Chapter 6 · Solved paper40 marks

Model Test – Solved ഉത്തരങ്ങൾ

Every question of the model paper, in the same order, with the full answer, the marking scheme, an animated figure or working for every numerical and explanation question, and a link to the page where the idea is taught. Try each one again before you open the answer, and mark your own paper honestly.

📐 How the figures are drawn

Forces are arrows: red for gravity (weight), green for the force that does positive work, amber for friction. Energy bars: blue = potential energy (സ്ഥിതികോർജം), red = kinetic energy (ഗതികോർജം), amber = heat or work on something else, green = total. Every number in the animations is worked out with W = F s, E_p = mgh, E_k = ½mv² and P = W/t, using the g given in the question.

SECTION AAnswer all · 8 × 1 = 8 marks

When a force of 1 N moves an object through 1 m in the direction of the force, the work done is .

(1 mark)

💡 W = F × s with F = 1 N and s = 1 m.

✅ Answer: 1 joule (1 J) (ജൂൾ (J))

W = F × s = 1 N × 1 m = 1 N m = 1 J. This is the definition of one joule.

Marking: 1 mark (the unit must be written).

📖 Review this on Page 5 · Key box 3: one joule

Which of the following is a scalar quantity?

(a) force(b) displacement(c) velocity(d) energy

(1 mark)

💡 Which one has no direction?

✅ Answer: (d) energy

Energy has only a size, no direction – it is a scalar (അദിശ അളവ്). Force, displacement and velocity have a direction – they are vectors (സദിശ അളവ്). Work is a scalar too.

Marking: 1 mark for (d).

📖 Review this on Page 9 · Similarities of work and energy · Page 4 · Scalars and vectors

A headload worker stands still for 5 minutes holding a 20 kg sack on his head. The work done by him on the sack during this time is

(a) 0 J(b) 200 J(c) 1000 J(d) 6000 J

(1 mark)

💡 Is the sack displaced?

✅ Answer: (a) 0 J

The sack is not displaced (സ്ഥാനാന്തരം = 0), so W = F × 0 = 0 J – even though he feels tired (Table 6.1 (d)).

Marking: 1 mark for (a).

📖 Review this on Page 5 · Table 6.1

Find the relation and fill in: kinetic energy : ½mv² :: potential energy due to position :

(1 mark)

💡 PE of an object raised to a height h.

✅ Answer: mgh

Kinetic energy (ഗതികോർജം) E_k = ½mv²; potential energy (സ്ഥിതികോർജം) due to position E_p = mgh.

Marking: 1 mark.

📖 Review this on Page 7 · E_p = mgh

A stone is thrown vertically upwards. While it rises, the work done on it by the force of gravity is

(a) positive(b) negative(c) zero(d) equal to its kinetic energy at the top

(1 mark)

💡 Compare the direction of the force with the direction of motion.

✅ Answer: (b) negative

Gravity acts downwards but the stone moves upwards – the displacement is opposite to the force, so the work is negative (നെഗറ്റീവ് പ്രവൃത്തി).

Marking: 1 mark for (b).

📖 Review this on Page 6 · Key box 5, positive and negative work

1 HP = W

(1 mark)

💡 The book value used for pumps and motors.

✅ Answer: 746

Horsepower (കുതിരശക്തി (HP)) is used for motors and vehicles: 1 HP = 746 W.

Marking: 1 mark.

📖 Review this on Page 10 · kW, MW and HP

The energy stored in a compressed spring is

(a) kinetic energy(b) potential energy due to position(c) potential energy due to configuration(d) heat energy

(1 mark)

💡 The spring is not raised – its shape is changed.

✅ Answer: (c) potential energy due to configuration

Compressing the spring changes its shape (arrangement), so it stores potential energy due to configuration (കോൺഫിഗറേഷൻ മൂലമുള്ള സ്ഥിതികോർജം) – Table 6.3.

Marking: 1 mark for (c).

📖 Review this on Page 7 · Table 6.3

The unit of power is named after the Scottish scientist , who built a much more efficient steam engine.

(1 mark)

💡 The unit of power is the W…

✅ Answer: James Watt

James Watt (1736–1819). While repairing a Newcomen engine in 1764 he built a much more efficient steam engine; the unit of power, the watt (വാട്ട് (W)), is named after him.

Marking: 1 mark.

📖 Review this on Page 10 · James Watt box

SECTION BAnswer all · 5 × 2 = 10 marks

A headload worker lifts a 40 kg sack of rice from the ground on to a lorry 1.5 m high in 4 s. (g = 10 m/s²) (a) Find the work he does against gravity. (b) Find his power.

(2 marks)

💡 Work against gravity = weight × height; power = work ÷ time.

🎬 Step by step

✅ Answer: (a) 600 J (b) 150 W

Given: m = 40 kg, h = 1.5 m, g = 10 m/s², t = 4 s

Formula: W = mgh; P = W ÷ t

Substitution: (a) W = 40 × 10 × 1.5; (b) P = 600 ÷ 4

Answer: (a) W = 600 J (b) P = 150 W

Marking: (a) 1 mark (formula ½, answer with unit ½); (b) 1 mark.

📖 Review this on Page 6 · Work against gravity W = mgh · Page 10 · Power P = W/t

10.

A ball of mass 2 kg moves with a velocity of 3 m/s. (a) Find its kinetic energy. (b) What will its kinetic energy be if its velocity is doubled?

(2 marks)

💡 Square the velocity first; then multiply by ½ and m.

🎬 Step by step

✅ Answer: (a) 9 J (b) 36 J (4 times)

Given: m = 2 kg, v = 3 m/s

Formula: E_k = ½mv²

Substitution: (a) ½ × 2 × 3² = 1 × 9; (b) ½ × 2 × 6² = 1 × 36

Answer: (a) 9 J (b) 36 J

Doubling v makes v² four times as big, so the kinetic energy becomes 4 times (Table 6.4).

Marking: (a) 1 mark; (b) 1 mark (½ value, ½ unit / “4 times”).

📖 Review this on Page 8 · E_k = ½mv², Table 6.4

11.

A girl pulls a box to the right along a table (figure). The forces on the box are: her pull, friction, the force of gravity and the upward force of the table. Which of these forces does (a) positive work, (b) negative work, (c) no work? Give a reason for (c).

(2 marks)

💡 Is the box displaced along, against or at right angles to each force?

🎬 Step by step

✅ Answer: (a) her pull (b) friction (c) gravity and the table’s force

(a) Her pull – the box moves in the direction of the pull: positive work (പോസിറ്റീവ് പ്രവൃത്തി).

(b) Friction (ഘർഷണബലം) – it acts opposite to the motion: negative work.

(c) Gravity and the upward force of the table do no work: they are vertical (at right angles to the motion), so the box is not displaced in their direction. They also balance each other.

Marking: (a) ½ mark; (b) ½ mark; (c) ½ for naming both forces, ½ for the reason.

📖 Review this on Page 6 · Fig 6.4 and Table 6.2

12.

A packet of biscuits gives 150 kcal of energy. (1 kcal = 4200 J) (a) Express this energy in joule. (b) If all of it could be used to climb, how high could a student of mass 50 kg rise? (g = 10 m/s²)

(2 marks)

💡 kcal × 4200 → J; then h = E ÷ (m × g).

🎬 Step by step

✅ Answer: (a) 630 000 J (b) 1260 m

Given: 150 kcal; 1 kcal = 4200 J

Formula: energy in J = kcal × 4200

Substitution: 150 × 4200

Answer: (a) 630 000 J (6.3 × 10⁵ J)

Given: E = 630 000 J, m = 50 kg, g = 10 m/s²

Formula: E = mgh, so h = E ÷ (mg)

Substitution: 630 000 ÷ (50 × 10)

Answer: (b) h = 1260 m

Our body is far from 100 % efficient – most food energy becomes heat – so a real climb would be much less. The idea: energy is the capacity to do work.

Marking: (a) 1 mark; (b) 1 mark (formula ½, answer ½).

📖 Review this on Page 7 · Calorie and the mango question · Page 3 · Rearranging h = E ÷ mg

13.

Water stored in the Idukki reservoir is used to light bulbs in our homes. Write the energy changes, in order, from the water in the dam to the light from a bulb.

(2 marks)

💡 Dam → penstock → turbine → generator → bulb.

🎬 Step by step

✅ Answer: PE → KE → mechanical (turbine) → electrical (generator) → light

Water in the dam: potential energy (സ്ഥിതികോർജം) → flowing down the penstock (പെൻസ്റ്റോക്ക്): kinetic energy (ഗതികോർജം) → turbine turning: mechanical energy (യാന്ത്രികോർജം) → generator: electrical energy (വൈദ്യുതോർജം) → bulb: light (and some heat) – Table 6.5.

Marking: ½ mark for each correct change in order (4 changes).

📖 Review this on Page 10 · Hydroelectric station, Table 6.5 · Page 14 · Idukki lab

SECTION CAnswer all · 4 × 3 = 12 marks

14.

A stone of mass 2 kg is dropped from a height of 20 m (figure). Ignore air resistance; g = 10 m/s². (a) Find its potential energy and kinetic energy at the top. (b) Find its potential energy and kinetic energy when it is 5 m above the ground. (c) Find its speed just before it hits the ground.

(3 marks)

💡 Total energy stays 400 J all the way down.

🎬 Step by step

✅ Answer: (a) 400 J, 0 J (b) 100 J, 300 J (c) 20 m/s

Given: m = 2 kg, h = 20 m, g = 10 m/s²; dropped, so v = 0 at the top

Formula: E_p = mgh; E_k = ½mv²

Substitution: 2 × 10 × 20

Answer: (a) E_p = 400 J, E_k = 0 J – total 400 J

(b) At 5 m: E_p = 2 × 10 × 5 = 100 J. The total stays 400 J (conservation of energy), so E_k = 400 − 100 = 300 J.

Given: just above the ground E_p = 0, so E_k = 400 J

Formula: ½mv² = E_k

Substitution: ½ × 2 × v² = 400 → v² = 400

Answer: (c) v = 20 m/s (check: v² = 2gh = 2 × 10 × 20 = 400)

Marking: 1 mark for each part.

📖 Review this on Page 9 · Pile driver and conservation · Page 8 · 0.5 kg ball question

15.

A motor pump lifts 600 kg of water into an overhead tank 10 m high in 2 minutes. (g = 10 m/s²) (a) Find the work done. (b) Find the power of the motor in watt. (c) Express it in horsepower.

(3 marks)

💡 Change 2 minutes into seconds before dividing.

🎬 Step by step

✅ Answer: (a) 60 000 J (b) 500 W (c) about 0.67 HP

Given: m = 600 kg, h = 10 m, g = 10 m/s², t = 2 min = 120 s

Formula: W = mgh

Substitution: 600 × 10 × 10

Answer: (a) W = 60 000 J

Given: W = 60 000 J, t = 120 s

Formula: P = W ÷ t

Substitution: 60 000 ÷ 120

Answer: (b) P = 500 W

Marking: 1 mark for each part (convert minutes to seconds!).

📖 Review this on Page 12 · Let us assess Q6 (motor pump) · Page 10 · Power and HP

16.

Anu (mass 40 kg) runs up a staircase 6 m high in 12 s. Binu (mass 50 kg) climbs the same staircase in 20 s. (g = 10 m/s²) (a) Find the work done by each. (b) Find the power of each. (c) Who did more work? Who is more powerful?

(3 marks)

💡 Work depends on weight and height; power also on the time.

🎬 Step by step

✅ Answer: (a) 2400 J, 3000 J (b) 200 W, 150 W (c) Binu; Anu

(a) Anu: W = mgh = 40 × 10 × 6 = 2400 J; Binu: 50 × 10 × 6 = 3000 J.

(b) Anu: P = 2400 ÷ 12 = 200 W; Binu: P = 3000 ÷ 20 = 150 W.

(c) Binu did more work (he is heavier), but Anu is more powerful – she does more work per second. More work does not always mean more power (Let us assess Q14).

Marking: (a) 1 mark; (b) 1 mark; (c) ½ + ½.

📖 Review this on Page 10 · Power race, Table 6.6 · Page 13 · Let us assess Q14

17.

A car of mass 1000 kg is moving at 72 km/h. (a) Express its speed in m/s. (b) Find its kinetic energy. (c) When the brakes are applied it stops after 40 m. Find the average braking force. Into which form does the kinetic energy change?

(3 marks)

💡 Work done by the brakes = the kinetic energy taken away.

🎬 Step by step

✅ Answer: (a) 20 m/s (b) 200 000 J (c) 5000 N; heat

(a) 72 km/h = 72 × 1000 m ÷ 3600 s = 20 m/s.

Given: m = 1000 kg, v = 20 m/s

Formula: E_k = ½mv²

Substitution: ½ × 1000 × 20² = 500 × 400

Answer: (b) E_k = 200 000 J

Given: the brakes do (negative) work equal to the kinetic energy: W = 200 000 J, s = 40 m

Formula: W = F × s, so F = W ÷ s

Substitution: 200 000 ÷ 40

Answer: (c) F = 5000 N; the kinetic energy becomes heat in the brakes and tyres (and some sound)

Marking: 1 mark for each part ((c): ½ force, ½ heat).

📖 Review this on Page 2 · km/h → m/s · Page 8 · Kinetic energy · Page 14 · Crumple zones lab

SECTION DAnswer all · 2 × 5 = 10 marks

18.

The hammer of a pile driver has a mass of 500 kg. It is raised to a height of 5 m above the top of a pile and released (figure). Take g = 10 m/s² and ignore air resistance. (a) How much work is done in raising the hammer? In which form is it stored? (b) What are the kinetic energy and the potential energy of the hammer just before it hits the pile (take the top of the pile as the zero level)? (c) Find its speed just before it hits the pile. (d) State the law that explains (b). (e) Assume all this kinetic energy is used to push the pile 0.25 m into the ground. Find the average resisting force of the soil.

(5 marks)

💡 PE at the top = KE at the bottom; then W = F × s.

🎬 Step by step

✅ Answer: (a) 25 000 J, PE (b) 25 000 J, 0 J (c) 10 m/s (d) conservation of energy (e) 100 000 N

Given: m = 500 kg, h = 5 m, g = 10 m/s²

Formula: W = mgh

Substitution: 500 × 10 × 5

Answer: (a) W = 25 000 J, stored as potential energy (സ്ഥിതികോർജം)

(b) Just before it hits: E_p = 0 J, E_k = 25 000 J – all the potential energy has become kinetic energy.

Given: E_k = 25 000 J, m = 500 kg

Formula: ½mv² = E_k (or v² = u² + 2gh with u = 0)

Substitution: ½ × 500 × v² = 25 000 → v² = 100

Answer: (c) v = 10 m/s

(d) Law of conservation of energy (ഊർജസംരക്ഷണ നിയമം): energy can neither be created nor destroyed; it can only be changed from one form to another without any loss or gain.

Given: W = 25 000 J, s = 0.25 m

Formula: W = F × s, so F = W ÷ s

Substitution: 25 000 ÷ 0.25

Answer: (e) F = 100 000 N (10⁵ N)

Marking: 1 mark for each part ((a): ½ value, ½ PE).

📖 Review this on Page 9 · Pile driver, key box 10 · Page 14 · Pile-driver lab · Page 11 · Pile driver solved

19.

A ball of mass 0.2 kg is thrown vertically upwards with a velocity of 20 m/s. Ignore air resistance; g = 10 m/s². (a) Find its kinetic energy when it is thrown. (b) Find the maximum height it reaches. (c) Find its potential energy and kinetic energy at a height of 15 m. (d) Which graph in the figure shows how its kinetic energy changes with height? (e) Find the work done by gravity on the ball while it rises. Is it positive or negative?

(5 marks)

💡 At the top all the KE has become PE.

🎬 Step by step

✅ Answer: (a) 40 J (b) 20 m (c) 30 J, 10 J (d) graph (c) (e) −40 J, negative

Given: m = 0.2 kg, v = 20 m/s

Formula: E_k = ½mv²

Substitution: ½ × 0.2 × 20² = 0.1 × 400

Answer: (a) E_k = 40 J

Given: at the top E_k = 0, so E_p = 40 J

Formula: mgh = 40 J, so h = 40 ÷ (mg)

Substitution: 40 ÷ (0.2 × 10)

Answer: (b) h = 20 m

(d) Graph (c): E_k = 40 − 2h falls in a straight line from 40 J at h = 0 to 0 J at h = 20 m (Graph 6.1 (a) of the textbook).

(e) Gravity (0.2 × 10 = 2 N) acts downwards while the ball moves 20 m upwards: W = −2 × 20 = −40 J – negative work. It takes away exactly the 40 J of kinetic energy.

Marking: 1 mark for each part ((d): ½ choice, ½ reason).

📖 Review this on Page 12 · Let us assess Q8, Graph 6.1 · Page 13 · Let us assess Q12 · Page 6 · Negative work

Marks summary

Section	Questions	Marks each	Total	Topics
A	1–8	1	8	one joule, scalars, no displacement – no work, mgh, negative work, 1 HP, PE of configuration, James Watt
B	9–13	2	10	work against gravity and power, kinetic energy when v doubles, positive/negative/zero work, kcal → J, hydroelectric energy chain
C	14–17	3	12	falling stone (conservation), motor pump and HP, power race on the stairs, car braking (km/h → m/s, F = W ÷ s)
D	18–19	5	10	pile driver, ball thrown up with the KE–height graph
Total			40

⏱️ Time management (15 min cool-off + 90 minutes)

Cool-off (15 min) – read the whole paper. Note which g each question uses and every unit that must be converted (minutes, km/h, kcal, kJ).
Section A – about 10 minutes. One word or one option; write the unit where one is asked.
Section B – about 20 minutes. Given, Formula, Substitution, Answer for Q9, Q10 and Q12; for Q11 and Q13 use the words “direction of the force” and name every form of energy.
Section C – about 25 minutes. Square only v in ½mv²; use the vertical height in mgh; change 2 minutes to 120 s and 72 km/h to 20 m/s.
Section D – about 25 minutes. Answer every part; state the law of conservation of energy in full.
Checking – keep 10 minutes: units J, W, N, m/s on every answer? total energy the same at the top and the bottom? work against gravity positive, gravity’s work on a rising ball negative?

🎉 Finished the chapter?

Count your marks: 32 or more – excellent; 24–31 – good, revise the pages linked under the questions you missed; below 24 – go back through pages 5 to 10 with their labs and do the scored quiz again. Then return to the chapter home and tick off what you have learned.