Chapter 3 · Let us assessTextbook pages 71–72

Let Us Assess Q10–Q17 & Extended Activities വിലയിരുത്താം · തുടർപ്രവർത്തനങ്ങൾ

The last eight end-of-chapter questions – two long jumpers, stopping a heavy vehicle, the same force on two bodies, a bullet in a log, the force–distance Graph 3.2 (animated), the four graphs of Graph 3.3, a block pulled both ways and two blocks pushed together – followed by the three extended activities. Try each one in your notebook first, then step through the solution.

📖 What you need (pages 2, 3, 8, 9 and 12 of this guide)

Momentum p = mv (unit kg m/s). Second law F = ma = m(v − u)/t (unit N = kg m/s²).
Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as. Retardation = the size of a negative acceleration.
Convert first: g → kg (÷ 1000), cm → m (÷ 100), km/h → m/s (× 5/18).
Always say which direction is positive. A minus sign on a force means “opposite to the positive direction”.
Third law: F₁₂ = −F₂₁ – the two forces act on different bodies.

Question 10 · Two long jumpers

10. 50 kg മാസുള്ള ഒരാൾ ലോങ് ജമ്പ് ചാടുന്നതിനു വേണ്ടി 8 m/s പ്രവേഗത്തിൽ ഓടി വന്ന് ചാടുന്നു. 60 kg മാസുള്ള മറ്റൊരാൾ 7 m/s പ്രവേഗത്തിൽ ഓടി വന്ന് ചാടുന്നു. ഇവരുടെ മൊമെന്റം താരതമ്യം ചെയ്യുക. A person of mass 50 kg runs up at a velocity of 8 m/s and takes a long jump. Another person of mass 60 kg runs up at 7 m/s and jumps. Compare their momenta (മൊമെന്റം).

Work out p = mv for each jumper separately, then compare the two numbers. The faster runner is not automatically the one with more momentum!

🎬 Solution animation – exam layout

🧪 Try other numbers – two runners

runner 1 mass (kg) runner 1 speed (m/s) runner 2 mass (kg) runner 2 speed (m/s)

Answer: p = mv.
First jumper: p₁ = 50 kg × 8 m/s = 400 kg m/s.
Second jumper: p₂ = 60 kg × 7 m/s = 420 kg m/s.
So the second jumper (60 kg) has more momentum, by 20 kg m/s, even though he runs more slowly – his larger mass more than makes up for the smaller velocity.

⚠️ Common trap

Comparing only the speeds (“8 m/s is faster, so the first has more momentum”). Momentum depends on both mass and velocity – always multiply.

Question 11 · Stopping a 14,000 kg vehicle

11. 14,000 kg മാസുള്ള ഒരു വാഹനത്തിന് 1.8 m/s² മന്ദീകരണം ലഭ്യമാക്കി നിശ്ചലമാക്കണമെങ്കിൽ പ്രയോഗിക്കേണ്ട ബലത്തിന്റെ അളവ് കണക്കാക്കുക. To stop a vehicle of mass 14,000 kg by giving it a retardation (മന്ദീകരണം) of 1.8 m/s², calculate the magnitude of the force that must be applied.

Retardation 1.8 m/s² means acceleration a = −1.8 m/s² (taking the direction of motion as positive). Use F = ma.

🎬 Solution animation – exam layout

🧪 Try other numbers – brake a KSRTC bus or a lorry

mass (kg) retardation (m/s²)

Answer: Take the direction of motion as positive. m = 14,000 kg, a = −1.8 m/s².
F = ma = 14,000 × (−1.8) = −25,200 N.
The force needed is 25,200 N (2.52 × 10⁴ N), acting opposite to the direction of motion – a braking force.

💡 Why this matters

Force = mass × retardation: a fully loaded lorry needs a much bigger braking force than a car to stop at the same rate. If the brakes cannot supply it, the lorry takes longer and travels farther before stopping – the physics behind extended activity 1 below.

Question 12 · The same force on a different body

12. 20 kg മാസുള്ള ഒരു വസ്തുവിൽ ഒരു ബലം 2 s സമയത്തേക്ക് പ്രയോഗിച്ചപ്പോൾ അതിന്റെ പ്രവേഗം 10 m/s ൽ നിന്നും 50 m/s ആയി മാറി. ഇതേ ബലം 10 kg മാസുള്ള 20 m/s പ്രവേഗത്തിൽ ചലിക്കുന്ന വസ്തുവിന്റെ ചലനദിശയിൽ 2 s സമയത്തേക്കു തന്നെ പ്രയോഗിച്ചാൽ അതിന്റെ അന്ത്യപ്രവേഗം കണക്കാക്കുക. When a force acted for 2 s on an object of mass 20 kg, its velocity changed from 10 m/s to 50 m/s. If the same force acts for 2 s on an object of mass 10 kg moving at 20 m/s, in its direction of motion (ചലനദിശ), calculate its final velocity (അന്ത്യപ്രവേഗം).

Two stages. First find the force from the 20 kg object: F = m(v − u)/t. Then use the same F on the 10 kg object: a = F/m, and v = u + at.

🎬 Solution animation – exam layout

🧪 Try other numbers – 400 N on another body

second mass (kg) its initial velocity (m/s) time (s)

Answer: Stage 1 (20 kg): F = m(v − u)/t = 20 × (50 − 10)/2 = 20 × 40/2 = 400 N.
Stage 2 (10 kg, same 400 N, in its direction of motion): a = F/m = 400/10 = 40 m/s².
v = u + at = 20 + 40 × 2 = 100 m/s.

⚠️ Common trap

Thinking “same force ⇒ same change of velocity (40 m/s) ⇒ 60 m/s”. The same force for the same time gives the same change of momentum (400 × 2 = 800 kg m/s). Half the mass gets twice the change of velocity: 80 m/s, so 20 + 80 = 100 m/s.

Question 13 · A bullet stopped by a log

13. 20 g മാസുള്ള ഒരു വെടിയുണ്ട 100 m/s പ്രവേഗത്തിൽ ഒരു മരത്തടിയിൽ തറയ്ക്കുന്നു. മരത്തടിയ്ക്കുള്ളിലേക്ക് 4 cm സഞ്ചരിച്ചപ്പോൾ അത് നിശ്ചലമായി. a) വെടിയുണ്ടയുടെ ത്വരണമെത്ര? b) വെടിയുണ്ടയുടെ മന്ദീകരണമെത്ര? c) ഈ വെടിയുണ്ട മരത്തടിയിൽ പ്രയോഗിച്ച ബലം കണക്കാക്കുക. A bullet of mass 20 g strikes a log of wood at 100 m/s. It comes to rest after travelling 4 cm into the log.

What is the acceleration (ത്വരണം) of the bullet?
What is the retardation (മന്ദീകരണം) of the bullet?
Calculate the force that the bullet applied on the log.

No time is given – but a distance is. Which equation of motion links u, v, a and s? Convert 20 g → kg and 4 cm → m first.

🎬 Solution animation – exam layout

🧪 Try other numbers – bullet and log

bullet mass (g) speed u (m/s) depth s (cm)

Double the speed: the force becomes four times bigger (it depends on u²). Double the depth (softer wood): the force halves.

Answer: Take the bullet's direction of motion as positive. m = 20 g = 0.02 kg, u = 100 m/s, v = 0, s = 4 cm = 0.04 m.
(a) v² = u² + 2as ⇒ 0 = 100² + 2 × a × 0.04 ⇒ 0.08a = −10,000 ⇒ a = −125,000 m/s² (= −1.25 × 10⁵ m/s²).
(b) Retardation = 125,000 m/s² (1.25 × 10⁵ m/s²).
(c) Force of the log on the bullet F = ma = 0.02 × (−125,000) = −2500 N (opposite to the motion). By the third law, the bullet pushes the log with an equal and opposite force: 2500 N, in the bullet's direction of motion.

⚠️ Common trap

Putting s = 4 (cm) and m = 20 (g) straight into the formulas gives answers wrong by factors of 100 and 1000. Convert to SI units before substituting.

Question 14 · Graph 3.2 – force against distance

14. 10 kg മാസുള്ള ഒരു വസ്തുവിൽ ബലം പ്രയോഗിക്കുന്നതുമായി ബന്ധപ്പെട്ട ഗ്രാഫ് നൽകിയിരിക്കുന്നു. ഗ്രാഫിൽ സൂചിപ്പിച്ചിരിക്കുന്ന പ്രകാരം ബലത്തിന്റെ അളവ് മാറുന്നുണ്ട് (ഘർഷണം പരിഗണിക്കേണ്ടതില്ല). The graph shows the force applied on an object of mass 10 kg. The size of the force changes as shown in the graph (ignore friction – ഘർഷണം).

വസ്തു 3 m ൽ ആയിരിക്കുമ്പോൾ അതിന്റെ ത്വരണമെത്ര? What is the acceleration of the object when it is at 3 m?
ഏതൊക്കെ അവസരങ്ങളിലാണ് വസ്തുവിന് സമപ്രവേഗം? In which parts does the object have uniform velocity (സമപ്രവേഗം)?
ഏതൊക്കെ അവസരങ്ങളിലാണ് വസ്തുവിന് സമത്വരണമുള്ളത്? In which parts does it have uniform acceleration (സമത്വരണം)?
ഏത് അവസരത്തിലാണ് മന്ദീകരണമുള്ളത്? In which part does it have retardation?

Read the force off the graph for each part and use a = F/m with m = 10 kg. F = 0 ⇒ a = 0 ⇒ uniform velocity (first law). Constant positive F ⇒ uniform acceleration. Negative F (opposite to the motion) ⇒ retardation. Assume the object is already moving to the right (+) at the start.

🎞️ Ride along Graph 3.2 – move the object along the distance axis

distance x (m)

The speed bar assumes the object was already moving at 2 m/s to the right at O (the question does not give a starting speed – any starting speed gives the same pattern: steady, rising, steady, falling, steady).

🎬 Solution animation – reading the graph part by part

Answer (m = 10 kg, a = F/m, direction of motion positive):
(a) At 3 m (part BC) F = 8 N, so a = 8/10 = 0.8 m/s².
(b) Uniform velocity where F = 0: OA (0–2 m), DE (6–8 m) and HI (11–13 m).
(c) Uniform acceleration where the force is constant and positive: BC (2–4 m), a = 0.8 m/s². (In CD the force falls from 8 N to 0, so the acceleration is decreasing – still speeding up, but not uniformly.)
(d) Retardation in FG (8–11 m): F = −6 N acts opposite to the motion, a = −6/10 = −0.6 m/s², i.e. a retardation of 0.6 m/s².

⚠️ Common trap

“F = 0, so the object is at rest.” No! Zero force means zero acceleration: the object keeps whatever velocity it already has (Newton's first law). And this is a force–distance graph, not a velocity–time graph – a flat line at 8 N means steady force, not steady speed.

Question 15 · Graph 3.3 – where is the resultant force zero?

15. പരിണതബലം പൂജ്യമായ ഗ്രാഫേത്? Which graph shows a resultant force (പരിണതബലം) of zero? (Graphs A and B: distance (ദൂരം) against time (സമയം); graphs C and D: speed (വേഗം) against time.)

Zero resultant force ⇒ zero acceleration ⇒ the velocity does not change. Which graph shows a body covering equal distances in equal times?

🗂️ Graph 3.3 – tap your choice for each graph

Score: 0 / 0

Answer: Graph A. A straight distance–time line means equal distances in equal times – uniform velocity, so no acceleration. By F = ma, a = 0 ⇒ resultant force = 0 (first law).
B: the distance–time curve gets steeper – the object speeds up, so there is a resultant force in the direction of motion.
C: speed increases steadily – uniform acceleration, a constant resultant force forwards.
D: speed decreases steadily – uniform retardation, a constant resultant force opposite to the motion.

💡 Why this matters

A slope that is steady on a distance–time graph means no force; a slope that is steady on a speed–time graph means a steady force. Always read the axis labels first.

Question 16 · A block pulled both ways (Fig 3.17)

16. നിശ്ചലാവസ്ഥയിലുള്ള ഒരു വസ്തുവിൽ ബലം പ്രയോഗിക്കുന്നത് ചിത്രീകരിച്ചിരിക്കുന്നു. ഈ വസ്തുവിന് ലഭിക്കുന്ന ത്വരണമെത്ര? 2 s കൊണ്ട് ഈ വസ്തുവിന് ഉണ്ടാകുന്ന സ്ഥാനാന്തരമെത്ര? The figure shows forces applied on an object at rest (നിശ്ചലാവസ്ഥ): a 10 kg block pulled with 16 N to the right and 7 N to the left. What acceleration does the object get? What is its displacement (സ്ഥാനാന്തരം) in 2 s?

First the resultant force (right +): 16 + (−7). Then a = F/m. Starting from rest, s = ut + ½at² with u = 0.

🎬 Solution animation – exam layout

🧪 Try other numbers – pull both ways and let go

pull to the right (N) pull to the left (N) mass (kg) time t (s)

Answer: Take “to the right” as positive. Resultant F = 16 + (−7) = 9 N to the right.
a = F/m = 9/10 = 0.9 m/s² (to the right).
s = ut + ½at² = 0 × 2 + ½ × 0.9 × 2² = ½ × 0.9 × 4 = 1.8 m to the right.

⚠️ Common trap

Adding 16 + 7 = 23 N. The forces are in opposite directions – one must be taken negative. Also, s = vt does not work here: the velocity is not constant (it grows from 0 to 1.8 m/s).

Question 17 · Two blocks pushed together (Fig 3.18)

17. ചിത്രം നിരീക്ഷിക്കുക. A, B എന്നിവ 6 kg, 4 kg എന്നിങ്ങനെ മാസുള്ള രണ്ട് വസ്തുക്കളാണ്. തമ്മിൽ സ്പർശിച്ചിരിക്കുന്ന ഇവ ഘർഷണമില്ലാത്ത ഒരു പ്രതലത്തിലാണ് വച്ചിരിക്കുന്നതെങ്കിൽ ഇവയിൽ 15 N ബലം പ്രയോഗിച്ചാൽ B എന്ന വസ്തു A എന്ന വസ്തുവിൽ പ്രയോഗിക്കുന്ന ബലം കണക്കാക്കുക. Observe the figure. A and B are two objects of mass 6 kg and 4 kg. They touch each other and rest on a frictionless surface. If a force of 15 N is applied on them (on A, as shown), calculate the force that object B exerts on object A.

The two blocks move together with one acceleration: a = 15/(6 + 4). The only horizontal force on B is the push of A – so that push is m_B × a. Then use the third law for the force of B on A.

🎬 Solution animation – exam layout

🧪 Contact-force lab – push the pair

mass of A (kg) mass of B (kg) push F (N) push

The contact force is always (mass of the block that is not pushed) × a. Push the same pair from the other side and the contact force changes from 6 N to 9 N!

Answer: Take “to the right” (the direction of the 15 N push) as positive. On a frictionless surface the blocks move together:
a = F/(m_A + m_B) = 15/(6 + 4) = 1.5 m/s².
Force of A on B: F_AB = m_B × a = 4 × 1.5 = 6 N to the right.
By the third law, F_BA = −F_AB: B pushes A with 6 N to the left.
Check on A: net force = 15 − 6 = 9 N, and m_A × a = 6 × 1.5 = 9 N ✓.

⚠️ Common trap

Answering 15 N (“B pushes back with the whole force”). If B pushed A back with 15 N, A would have no net force and could not accelerate. The contact force only has to accelerate B: 4 × 1.5 = 6 N.

Extended activities തുടർപ്രവർത്തനങ്ങൾ

1. Seminar paper – overloading, overspeeding and road safety

വാഹനങ്ങളിലെ അമിതഭാരവും അമിതവേഗവും റോഡ് സുരക്ഷയെ എപ്രകാരം ബാധിക്കും എന്നതിനെക്കുറിച്ച് ഒരു സെമിനാർ പേപ്പർ തയ്യാറാക്കി അവതരിപ്പിക്കുക. Prepare and present a seminar paper on how overloading and overspeeding of vehicles affect road safety.

🛠️ Guidance card

🚚 Stopping-distance lab – a lorry with the same brakes

lorry + load (tonnes) speed (km/h)

Model: the brakes can supply at most 80,000 N; the driver needs 1 s to react. Allowed load: 16 tonnes in total. Braking distance = v²/2a with a = F/m.

The physics to explain

Momentum p = mv grows with both load and speed: a 24 t overloaded lorry at 40 km/h carries 1.5 times the momentum of a 16 t lorry.
Impulse: to stop, F × t = change of momentum. More momentum with the same brakes ⇒ longer stopping time and distance; in a crash, a far bigger force.
F = ma: the same brake force gives a smaller retardation to a larger mass.
Speed counts twice: braking distance = v²/2a – double the speed, four times the braking distance. The reaction distance (v × reaction time) also doubles.
Inertia of motion: passengers without seat belts and loose cargo keep moving forward when the vehicle stops suddenly; an overloaded vehicle also topples more easily on bends.

Suggested outline

Introduction – road accident news from Kerala; why trucks, buses and autorickshaws are involved.
Laws of motion behind it – inertia, momentum, F = ma, impulse (use the lab numbers above as a table).
Examples – an overloaded autorickshaw tipping on a turn, a sand lorry on a ghat road, a speeding KSRTC bus braking at a junction.
Rules – speed limits, load limits, seat belts and helmets, speed governors.
Conclusion and suggestions; list of sources.

⚠ Collect data from newspapers, the motor vehicles department or the police – never by riding or standing near speeding vehicles.

🌐 Explore online needs internet · opens in a new tab

📄 Road traffic injuries – fact sheet (WHO) WHO
Facts on how speeding raises crash risk and how seat belts and helmets save lives.
▶️ front seat belt vs rear seat belt | ഒരു അപകട വിശകലനം | MVD Kerala YouTube – MVD KERALA (official Kerala Motor Vehicles Dept. channel) · 1:35
Kerala Motor Vehicles Department's Malayalam analysis of a real crash and seat belts.
▶️ Crash Test - Belted vs Unbelted Passengers YouTube – Transport For NSW (Australian state govt.) · 0:27
A 40 km/h crash test: see what happens to passengers with and without seat belts.
▶️ No need to speed - explanimation video YouTube – Brake Charity · 3:10
Animation using physics to show why a little extra speed makes crashes much worse.

2. Science-club report – shock absorbers and the discs of the spine

വാഹനങ്ങളിലെ ഷോക്ക് അബ്സോർബർ, നമ്മുടെ നട്ടെല്ലിലെ ഡിസ്കുകൾ എന്നിവയുടെ പ്രവർത്തനം വിശദീകരിക്കാൻ ആവേഗം എന്ന ആശയം എങ്ങനെ പ്രയോജനപ്പെടുത്താം എന്നതിനെക്കുറിച്ച് റിപ്പോർട്ട് തയ്യാറാക്കി സയൻസ് ക്ലബ്ബിൽ അവതരിപ്പിക്കുക. Prepare a report on how the idea of impulse (ആവേഗം) can be used to explain the working of shock absorbers in vehicles and the discs in our spine, and present it in the science club.

🛠️ Guidance card

🛞 Same impulse, longer time, smaller force

cushioning time t (s)

Key idea (impulse–momentum principle): F × t = mv − mu. A bump on the road (or landing after a jump) produces a fixed change of momentum; if it happens over a longer time, the force is smaller.
Shock absorber: a spring with an oil-filled damper between the wheel and the body. The spring lets the wheel move up slowly instead of jerking the body; the damper stops the bouncing. Example in the lab: a 40 kg wheel unit whose vertical velocity changes by 2 m/s ⇒ Δp = 80 kg m/s. In 0.01 s the force is 8000 N; spread over 0.2 s it is only 400 N.
Discs of the spine: soft, jelly-like cushions between the vertebrae. When we walk, run or jump, they squash slightly, increasing the time of each jolt and reducing the force on the bones and the brain.
Report structure: aim → the principle (I = Ft = Δp) → diagrams of the shock absorber and the spine → a worked example → other examples (helmets, airbags, sports shoes, sand pit) → conclusion. Tip: bend your knees when you land – you are using your own “shock absorbers”.

3. Seminar – the laws of motion in everyday life

ബലവുമായി ബന്ധപ്പെട്ട് ഈ യൂണിറ്റിൽ പഠിച്ച ആശയങ്ങൾ നിത്യ ജീവിതത്തിൽ പ്രയോജനപ്പെടുത്തിയിരിക്കുന്ന സന്ദർഭങ്ങളെക്കുറിച്ച് സെമിനാർ അവതരിപ്പിക്കുക. Present a seminar on situations in everyday life where the ideas about force learnt in this unit are put to use.

🛠️ Starter list – pick 8 to 10, explain each with a sketch

Situation	Idea used	One-line explanation
Seat belts in cars and buses	Inertia of motion (first law)	The belt supplies the force that stops the body when the vehicle stops suddenly.
Shaking a mango tree; beating a carpet	Inertia of rest	Branch or carpet moves suddenly; mangoes or dust tend to stay and separate.
Tug of war at Onam	Resultant of unbalanced forces	The rope moves towards the team whose force is bigger.
Catching a cricket ball with hands moving back	Impulse I = Ft = Δp	Longer stopping time ⇒ smaller force on the hands.
Airbags, helmets, crumple zones	Impulse	Increase the impact time, reduce the force on the body.
Sand pit for long jump, foam bed for pole vault	Impulse	Soft landing stretches the stopping time.
Packing glassware in straw or sponge	Impulse	Jolts are spread over a longer time ⇒ less breakage.
Hammering a nail	Impulsive force, momentum	A moving hammer's momentum is removed in a tiny time ⇒ a large force.
Speed and load limits for lorries	Momentum p = mv, F = ma	More momentum needs more force or more distance to stop.
Rowing a vallam in a boat race	Third law	Oars push water back; water pushes the boat forward.
Rocket launch from Sriharikota	Third law	Gases are pushed down; they push the rocket up.
Walking; why walking on wet sand is hard	Third law, friction	We push the ground back; the ground pushes us forward.
A gun's recoil; jumping from a boat	Third law	Bullet/person forward, gun/boat backward.
Bowling vs batting: heavier bat hits harder	Momentum, second law	Bigger mass × speed ⇒ bigger change of momentum given to the ball.

Seminar tip: for each situation say which law, show the force arrows, and give one number (e.g. “stopping time doubled ⇒ force halved”).

Check yourself

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