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A Level H2 Physics Practice Paper 5

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TuitionGoWhere Practice Paper — Physics H2 A-Level

TuitionGoWhere Secondary School (AI)


Subject:	Physics (H2)
Level:	A-Level
Paper:	Practice Paper — Mechanics (Version 5 of 5)
Duration:	1 hour 30 minutes
Total Marks:	60
Name:	________________________
Class:	________________________
Date:	________________________

Instructions to Candidates:

Write your name, class, and date in the spaces provided above.
Answer all questions in the spaces provided.
Write in dark blue or black pen.
You may use a pencil for any diagrams, graphs, or working.
Non-programmable calculators are permitted.
The total mark for this paper is 60.
The number of marks for each question or part question is shown in brackets [ ].
Unless otherwise stated, the value of the acceleration due to gravity is $g = 9.81 \text{ m s}^{-2}$ .

Section A: Short Answer Questions (20 marks)

Answer all questions 1–10.

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

.......................................................................................................................

2. A car accelerates uniformly from rest to a speed of $28 \text{ m s}^{-1}$ in $7.0 \text{ s}$ . Calculate the acceleration of the car. [2]

.......................................................................................................................

3. Define work done by a force. State the SI unit for work. [2]

.......................................................................................................................

4. A ball is thrown vertically upwards with an initial speed of $15 \text{ m s}^{-1}$ . Calculate the maximum height reached by the ball. [3]

.......................................................................................................................

5. State Newton's first law of motion. [2]

.......................................................................................................................

6. A $2.0 \text{ kg}$ object moves in a horizontal circle of radius $3.0 \text{ m}$ at a constant speed of $6.0 \text{ m s}^{-1}$ . Calculate the magnitude of the centripetal acceleration. [2]

.......................................................................................................................

7. Distinguish between elastic and inelastic collisions. Give one example of each. [3]

.......................................................................................................................

8. State the SI base units of force (the newton) in terms of kilograms, metres, and seconds. [1]

.......................................................................................................................

9. A projectile is launched horizontally from a cliff $45 \text{ m}$ above level ground with a horizontal speed of $12 \text{ m s}^{-1}$ . Calculate the time taken for the projectile to reach the ground. [3]

.......................................................................................................................

10. Define power and state its SI unit. [2]

.......................................................................................................................

Section B: Structured Questions (25 marks)

Answer all questions 11–15.

11. A $5.0 \text{ kg}$ block is pulled along a rough horizontal surface by a constant horizontal force of $30 \text{ N}$ . The block accelerates from rest and travels $8.0 \text{ m}$ in $4.0 \text{ s}$ .

(a) Show that the acceleration of the block is $1.0 \text{ m s}^{-2}$ . [2]

.......................................................................................................................

(b) Calculate the resultant force acting on the block. [1]

.......................................................................................................................

(c) Determine the magnitude of the frictional force acting on the block. [2]

.......................................................................................................................

(d) Calculate the work done by the applied $30 \text{ N}$ force over the $8.0 \text{ m}$ displacement. [1]

.......................................................................................................................

(e) Calculate the average power delivered by the applied force during the $4.0 \text{ s}$ interval. [2]

.......................................................................................................................

12. A $0.40 \text{ kg}$ trolley A moves at $2.5 \text{ m s}^{-1}$ on a frictionless horizontal track and collides with a stationary trolley B of mass $0.60 \text{ kg}$ . After the collision, trolley A moves in the opposite direction at $0.50 \text{ m s}^{-1}$ .

(a) Calculate the total momentum of the system before the collision. [1]

.......................................................................................................................

(b) Using the principle of conservation of linear momentum, calculate the velocity of trolley B after the collision. State its direction. [3]

.......................................................................................................................

(c) Determine whether the collision is elastic or inelastic. Show your working clearly. [3]

.......................................................................................................................

13. A small ball of mass $0.20 \text{ kg}$ is attached to a light inextensible string of length $0.80 \text{ m}$ and made to move in a vertical circle. The ball passes through the lowest point of the circle with a speed of $5.0 \text{ m s}^{-1}$ .

(a) Calculate the kinetic energy of the ball at the lowest point. [1]

.......................................................................................................................

(b) By applying the principle of conservation of energy, calculate the speed of the ball at the highest point of the circle. [4]

.......................................................................................................................

(c) Calculate the tension in the string when the ball is at the lowest point. [3]

.......................................................................................................................

14. A car of mass $1200 \text{ kg}$ travels along a straight horizontal road. The engine exerts a driving force of $2400 \text{ N}$ and the total resistive force acting on the car is $600 \text{ N}$ .

(a) Calculate the acceleration of the car. [2]

.......................................................................................................................

(b) The car starts from rest. Calculate the kinetic energy of the car after it has travelled $50 \text{ m}$ . [2]

.......................................................................................................................

(c) The car now travels up a slope inclined at $5.0^{\circ}$ to the horizontal at a constant speed. The driving force remains $2400 \text{ N}$ and the resistive force (excluding the component of weight) remains $600 \text{ N}$ . Calculate the power developed by the engine when the speed is $15 \text{ m s}^{-1}$ . [3]

.......................................................................................................................

15. A ball is projected from ground level at an angle of $35^{\circ}$ above the horizontal with an initial speed of $25 \text{ m s}^{-1}$ . Air resistance is negligible.

(a) Show that the horizontal component of the initial speed is approximately $20.5 \text{ m s}^{-1}$ . [1]

.......................................................................................................................

(b) Calculate the vertical component of the initial speed. [1]

.......................................................................................................................

(c) Calculate the maximum height reached by the ball above the ground. [3]

.......................................................................................................................

(d) Calculate the horizontal range of the ball. [3]

.......................................................................................................................

Section C: Long Structured Question (15 marks)

Answer question 16.

16. A student investigates the motion of a ball bouncing on a horizontal surface. The ball is dropped from a height $h_0 = 2.0 \text{ m}$ above the surface. After each bounce, the ball reaches a maximum height that is $64\%$ of the maximum height after the previous bounce. Air resistance is negligible throughout.

(a) Explain, using Newton's laws of motion, why the ball accelerates downwards after it is released. [2]

.......................................................................................................................

.......................................................................................................................................................................................................................................

(b) Calculate the speed of the ball just before it hits the surface for the first time. [2]

.......................................................................................................................

(c) Calculate the height reached by the ball after the first bounce. [1]

.......................................................................................................................

(d) The student suggests that the total distance travelled by the ball before it comes to rest can be calculated using the formula for the sum of an infinite geometric series.

(i) Write an expression for the total distance $D$ travelled by the ball in terms of $h_0$ and the coefficient of restitution factor. [2]

.......................................................................................................................

(ii) Hence, calculate the total distance travelled by the ball. [2]

.......................................................................................................................

(e) The student also measures the time between the first and second bounce. Calculate this time interval. [3]

.......................................................................................................................

(f) Sketch a velocity–time graph for the ball from the moment it is dropped until just after the second bounce. Show clearly the magnitudes of key velocities and the times at which bounces occur. [3]

<image_placeholder> id: Q16-fig1 type: graph linked_question: Q16(f) description: Velocity-time graph for a bouncing ball from release until just after the second bounce. The ball is dropped from rest, accelerates downward (negative velocity), hits the ground, rebounds with reduced upward (positive) velocity, decelerates to zero at the apex, then accelerates downward again to hit the ground a second time. labels: Time axis (t/s), Velocity axis (v/m s^-1), labels for: initial drop (v=0), first impact velocity (v1 downward, negative), rebound velocity (v2 upward, positive, magnitude 0.8 of v1), second impact velocity (v2 downward, negative), apex after first bounce (v=0) values: Drop height h0=2.0 m, g=9.81 m s^-2. v1 = -sqrt(29.812.0) ≈ -6.26 m/s. Rebound speed = 0.8 × 6.26 ≈ 5.01 m/s. Time to first impact = sqrt(2*2.0/9.81) ≈ 0.639 s. Time from first bounce to apex = 5.01/9.81 ≈ 0.511 s. Time from apex to second impact = 0.511 s. Total time to second impact ≈ 0.639 + 0.511 + 0.511 = 1.661 s. must_show: Axes with units, initial condition v=0, linear negative slope to first bounce, discontinuous jump to reduced positive velocity, linear negative slope to v=0 (apex), linear negative slope to second bounce, all key velocity magnitudes and times labelled. </image_placeholder>

.......................................................................................................................

End of Paper

Answers

TuitionGoWhere Practice Paper — Physics H2 A-Level

Answer Key — Practice Paper, Mechanics (Version 5 of 5)

Section A: Short Answer Questions

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

Answer:

The total momentum of a system remains constant (or is conserved) provided that no net external force acts on the system.

Marking notes:

1 mark for stating that total momentum remains constant / is conserved.
1 mark for stating the condition: no external force / closed (isolated) system.
Accept equivalent wording such as "In a closed system, the total momentum before an interaction equals the total momentum after the interaction."
Common mistake: Stating only "momentum is conserved" without any condition — award 1 mark only.

2. A car accelerates uniformly from rest to a speed of $28 \text{ m s}^{-1}$ in $7.0 \text{ s}$ . Calculate the acceleration of the car. [2]

Answer:

Using $a = \frac{v - u}{t}$ :

$a = \frac{28 - 0}{7.0} = 4.0 \text{ m s}^{-2}$

Marking notes:

1 mark for correct formula or method.
1 mark for correct answer with unit.
Accept $4 \text{ m s}^{-2}$ or $4.0 \text{ m s}^{-2}$ .

3. Define work done by a force. State the SI unit for work. [2]

Answer:

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

$W = F \cdot s \cdot \cos\theta$

The SI unit of work is the joule (J), where $1 \text{ J} = 1 \text{ N m}$ .

Marking notes:

1 mark for correct definition (force × displacement in direction of force).
1 mark for correct unit: joule or J.

4. A ball is thrown vertically upwards with an initial speed of $15 \text{ m s}^{-1}$ . Calculate the maximum height reached by the ball. [3]

Answer:

At maximum height, final velocity $v = 0$ .

Using $v^2 = u^2 + 2as$ :

$0 = (15)^2 + 2(-9.81)(h)$

$h = \frac{225}{2 \times 9.81} = \frac{225}{19.62} = 11.5 \text{ m}$

Marking notes:

1 mark for selecting the correct kinematic equation.
1 mark for correct substitution (including correct sign convention).
1 mark for correct answer: $h = 11.5 \text{ m}$ (accept $11.47 \text{ m}$ or $11 \text{ m}$ to 2 s.f.).
Award full marks for energy method: $\frac{1}{2}mv^2 = mgh \Rightarrow h = \frac{v^2}{2g}$ .

5. State Newton's first law of motion. [2]

Answer:

An object remains at rest or continues to move at a constant velocity (in a straight line) unless acted upon by a resultant (net) external force.

Marking notes:

1 mark for stating the object remains at rest or moves with constant velocity.
1 mark for stating the condition: unless a resultant/net external force acts.
Accept equivalent phrasing.

6. A $2.0 \text{ kg}$ object moves in a horizontal circle of radius $3.0 \text{ m}$ at a constant speed of $6.0 \text{ m s}^{-1}$ . Calculate the magnitude of the centripetal acceleration. [2]

Answer:

$a_c = \frac{v^2}{r} = \frac{(6.0)^2}{3.0} = \frac{36}{3.0} = 12.0 \text{ m s}^{-2}$

Marking notes:

1 mark for correct formula.
1 mark for correct answer with unit: $12 \text{ m s}^{-2}$ .

7. Distinguish between elastic and inelastic collisions. Give one example of each. [3]

Answer:

In an elastic collision, both total kinetic energy and total momentum of the system are conserved. Example: collision between two billiard balls (approximately elastic).
In an inelastic collision, total momentum is conserved but total kinetic energy is not conserved (some kinetic energy is transformed into other forms such as heat, sound, or deformation energy). Example: a car crash where the vehicles crumple together.

Marking notes:

1 mark for stating elastic collision conserves kinetic energy (and momentum).
1 mark for stating inelastic collision does not conserve kinetic energy (but momentum is conserved).
1 mark for a valid example of each.
Common mistake: Stating that momentum is not conserved in inelastic collisions — this is incorrect; momentum is always conserved in collisions (in the absence of external forces).

8. State the SI base units of force (the newton) in terms of kilograms, metres, and seconds. [1]

Answer:

$1 \text{ N} = 1 \text{ kg m s}^{-2}$

Marking notes:

1 mark for $\text{kg m s}^{-2}$ (order does not matter).

Answer:

For the vertical motion, initial vertical velocity $u_y = 0$ :

$s = u_y t + \frac{1}{2}g t^2$

$45 = 0 + \frac{1}{2}(9.81)t^2$

$t^2 = \frac{45 \times 2}{9.81} = \frac{90}{9.81} = 9.174$

$t = \sqrt{9.174} = 3.03 \text{ s}$

Marking notes:

1 mark for identifying that vertical motion determines the time.
1 mark for correct substitution into $s = \frac{1}{2}gt^2$ .
1 mark for correct answer: $t = 3.03 \text{ s}$ (accept $3.0 \text{ s}$ to 2 s.f.).
Note: The horizontal speed is irrelevant for the time calculation.

10. Define power and state its SI unit. [2]

Answer:

Power is the rate of doing work (or the rate of energy transfer).

$P = \frac{W}{t} = \frac{\Delta E}{t}$

The SI unit of power is the watt (W), where $1 \text{ W} = 1 \text{ J s}^{-1}$ .

Marking notes:

1 mark for correct definition (rate of doing work / rate of energy transfer).
1 mark for correct unit: watt or W.

Section B: Structured Questions

(a) Show that the acceleration of the block is $1.0 \text{ m s}^{-2}$ . [2]

Answer:

Using $s = ut + \frac{1}{2}at^2$ with $u = 0$ :

$8.0 = 0 + \frac{1}{2} \times a \times (4.0)^2$

$8.0 = \frac{1}{2} \times a \times 16$

$8.0 = 8.0a$

$a = 1.0 \text{ m s}^{-2} \quad \text{(shown)}$

Marking notes:

1 mark for correct equation and substitution.
1 mark for correct derivation of $a = 1.0 \text{ m s}^{-2}$ .

(b) Calculate the resultant force acting on the block. [1]

Answer:

$F_{\text{resultant}} = ma = 5.0 \times 1.0 = 5.0 \text{ N}$

Marking notes:

1 mark for correct answer: $5.0 \text{ N}$ .

(c) Determine the magnitude of the frictional force acting on the block. [2]

Answer:

$F_{\text{resultant}} = F_{\text{applied}} - f$

$5.0 = 30 - f$

$f = 30 - 5.0 = 25 \text{ N}$

Marking notes:

1 mark for correct equation (resultant = applied − friction).
1 mark for correct answer: $25 \text{ N}$ .

(d) Calculate the work done by the applied $30 \text{ N}$ force over the $8.0 \text{ m}$ displacement. [1]

Answer:

$W = F \times s = 30 \times 8.0 = 240 \text{ J}$

Marking notes:

1 mark for correct answer: $240 \text{ J}$ .

(e) Calculate the average power delivered by the applied force during the $4.0 \text{ s}$ interval. [2]

Answer:

$P = \frac{W}{t} = \frac{240}{4.0} = 60 \text{ W}$

Marking notes:

1 mark for correct formula.
1 mark for correct answer: $60 \text{ W}$ .

(a) Calculate the total momentum of the system before the collision. [1]

Answer:

$p_{\text{before}} = m_A u_A + m_B u_B = 0.40 \times 2.5 + 0.60 \times 0 = 1.0 \text{ kg m s}^{-1}$

Marking notes:

1 mark for correct answer: $1.0 \text{ kg m s}^{-1}$ .

(b) Using the principle of conservation of linear momentum, calculate the velocity of trolley B after the collision. State its direction. [3]

Answer:

Taking the initial direction of A as positive:

$m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$1.0 = 0.40 \times (-0.50) + 0.60 \times v_B$

$1.0 = -0.20 + 0.60 v_B$

$1.20 = 0.60 v_B$

$v_B = 2.0 \text{ m s}^{-1}$

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

Marking notes:

1 mark for correct conservation of momentum equation.
1 mark for correct substitution (note: $v_A = -0.50$ because it reverses direction).
1 mark for correct answer with direction: $2.0 \text{ m s}^{-1}$ in the original direction of A.

(c) Determine whether the collision is elastic or inelastic. Show your working clearly. [3]

Answer:

Calculate total kinetic energy before:

$KE_{\text{before}} = \frac{1}{2}(0.40)(2.5)^2 + 0 = \frac{1}{2}(0.40)(6.25) = 1.25 \text{ J}$

Calculate total kinetic energy after:

$KE_{\text{after}} = \frac{1}{2}(0.40)(-0.50)^2 + \frac{1}{2}(0.60)(2.0)^2$

$= \frac{1}{2}(0.40)(0.25) + \frac{1}{2}(0.60)(4.0)$

$= 0.05 + 1.20 = 1.25 \text{ J}$

Since $KE_{\text{before}} = KE_{\text{after}} = 1.25 \text{ J}$ , kinetic energy is conserved.

The collision is elastic.

Marking notes:

1 mark for correct calculation of KE before.
1 mark for correct calculation of KE after.
1 mark for correct conclusion: elastic, with justification that KE is conserved.
If a student gets the wrong KE values but correctly compares them and states the conclusion consistently, award the conclusion mark.

(a) Calculate the kinetic energy of the ball at the lowest point. [1]

Answer:

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.20)(5.0)^2 = \frac{1}{2}(0.20)(25) = 2.5 \text{ J}$

Marking notes:

1 mark for correct answer: $2.5 \text{ J}$ .

(b) By applying the principle of conservation of energy, calculate the speed of the ball at the highest point of the circle. [4]

Answer:

The height difference between the lowest and highest points is $2r = 2 \times 0.80 = 1.60 \text{ m}$ .

Using conservation of energy (lowest → highest):

$KE_{\text{lowest}} = KE_{\text{highest}} + PE_{\text{gain}}$

$\frac{1}{2}mv_{\text{low}}^2 = \frac{1}{2}mv_{\text{high}}^2 + mg(2r)$

$\frac{1}{2}(0.20)(5.0)^2 = \frac{1}{2}(0.20)v_{\text{high}}^2 + (0.20)(9.81)(1.60)$

$2.5 = 0.10 \, v_{\text{high}}^2 + 3.1392$

$0.10 \, v_{\text{high}}^2 = 2.5 - 3.1392 = -0.6392$

Since this gives a negative value for $v_{\text{high}}^2$ , the ball does not reach the highest point of the circle.

Alternative interpretation (if the question assumes the ball does reach the top):

If we proceed formally:

$v_{\text{high}}^2 = v_{\text{low}}^2 - 4gr = 25 - 4(9.81)(0.80) = 25 - 31.392 = -6.392$

Since $v^2 < 0$ , the ball cannot complete the full vertical circle. It will not reach the highest point.

Marking notes:

1 mark for correct energy conservation equation.
1 mark for identifying the height difference as $2r = 1.60 \text{ m}$ .
1 mark for correct substitution.
1 mark for correct conclusion: the ball does not reach the highest point because $v^2$ would be negative (insufficient speed).
Accept the answer: "The ball does not have sufficient speed to reach the top of the circle."

(c) Calculate the tension in the string when the ball is at the lowest point. [3]

Answer:

At the lowest point, applying Newton's second law radially (taking upward as positive):

$T - mg = \frac{mv^2}{r}$

$T = mg + \frac{mv^2}{r}$

$T = (0.20)(9.81) + \frac{(0.20)(5.0)^2}{0.80}$

$T = 1.962 + \frac{(0.20)(25)}{0.80}$

$T = 1.962 + \frac{5.0}{0.80}$

$T = 1.962 + 6.25 = 8.21 \text{ N}$

Marking notes:

1 mark for correct equation: $T - mg = \frac{mv^2}{r}$ .
1 mark for correct substitution.
1 mark for correct answer: $T = 8.21 \text{ N}$ (accept $8.2 \text{ N}$ ).

(a) Calculate the acceleration of the car. [2]

Answer:

$F_{\text{net}} = 2400 - 600 = 1800 \text{ N}$

$a = \frac{F_{\text{net}}}{m} = \frac{1800}{1200} = 1.5 \text{ m s}^{-2}$

Marking notes:

1 mark for correct net force.
1 mark for correct answer: $1.5 \text{ m s}^{-2}$ .

(b) The car starts from rest. Calculate the kinetic energy of the car after it has travelled $50 \text{ m}$ . [2]

Answer:

Using $v^2 = u^2 + 2as$ :

$v^2 = 0 + 2(1.5)(50) = 150$

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(150) = 90\,000 \text{ J} = 90 \text{ kJ}$

Marking notes:

1 mark for finding $v^2 = 150$ (or equivalent).
1 mark for correct answer: $90\,000 \text{ J}$ or $90 \text{ kJ}$ .
Alternative: Work-energy theorem: $KE = F_{\text{net}} \times s = 1800 \times 50 = 90\,000 \text{ J}$ . Award full marks.

Answer:

Since the car travels at constant speed, the net force is zero. The driving force balances the resistive force plus the component of weight down the slope:

$F_{\text{drive}} = f + mg\sin\theta$

The power developed by the engine is:

$P = F_{\text{drive}} \times v = 2400 \times 15 = 36\,000 \text{ W} = 36 \text{ kW}$

Marking notes:

1 mark for understanding that power = driving force × velocity (since we want the power developed by the engine, not the net power).
1 mark for correct substitution: $P = 2400 \times 15$ .
1 mark for correct answer: $36\,000 \text{ W}$ or $36 \text{ kW}$ .
Note: The slope angle and resistive force are extra information needed to verify that the driving force is sufficient, but the power developed by the engine is simply $F_{\text{drive}} \times v$ .

15. A ball is projected from ground level at an angle of $35^{\circ}$ above the horizontal with an initial speed of $25 \text{ m s}^{-1}$ . Air resistance is negligible.

(a) Show that the horizontal component of the initial speed is approximately $20.5 \text{ m s}^{-1}$ . [1]

Answer:

$u_x = u\cos\theta = 25\cos 35^{\circ} = 25 \times 0.8192 = 20.48 \approx 20.5 \text{ m s}^{-1} \quad \text{(shown)}$

Marking notes:

1 mark for correct calculation: $25\cos 35^{\circ} \approx 20.5 \text{ m s}^{-1}$ .

(b) Calculate the vertical component of the initial speed. [1]

Answer:

$u_y = u\sin\theta = 25\sin 35^{\circ} = 25 \times 0.5736 = 14.3 \text{ m s}^{-1}$

Marking notes:

1 mark for correct answer: $14.3 \text{ m s}^{-1}$ (accept $14.34 \text{ m s}^{-1}$ ).

(c) Calculate the maximum height reached by the ball above the ground. [3]

Answer:

At maximum height, $v_y = 0$ :

$v_y^2 = u_y^2 - 2gH$

$0 = (14.34)^2 - 2(9.81)H$

$H = \frac{(14.34)^2}{2 \times 9.81} = \frac{205.6}{19.62} = 10.5 \text{ m}$

Marking notes:

1 mark for correct kinematic equation.
1 mark for correct substitution.
1 mark for correct answer: $H = 10.5 \text{ m}$ (accept $10.48 \text{ m}$ ).

(d) Calculate the horizontal range of the ball. [3]

Answer:

Time of flight: The ball returns to ground level, so vertical displacement $s_y = 0$ :

$s_y = u_y t - \frac{1}{2}g t^2 = 0$

$t\left(u_y - \frac{1}{2}g t\right) = 0$

$t = 0 \quad \text{or} \quad t = \frac{2u_y}{g} = \frac{2 \times 14.34}{9.81} = \frac{28.68}{9.81} = 2.924 \text{ s}$

Range:

$R = u_x \times t = 20.48 \times 2.924 = 59.9 \text{ m}$

Marking notes:

1 mark for correct time of flight calculation.
1 mark for correct range formula.
1 mark for correct answer: $R \approx 60 \text{ m}$ (accept $59.9 \text{ m}$ or $60 \text{ m}$ to 2 s.f.).

Section C: Long Structured Question

(a) Explain, using Newton's laws of motion, why the ball accelerates downwards after it is released. [2]

Answer:

When the ball is released, the only force acting on it is its weight (gravitational force) directed downwards. By Newton's second law ( $F = ma$ ), a resultant downward force produces a downward acceleration. Since the ball was initially at rest, it begins to move in the direction of the resultant force (downwards), consistent with Newton's first law — the ball would remain at rest unless acted upon by a resultant force.

Marking notes:

1 mark for identifying that weight/gravitational force is the only force acting (after release).
1 mark for linking to Newton's second law: resultant force causes acceleration in the direction of the force.

(b) Calculate the speed of the ball just before it hits the surface for the first time. [2]

Answer:

Using conservation of energy:

$mgh_0 = \frac{1}{2}mv^2$

$v = \sqrt{2gh_0} = \sqrt{2 \times 9.81 \times 2.0} = \sqrt{39.24} = 6.26 \text{ m s}^{-1}$

Marking notes:

1 mark for correct energy conservation equation or kinematic equation.
1 mark for correct answer: $6.26 \text{ m s}^{-1}$ .

(c) Calculate the height reached by the ball after the first bounce. [1]

Answer:

$h_1 = 0.64 \times h_0 = 0.64 \times 2.0 = 1.28 \text{ m}$

Marking notes:

1 mark for correct answer: $1.28 \text{ m}$ .

(d) The student suggests that the total distance travelled by the ball before it comes to rest can be calculated using the formula for the sum of an infinite geometric series.

(i) Write an expression for the total distance $D$ travelled by the ball in terms of $h_0$ and the coefficient of restitution factor. [2]

Answer:

The ball first falls $h_0$ , then bounces up $h_1 = 0.64h_0$ and falls $h_1$ , then bounces up $h_2 = 0.64h_1 = (0.64)^2 h_0$ and falls $h_2$ , and so on.

$D = h_0 + 2h_1 + 2h_2 + 2h_3 + \cdots$

$D = h_0 + 2(0.64h_0) + 2(0.64)^2 h_0 + 2(0.64)^3 h_0 + \cdots$

$D = h_0 + 2h_0\left[0.64 + (0.64)^2 + (0.64)^3 + \cdots\right]$

The series in brackets is an infinite geometric series with first term $a = 0.64$ and common ratio $r = 0.64$ :

$S = \frac{a}{1-r} = \frac{0.64}{1 - 0.64} = \frac{0.64}{0.36}$

$D = h_0 + 2h_0 \times \frac{0.64}{0.36} = h_0\left(1 + \frac{2 \times 0.64}{0.36}\right) = h_0\left(1 + \frac{1.28}{0.36}\right) = h_0\left(\frac{0.36 + 1.28}{0.36}\right) = h_0\left(\frac{1.64}{0.36}\right)$

$\boxed{D = h_0\left(\frac{1 + e^2}{1 - e^2}\right)}$

where $e = 0.64$ (coefficient of restitution factor), or equivalently:

$D = h_0\left(1 + \frac{2r}{1-r}\right)$

where $r = 0.64$ .

Marking notes:

1 mark for correct identification of the geometric series pattern (initial drop + twice each bounce height).
1 mark for correct expression for total distance.

(ii) Hence, calculate the total distance travelled by the ball. [2]

Answer:

$D = 2.0 \times \frac{1.64}{0.36} = 2.0 \times 4.556 = 9.11 \text{ m}$

Marking notes:

1 mark for correct substitution.
1 mark for correct answer: $D = 9.11 \text{ m}$ (accept $9.1 \text{ m}$ ).

(e) The student also measures the time between the first and second bounce. Calculate this time interval. [3]

Answer:

After the first bounce, the ball rebounds with speed:

$v_{\text{rebound}} = 0.8 \times 6.26 = 5.01 \text{ m s}^{-1}$

(Note: Since $h_1/h_0 = 0.64 = e^2$ , the coefficient of restitution $e = \sqrt{0.64} = 0.8$ , so the speed after bounce is $0.8 \times$ speed before bounce.)

Time to reach maximum height after first bounce:

$t_{\text{up}} = \frac{v_{\text{rebound}}}{g} = \frac{5.01}{9.81} = 0.511 \text{ s}$

By symmetry (no air resistance), time to fall back down = time to go up:

$t_{\text{down}} = 0.511 \text{ s}$

Total time between first and second bounce:

$\Delta t = t_{\text{up}} + t_{\text{down}} = 0.511 + 0.511 = 1.02 \text{ s}$

Marking notes:

1 mark for calculating rebound speed ( $e \times$ impact speed = $0.8 \times 6.26 = 5.01 \text{ m s}^{-1}$ ).
1 mark for calculating time up (or time down) using $t = v/g$ .
1 mark for correct total time: $1.02 \text{ s}$ .

Answer:

The graph should show:

From $t = 0$ to $t = 0.639$ s: A straight line with negative slope ( $-g = -9.81 \text{ m s}^{-2}$ ) starting from $v = 0$ and ending at $v = -6.26 \text{ m s}^{-1}$ (downward velocity just before first impact).
At $t = 0.639$ s: A discontinuous jump from $v = -6.26 \text{ m s}^{-1}$ to $v = +5.01 \text{ m s}^{-1}$ (rebound, velocity reverses and reduces in magnitude).
From $t = 0.639$ s to $t = 1.150$ s: A straight line with negative slope ( $-g$ ) from $v = +5.01 \text{ m s}^{-1}$ to $v = 0$ (at maximum height after first bounce).
From $t = 1.150$ s to $t = 1.661$ s: A straight line with negative slope from $v = 0$ to $v = -5.01 \text{ m s}^{-1}$ (ball falling back down).
At $t = 1.661$ s: A discontinuous jump from $v = -5.01 \text{ m s}^{-1}$ to $v = +4.01 \text{ m s}^{-1}$ (second bounce, $0.8 \times 5.01$ ).

Key values to label:

Time axis: $t_1 = 0.639$ s (first bounce), $t_{\text{apex}} = 1.150$ s, $t_2 = 1.661$ s (second bounce)
Velocity values: $-6.26$ , $+5.01$ , $0$ , $-5.01$ , $+4.01$ (all in $\text{m s}^{-1}$ )
Slope of each straight-line segment = $-9.81 \text{ m s}^{-2}$

Marking notes:

1 mark for correct shape: straight-line segments with negative slope, discontinuous jumps at bounces.
1 mark for correct velocity magnitudes labelled ( $-6.26$ , $+5.01$ , etc.).
1 mark for correct time values labelled at bounce points.
The graph should show velocity on the y-axis (positive upward) and time on the x-axis.

End of Answer Key

Marks Summary:

Section	Marks
A: Q1–Q10	20
B: Q11	8
B: Q12	7
B: Q13	8
B: Q14	7
B: Q15	8
C: Q16	15
Total	60 ✓