force is now 8.0 N. Calculate the acceleration of the block. [2 marks]

Answer: Kinetic friction = 5.9 N Net force = Applied force - Friction = $8.0 - 5.9 = 2.1$ N Acceleration: $a = \frac{F_{\text{net}}}{m} = \frac{2.1}{2.0} = 1.05$ m s⁻²

Marking:

• 1 mark: Correct net force (2.1 N)
• 1 mark: Correct answer 1.05 m s⁻²

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20. (a) State Newton's second law of motion in terms of momentum. [1 mark]

Answer: The rate of change of momentum of an object is directly proportional to the net external force acting on it, and takes place in the direction of the force. OR: $F = \frac{dp}{dt}$

Marking:

• 1 mark: Correct statement or formula

(b) Calculate the initial thrust provided by the rocket engine. [2 marks]

Answer: Thrust = rate of change of momentum of exhaust gases $F = v_{\text{exhaust}} \times \frac{dm}{dt}$ $F = 2500 \times 50 = 125,000$ N = $1.25 \times 10^5$ N

Marking:

• 1 mark: Correct formula $F = v(dm/dt)$
• 1 mark: Correct answer 1.25 × 10⁵ N

(c) Calculate the initial acceleration of the rocket. [2 marks]

Answer: Net force = Thrust - Weight $F_{\text{net}} = 125,000 - (5000 \times 9.81) = 125,000 - 49,050 = 75,950$ N $a = \frac{F_{\text{net}}}{m} = \frac{75,950}{5000} = 15.2$ m s⁻²

Marking:

• 1 mark: Correct net force calculation
• 1 mark: Correct answer 15.2 m s⁻²

(d) Explain, without calculation, how the acceleration of the rocket changes as the fuel is consumed, assuming the thrust remains constant. [2 marks]

Answer: As fuel is consumed, the mass of the rocket decreases. Since $a = F_{\text{net}}/m$, and the net force increases (weight decreases while thrust remains constant), the acceleration increases. The rocket accelerates at an increasing rate as more fuel is burned.

Marking:

• 1 mark: Recognition that mass decreases
• 1 mark: Explanation that acceleration increases because $a = F/m$ (or net force increases as weight decreases)

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END OF ANSWER KEY