PhysicsCBSE Class 11 · Chapter 4

Laws of Motion Class 11: Notes, Friction, Banking and Solved Numericals

Updated 14 October 2026 · Fact-checked

Laws of Motion is Chapter 4 of CBSE Class 11 Physics. It covers inertia and Newton's first law, momentum and the second law, impulse, the third law, conservation of linear momentum, equilibrium of concurrent forces, static, kinetic and rolling friction, and the dynamics of circular motion on level and banked roads.

What Laws of Motion is about

Kinematics told you how things move; this chapter tells you why. Aristotle thought a force was needed to keep something moving. Galileo showed the opposite: an object keeps moving at constant velocity unless something (usually friction) stops it. This property is inertia, and Newton built it into his first law.

The second law says the net force equals the rate of change of momentum, $F = \frac{d p}{d t}$ , which becomes $F = m a$ for constant mass. A large force for a short time is described by its impulse, $F Δ t = Δ p$ . The third law says forces come in equal and opposite pairs acting on different bodies; together with the second law it gives the conservation of linear momentum, which explains the recoil of a gun and rockets.

You then apply the laws: equilibrium of a particle under several forces, friction (static, kinetic and rolling) and circular motion, where friction or a banked road supplies the centripetal force $\frac{m v ^{2}}{r}$ . The key skill is drawing a free-body diagram for each body. The old additional exercises have been dropped, but every idea here is used in Work, Energy and Power, rotation and gravitation, and it is core JEE and NEET material.

Weightage: This chapter is part of Unit III, Laws of Motion. The syllabus groups Unit III with Unit I, Physical World and Measurement and Unit II, Kinematics: together they carry 23 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit III Laws of Motion

Key concepts

Inertia and Newton's first law

NCERT section 4.2 to 4.4

Aristotle's fallacy was that a force is needed to keep a body moving. Galileo, using balls on double inclined planes, concluded that on a frictionless plane a ball would roll on for ever: a body keeps its state of motion unless an external force acts. This tendency is inertia, and mass is its measure.

Newton's first law: a body stays at rest, or keeps moving in a straight line at constant speed, until a net external force makes it change that state.

If the net external force is zero, the acceleration is zero (and the converse).
Inertia of rest: a passenger falls backwards when a bus starts suddenly. Inertia of motion: she lurches forward when it brakes. Inertia of direction: you are pushed outwards when a car turns sharply.
The first law also defines an inertial frame: one in which a body with no net force on it has no acceleration.

Worked example

Why do passengers in a moving bus fall forward when the driver brakes suddenly? [1 mark]

Try it first, then show the solutionSolution

The bus is stopped by friction at the tyres, but the passengers' upper bodies have no such force on them.
By inertia of motion they keep moving forward, so they lurch forward.

Answer: Their bodies tend to continue moving forward (inertia of motion) while the bus stops.

Momentum and Newton's second law

NCERT section 4.5

Linear momentum $p = m v$ is a vector (SI unit kg m s $^{- 1}$ ). The same force changes the velocity of a light body more than that of a heavy one, and stopping a fast or heavy body needs more force: momentum captures both ideas.

Second law: the rate of change of momentum of a body is proportional to the net external force and takes place in the direction of the force:
$F = \frac{d p}{d t} = m \frac{d v}{d t} = m a$ (for constant mass).

The unit of force is the newton: 1 N gives 1 kg an acceleration of 1 m s $^{- 2}$ .

It is a vector law: $F_{x} = m a_{x}$ , $F_{y} = m a_{y}$ , $F_{z} = m a_{z}$ separately.
It is a local law: the force at a point at an instant decides the acceleration there and then.
F is the net external force; internal forces of a system cancel.
If F = 0, a = 0: the first law is consistent with the second.

Worked example

A force acts on a 2 kg body and changes its velocity from 3 m s $^{- 1}$ to 7 m s $^{- 1}$ in 2 s, in a straight line. Find the force. [1 mark]

Try it first, then show the solutionSolution

$Δ p = 2 \times (7 - 3) = 8 k g m s^{- 1}$
$F = \frac{Δ p}{Δ t} = \frac{8}{2} = 4 N$

Answer: 4 N, in the direction of motion

Impulse

NCERT section 4.5 (impulse)

When a large force acts for a short time (a bat hitting a ball, a hammer on a nail), it is hard to measure the force itself but easy to measure the change of momentum. We define impulse:
$J = F Δ t = Δ p$ (for a varying force, the area under the F-t graph).

Impulse has the units of momentum (N s = kg m s $^{- 1}$ ).

Since $F = \frac{Δ p}{Δ t}$ , increasing the time of impact reduces the force for the same change in momentum. That is why:

a cricketer draws the hands back while catching a fast ball;
vehicles have crumple zones and airbags;
athletes land on sand or foam mats.

Worked example

A cricket ball of mass 0.15 kg moving at 20 m s $^{- 1}$ is hit straight back at 30 m s $^{- 1}$ . Find the impulse, and the average force if the contact lasts 0.01 s. [2 marks]

Try it first, then show the solutionSolution

Take the original direction as negative: $Δ p = 0.15 \times 30 - 0.15 \times (- 20) = 0.15 \times 50 = 7.5 N s$
$F = \frac{7.5}{0.01} = 750 N$

Answer: Impulse 7.5 N s (towards the bowler); average force 750 N

Newton's third law

NCERT section 4.6

Third law: to every action there is always an equal and opposite reaction. Forces always occur in pairs: if A pushes B with force $F_{A B}$ , then B pushes A with $F_{B A} = - F_{A B}$ .

Action and reaction act at the same instant; neither is the 'cause' of the other.
They act on different bodies, so they never cancel each other. (They cancel only when you consider the two bodies as one system, where they are internal forces.)
Examples: walking (you push the ground back, the ground pushes you forward), rowing, swimming, a rocket pushing gases back.

Bodies in contact: if a force F pushes two blocks $m_{1}$ and $m_{2}$ on a smooth floor, $a = \frac{F}{m _{1} + m _{2}}$ and the contact force on $m_{2}$ is $m_{2} a$ . By the third law, $m_{2}$ pushes back on $m_{1}$ with the same size.

Worked example

Two blocks of mass 3 kg and 2 kg are in contact on a smooth floor. A 10 N force pushes the 3 kg block. Find the acceleration and the force the blocks exert on each other. [2 marks]

Try it first, then show the solutionSolution

$a = \frac{10}{3 + 2} = 2 m s^{- 2}$
Force on the 2 kg block $= 2 \times 2 = 4 N$ forwards; the 2 kg block pushes the 3 kg block back with 4 N (third law)

Answer: 2 m s $^{- 2}$ ; contact force 4 N

Conservation of linear momentum

NCERT section 4.7

Law: the total momentum of an isolated system of interacting particles (no net external force) remains constant.

From the second and third laws: when A and B interact for time Δt, $F_{A B} Δ t = Δ p_{B}$ and $F_{B A} Δ t = Δ p_{A}$ . Since $F_{B A} = - F_{A B}$ , $Δ p_{A} + Δ p_{B} = 0$ : total momentum does not change.

Applications:

Recoil of a gun: before firing total momentum is zero, so $m_{b} v_{b} + m_{g} v_{g} = 0$ and $v_{g} = - \frac{m _{b} v _{b}}{m _{g}}$ .
Explosions: fragments fly apart with total momentum equal to that before (zero if at rest).
Collisions: total momentum is the same before and after, whatever happens to kinetic energy.
Rockets: exhaust gases are pushed back, the rocket gains forward momentum.

Worked example

A gun of mass 4 kg fires a bullet of mass 20 g at 400 m s $^{- 1}$ . Find the recoil speed of the gun. [2 marks]

Try it first, then show the solutionSolution

Initial momentum = 0, so $0.020 \times 400 + 4 \times v = 0$
$v = - \frac{8}{4} = - 2 m s^{- 1}$ (backwards)

Answer: 2 m s $^{- 1}$ backwards

Equilibrium of concurrent forces

NCERT section 4.8

A particle is in equilibrium when the net external force on it is zero: it stays at rest or moves at constant velocity.

Two forces: equal and opposite.
Three concurrent forces: $F_{1} + F_{2} + F_{3} = 0$ ; they can be represented by the sides of a closed triangle.
In general, resolve: $\sum F_{x} = 0$ , $\sum F_{y} = 0$ , $\sum F_{z} = 0$ .

Method: isolate the body, draw a free-body diagram with every force on it (weight, tension, normal reaction, friction, applied forces), choose axes and write the component equations.

Worked example

A 5 kg mass hangs from a rope. It is pulled aside by a horizontal force F until the rope makes 30° with the vertical. Find the tension and F. (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Vertical: $T cos 30° = m g = 50 N \Rightarrow T = \frac{50}{0.866} \approx 57.7 N$
Horizontal: $F = T sin 30° = 57.7 \times 0.5 \approx 28.9 N$
Check: $\frac{F}{m g} = tan 30°$ ✓

Answer: T ≈ 57.7 N, F ≈ 28.9 N

Friction: static, kinetic and rolling; laws and lubrication

NCERT section 4.9 and 4.9.1

Friction opposes relative motion (or impending relative motion) between surfaces in contact. It acts along the surface.

Static friction $f_{s}$ adjusts itself to the applied force up to a maximum, limiting friction: $f_{s} \leq (f_{s})_{ma x} = μ_{s} N$ .
Kinetic friction acts once sliding starts: $f_{k} = μ_{k} N$ , nearly independent of speed and of the area of contact. Usually $μ_{k} < μ_{s}$ , which is why it is harder to start a box moving than to keep it moving.
Laws: limiting friction is proportional to the normal reaction N, does not depend on the apparent area of contact, and depends on the nature of the surfaces.
Angle of repose: a block just begins to slide on an incline when $tan θ = μ_{s}$ .
Rolling friction is much smaller than sliding friction, which is why wheels and ball bearings help.
Lubrication (oil, grease, air cushions) reduces friction; friction is also useful: walking, brakes and tyres all need it.

Pulling vs pushing at an angle: pulling at angle θ above the horizontal lowers N to $m g - F sin θ$ , reducing friction; pushing at θ below raises it.

Worked example

A 4 kg box rests on a floor with $μ_{s} = 0.5$ and $μ_{k} = 0.4$ . Find the friction force when a horizontal push of (a) 15 N, (b) 25 N is applied, and the acceleration in (b). (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

N = 40 N, limiting friction $= 0.5 \times 40 = 20 N$
(a) 15 N < 20 N: no sliding; static friction = 15 N (just balances)
(b) 25 N > 20 N: it slides; $f_{k} = 0.4 \times 40 = 16 N$ , $a = \frac{25 - 16}{4} = 2.25 m s^{- 2}$

Answer: (a) 15 N, at rest (b) 16 N; 2.25 m s $^{- 2}$

Dynamics of circular motion: level and banked roads

NCERT section 4.10

A body in uniform circular motion needs a net force $\frac{m v ^{2}}{r}$ towards the centre: the centripetal force. It is not a new kind of force; tension, gravity, friction or a component of the normal reaction supplies it.

Car on a level road: static friction provides it: $\frac{m v ^{2}}{r} \leq μ_{s} m g$ , so $v_{ma x} = μ_{s} r g$ , independent of the mass.

Banked road (outer edge raised, angle θ): the horizontal component of the normal reaction helps.

Without friction: $N cos θ = m g$ , $N sin θ = \frac{m v ^{2}}{r}$ , giving the optimum speed $v_{0} = r g tan θ$ ; at this speed no friction is needed and tyre wear is least.
With friction (maximum speed, friction acting down the slope):

$v_{ma x} = r g \frac{μ _{s} + t a n θ}{1 - μ _{s} t a n θ}$

A cyclist on a level curve leans inwards at angle θ with $tan θ = \frac{v ^{2}}{r g}$ .

Worked example

What is the maximum speed at which a car can go round a level curve of radius 50 m if $μ_{s} = 0.4$ ? (g = 10 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$v_{ma x} = μ_{s} r g = 0.4 \times 50 \times 10 = 200$
$v_{ma x} \approx 14.1 m s^{- 1}$ (about 51 km h $^{- 1}$ )

Answer: About 14.1 m s $^{- 1}$

Physics in real life: Laws of Motion

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Impulse and landing

Suppose a 50 kg student jumps down from a wall 1.25 m high and lands either on a sand pit or on a concrete floor.

Her landing speed is 5 m s $^{- 1}$ (g = 10 m s $^{- 2}$ ). Find the average force on her if she comes to rest in 0.5 s on sand and in 0.05 s on concrete.

Try it first, then show the solutionSolution

Change in momentum $= 50 \times 5 = 250 N s$ in both cases
Sand: $F = \frac{250}{0.5} = 500 N$
Concrete: $F = \frac{250}{0.05} = 5000 N$ , ten times larger

Answer: 500 N on sand, 5000 N on concrete

Memory hook: Same momentum to lose; stretch the time, shrink the force.

Speed limit on a curve

Imagine a car taking an unbanked turn of radius 80 m on a rainy day when $μ_{s}$ between tyres and road has fallen to 0.4.

Suppose the driver goes round at 72 km h $^{- 1}$ . Will the car skid? (g = 10 m s $^{- 2}$ )

Try it first, then show the solutionSolution

$v_{ma x} = 0.4 \times 80 \times 10 = 320 \approx 17.9 m s^{- 1}$
72 km h $^{- 1}$ = 20 m s $^{- 1}$ , which is more than 17.9 m s $^{- 1}$
Friction cannot supply the needed centripetal force, so the car skids outwards

Answer: Yes; the safe limit is about 17.9 m s $^{- 1}$ (64 km h $^{- 1}$ )

Memory hook: Wet road, smaller μ, lower safe speed.

Conservation of momentum

Suppose a 60 kg boy standing in a 40 kg boat at rest on still water jumps horizontally on to the bank at 2 m s $^{- 1}$ . Ignore water resistance.

With what speed does the boat move, and in which direction?

Try it first, then show the solutionSolution

Total momentum before = 0
$60 \times 2 + 40 \times v = 0 \Rightarrow v = - 3 m s^{- 1}$

Answer: 3 m s $^{- 1}$ , away from the bank

Memory hook: Push forward, get pushed back: the lighter body moves faster.

Topics in this chapter

Laws of Motion formula sheet

Momentum: $p = m v$
Second law: $F = \frac{d p}{d t} = m a$
Impulse: $J = F Δ t = Δ p$
Recoil of a gun: $m_{b} v_{b} + m_{g} v_{g} = 0 \Rightarrow v_{g} = - \frac{m _{b} v _{b}}{m _{g}}$
Friction: $f_{s} \leq μ_{s} N, f_{k} = μ_{k} N, tan θ_{r e p ose} = μ_{s}$
Block sliding down a rough incline: $a = g (sin θ - μ_{k} cos θ)$
Apparent weight in a lift: $R = m (g + a) (accelerating up), R = m (g - a) (accelerating down)$
Atwood machine: $a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}$
Level road: $v_{ma x} = μ_{s} r g$
Banked road: $v_{0} = r g tan θ, v_{ma x} = r g \frac{μ _{s} + t a n θ}{1 - μ _{s} t a n θ}$

How the school exam asks Laws of Motion

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Explaining everyday effects	Why passengers jerk, why a cricketer pulls hands back, why a horse can pull a cart, why it is easier to pull than push a roller.	Name the law (inertia, impulse, third law, friction) and link it to the effect in two sentences.
Second law and impulse numericals	Force from change in momentum; impulse from an F-t graph; force when p(t) is given.	Watch signs when a ball reverses; impulse = area under F-t; differentiate p(t).
Conservation of momentum	Recoil, explosions into fragments, a person jumping off a boat; proof from Newton's laws.	Write total momentum before = after, with a sign convention.
Connected bodies and lifts	Blocks in contact or joined by strings; Atwood machine; reading of a weighing scale in a lift.	Free-body diagram for each body; same acceleration for connected bodies; normal reaction = apparent weight.
Friction	Static vs kinetic friction for a given push; motion on a rough incline; angle of repose; minimum force.	First compare the applied force with $μ_{s} N$ ; only if it is larger, use $μ_{k} N$ .
Circular motion	Derive the speed limits on level and banked roads; numericals on curves and cyclists.	Resolve N and f along the horizontal (towards the centre) and vertical; set the horizontal sum equal to $m v^{2} / r$ .

Tricks to remember Laws of Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

1st: lazy, 2nd: measured, 3rd: paired
First law: bodies are lazy (inertia). Second law: how much they change ( $F = d p / d t$ ). Third law: forces come in pairs on different bodies.
Use it when: Stating the three laws.
Different bodies, never cancel
Action-reaction pairs act on two different objects, so you never add them in one free-body diagram.
Use it when: Horse-cart and assertion-reason questions.
Lift: up-accelerating feels heavy
Accelerating up (or slowing while going down): R = m(g + a). Accelerating down: R = m(g − a). Free fall: R = 0.
Use it when: Weighing-scale questions.
Check static first
Applied force less than $μ_{s} N$ ? Friction equals the applied force. More? Switch to $μ_{k} N$ .
Use it when: Every friction numerical.
Banking: tan theta equals v² over rg
Same formula for the banking angle with no friction and for a cyclist's lean.
Use it when: Circular motion numericals.

Laws of Motion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Lurching passengers · 1 mark · MCQExam level

When a moving bus stops suddenly, passengers lurch forward. This is explained by:
(a) inertia of rest
(b) inertia of motion
(c) Newton's third law
(d) conservation of energy

Try it first, then show the solutionSolution

Their bodies tend to keep moving forward as the bus stops.

Answer: (b) inertia of motion

Example 2: Do action and reaction cancel? · 1 mark · Assertion-reasonExam level

Assertion (A): Action and reaction forces do not cancel each other.
Reason (R): Action and reaction act on different bodies.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Forces cancel only when they act on the same body. R gives exactly that reason.

Answer: (a)

Example 3: Weighing scale in a lift · 2 marks · Very short answerHarder exam level

A 60 kg man stands on a weighing scale in a lift. What does the scale read (in newtons) when the lift (a) accelerates upwards at 2 m s $^{- 2}$ , (b) accelerates downwards at 2 m s $^{- 2}$ , (c) falls freely? (g = 10 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

(a) $R = m (g + a) = 60 \times 12 = 720 N$
(b) $R = m (g - a) = 60 \times 8 = 480 N$ ; (c) $R = m (g - g) = 0$

Answer: (a) 720 N (b) 480 N (c) 0

Example 4: Block down a rough incline · 3 marks · Short answerHarder exam level

A 2 kg block slides down a 30° incline with $μ_{k} = 0.2$ . Find its acceleration and the friction force. (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

$N = m g cos 30° = 20 \times 0.866 = 17.3 N$ ; $f_{k} = 0.2 \times 17.3 \approx 3.46 N$
Along the slope: $m g sin 30° - f_{k} = ma \Rightarrow 10 - 3.46 = 2 a$
$a \approx 3.27 m s^{- 2}$ (check: $g (sin 30° - 0.2 cos 30°) = 3.27$ )

Answer: a ≈ 3.27 m s $^{- 2}$ ; friction ≈ 3.5 N up the slope

Example 5: Three blocks pulled by a string · 3 marks · Short answerHarder exam level

Blocks A (2 kg), B (3 kg) and C (5 kg) lie in a line on a smooth table, joined by light strings A-B and B-C. A force of 30 N pulls C horizontally away from the others. Find the acceleration and the tensions in both strings. [3 marks]

Try it first, then show the solutionSolution

Whole system: $a = \frac{30}{2 + 3 + 5} = 3 m s^{- 2}$
String A-B pulls only A: $T_{1} = 2 \times 3 = 6 N$
String B-C pulls A and B: $T_{2} = (2 + 3) \times 3 = 15 N$

Answer: 3 m s $^{- 2}$ ; 6 N and 15 N

Example 6: Banked road · 5 marks · Long answerHarder exam level

(a) Derive the optimum speed for a car on a frictionless road of radius r banked at angle θ. (b) A circular track of radius 60 m is banked at 20°. Find the optimum speed. (c) Explain what happens if the car goes faster than this on a wet (nearly frictionless) track. (d) Why is the optimum speed independent of the mass of the car? (g = 10 m s $^{- 2}$ , tan 20° = 0.364) [5 marks]

Try it first, then show the solutionSolution

(a) Only weight mg and normal reaction N act. Vertical: $N cos θ = m g$
Horizontal (towards the centre): $N sin θ = \frac{m v ^{2}}{r}$
Dividing: $tan θ = \frac{v ^{2}}{r g} \Rightarrow v_{0} = r g tan θ$
(b) $v_{0} = 60 \times 10 \times 0.364 = 218.4 \approx 14.8 m s^{- 1}$
(c) The horizontal part of N is too small to supply $m v^{2} / r$ ; with no friction to help, the car slides up and outwards. (d) m cancels in the division.

Answer: $v_{0} = r g tan θ$ ; about 14.8 m s $^{- 1}$ ; it slides outward; mass cancels

Example 7: Impulse from a force-time graph · 2 marks · Very short answerHOTS / competency

A 2 kg body at rest is acted on by a force that rises uniformly from 0 to 20 N in 2 s and then falls uniformly to 0 in the next 2 s. Find the impulse and the final speed of the body. [2 marks]

Try it first, then show the solutionSolution

Impulse = area of the triangle $= \frac{1}{2} \times 4 \times 20 = 40 N s$
$Δ p = 40 \Rightarrow v = \frac{40}{2} = 20 m s^{- 1}$

Answer: 40 N s; 20 m s $^{- 1}$

Example 8: Holding a block against a wall · 3 marks · Short answerHOTS / competency

A 2 kg block is pressed against a vertical wall by a horizontal force F. The coefficient of static friction is 0.5. Find the least F that stops the block from slipping down. If F is doubled, what is the friction force? (g = 10 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

Normal reaction N = F (horizontal balance); friction acts upwards and must balance mg = 20 N
Need $μ_{s} F \geq 20 \Rightarrow F \geq \frac{20}{0.5} = 40 N$
If F = 80 N, limiting friction is 40 N, but static friction only needs to be 20 N: it stays 20 N

Answer: 40 N; still 20 N (static friction just balances the weight)

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Weighing yourself in a lift

Suppose a 50 kg girl stands on a bathroom scale (marked in newtons) in a lift of a tall building. Take g = 10 m s $^{- 2}$ .

(i) · 1 mark
What does the scale read when the lift starts upwards with an acceleration of 1.5 m s $^{- 2}$ ?
Answer
Answer: 575 N
$50 \times (10 + 1.5)$
(ii) · 1 mark
What does it read when the lift moves up at constant speed?
Answer
Answer: 500 N
a = 0, so R = mg
(iii) · 1 mark
What does it read when the lift, still moving up, slows down at 1.5 m s $^{- 2}$ ?
Answer
Answer: 425 N
Acceleration is downwards: $50 \times (10 - 1.5)$
(iv) · 1 mark
What would it read if the cable broke?
Answer
Answer: 0
Free fall: R = m(g − g)

Friction on a ramp

Suppose a 10 kg crate rests on a loading ramp inclined at 30°, with $μ_{s} = 0.3$ (take $cos 30° = 0.866$ , g = 10 m s $^{- 2}$ ).

(i) · 1 mark
Find the normal reaction on the crate.
Answer
Answer: About 86.6 N
$m g cos 30°$
(ii) · 1 mark
Will the crate slide down on its own? Explain.
Answer
Answer: Yes
Down-slope pull $m g sin 30° = 50 N$ exceeds the limiting friction $0.3 \times 86.6 \approx 26 N$ (also tan 30° = 0.58 > 0.3).
(iii) · 1 mark
Find the least force along the ramp to just start pushing it up.
Answer
Answer: About 76 N
$m g sin 30° + μ_{s} m g cos 30° = 50 + 26$
(iv) · 1 mark
What is the steepest angle at which the crate would stay at rest?
Answer
Answer: About 16.7°
Angle of repose: $tan θ = 0.3$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A ball of mass 0.2 kg moving at 10 m s $^{- 1}$ hits a wall and rebounds at 8 m s $^{- 1}$ . Find the impulse on the ball. [2 marks] (2 marks)

Step	Marks
Take the rebound direction as positive: $Δ p = 0.2 \times 8 - 0.2 \times (- 10)$	1
$Δ p = 1.6 + 2.0 = 3.6 N s$ , directed away from the wall	1

Using Newton's second and third laws, prove the law of conservation of linear momentum for two colliding bodies. [3 marks] (3 marks)

Step	Marks
During contact time Δt, A exerts $F_{A B}$ on B and B exerts $F_{B A}$ on A; by the third law $F_{B A} = - F_{A B}$	1
By the second law, $Δ p_{B} = F_{A B} Δ t$ and $Δ p_{A} = F_{B A} Δ t$	1
Adding: $Δ p_{A} + Δ p_{B} = 0$ , so $p_{A} + p_{B}$ is constant	1

(a) A car goes round a road of radius r banked at angle θ, with coefficient of static friction $μ_{s}$ . Draw the free-body diagram and derive the maximum safe speed. (b) Find the optimum speed and the maximum speed for r = 100 m, $tan θ = 0.25$ , $μ_{s} = 0.2$ . (g = 10 m s $^{- 2}$ ) [5 marks] (5 marks)

Step	Marks
(a) Forces: weight mg down, N perpendicular to the road, friction f down the slope (at maximum speed)	1
Vertical: $N cos θ = m g + f sin θ$ ; horizontal: $N sin θ + f cos θ = \frac{m v ^{2}}{r}$ , with $f = μ_{s} N$	1
Dividing: $v_{ma x}^{2} = r g \frac{μ _{s} + t a n θ}{1 - μ _{s} t a n θ}$	1
(b) $v_{0} = r g tan θ = 100 \times 10 \times 0.25 \approx 15.8 m s^{- 1}$	1
$v_{ma x} = 1000 \times \frac{0.45}{0.95} \approx 21.8 m s^{- 1}$	1

Where students lose marks in Laws of Motion

The slips examiners see most often, and what to write instead.

Mistake: Taking kinetic friction even when the body does not move
Using $μ N$ automatically.
Wrong working
Push 15 N, $μ_{s} N$ = 20 N: friction = 20 N, so the box moves backwards
Right working
Friction = 15 N (static friction only balances the push)
Do this: Static friction is self-adjusting up to $μ_{s} N$ .
Usually costs: 1-2 marks.
Mistake: Forgetting the sign change when a ball rebounds
Subtracting speeds.
Wrong working
Δp = 0.15(30 − 20) = 1.5 N s
Right working
Δp = 0.15(30 − (−20)) = 7.5 N s
Do this: Choose a positive direction and give both velocities signs.
Usually costs: 1 mark.
Mistake: Writing centripetal force as an extra force on the free-body diagram
Treating it as a new type of force.
Do this: Draw only real forces (weight, N, friction, tension); their resultant towards the centre equals $m v^{2} / r$ .
Usually costs: 1 mark in derivations.
Mistake: Saying action and reaction cancel
Equal and opposite sounds like cancelling.
Do this: They act on different bodies, so they cannot cancel.
Usually costs: 1 mark.
Mistake: Using N = mg on an incline or with an angled force
Habit from horizontal surfaces.
Do this: On an incline N = mg cos θ; with a pull F at angle θ above horizontal, N = mg − F sin θ.
Usually costs: Whole numerical.

NCERT exercise guide: Laws of Motion

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 4.1-4.4 (net force and free-body thinking)

What it tests: Net force on raindrops, pebbles, stones in moving trains and particles on strings in different situations.

How to approach it: Identify every real force; if the velocity is constant the net force is zero; after release only gravity acts.

Same kind as Exercise 4.2

A ball of mass 0.1 kg is thrown vertically upwards. What is the net force on it (a) on the way up, (b) at the highest point, (c) on the way down? Ignore air resistance; g = 10 m s $^{- 2}$ .

Show the solutionSolution

Once released, the only force is its weight: $m g = 0.1 \times 10 = 1 N$ downwards.
This is true on the way up, at the top (where v = 0 but a = g) and on the way down.

Answer: 1 N vertically downwards in all three cases

Exercises 4.5-4.16 (second law, lifts and connected bodies)

What it tests: Retarding forces, braking, rockets, a man on a scale in a lift, F from x-t graphs, blocks joined by strings and masses over pulleys.

How to approach it: Free-body diagram for each body; $\sum F = ma$ along the motion; connected bodies share a and the string tension.

Same kind as Exercise 4.16

Masses of 5 kg and 3 kg hang at the two ends of a light string over a smooth pulley. Find their acceleration and the tension in the string. (g = 10 m s $^{- 2}$ )

Show the solutionSolution

5 kg: $50 - T = 5 a$ ; 3 kg: $T - 30 = 3 a$
Adding: $20 = 8 a \Rightarrow a = 2.5 m s^{- 2}$
$T = 30 + 3 \times 2.5 = 37.5 N$

Answer: a = 2.5 m s $^{- 2}$ ; T = 37.5 N

Exercises 4.17-4.23 (momentum, impulse and circular motion)

What it tests: Disintegration of a nucleus, billiard-ball impulse, gun recoil, a ball deflected by a bat, a stone whirled on a string, and explanations of everyday effects.

How to approach it: Momentum conservation for isolated systems; impulse = Δp as a vector; tension supplies $m v^{2} / r$ ; once the string breaks the stone moves along the tangent.

Same kind as Exercise 4.19

A gun of mass 5 kg fires a shell of 50 g with a muzzle speed of 300 m s $^{- 1}$ . Find the recoil speed of the gun.

Show the solutionSolution

$0.05 \times 300 + 5 v = 0$
$v = - 3 m s^{- 1}$

Answer: 3 m s $^{- 1}$ backwards

Practice questions on Laws of Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The SI unit of impulse is the same as that of:
(a) force
(b) momentum
(c) energy
(d) power
Hint
J = Δp.
Answer
(b) momentum
Q2 · 1 mark · MCQExam level
A block just begins to slide when an incline is raised to angle θ. The coefficient of static friction is:
(a) sin θ
(b) cos θ
(c) tan θ
(d) cot θ
Hint
Angle of repose.
Answer
(c) tan θ
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): It is easier to pull a lawn roller than to push it.
Reason (R): Pulling at an angle above the horizontal reduces the normal reaction and so the friction.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
N = mg − F sin θ when pulling.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Why does a cricketer move the hands backwards while catching a fast ball?
Hint
F = Δp/Δt.
Answer
The ball's change in momentum is fixed. Moving the hands back increases the time over which it is stopped, so the average force on the hands ( $Δ p /Δ t$ ) is smaller and the catch hurts less.
Q5 · 2 marks · Very short answerHarder exam level
The momentum of a body varies with time as $p = 2 t^{2} + 3 t$ (SI units). Find the force on it at t = 2 s.
Hint
F = dp/dt.
Answer
$F = 4 t + 3 = 11 N$ at t = 2 s.
Q6 · 2 marks · Very short answerHarder exam level
A shell of mass 3 kg at rest explodes into two pieces of 1 kg and 2 kg. The 1 kg piece flies off at 30 m s $^{- 1}$ . Find the velocity of the other piece.
Hint
Total momentum stays zero.
Answer
$1 \times 30 + 2 v = 0 \Rightarrow v = - 15$ m s $^{- 1}$ : 15 m s $^{- 1}$ in the opposite direction.
Q7 · 2 marks · Very short answerHarder exam level
Find the maximum speed with which a car can safely take a level curve of radius 45 m if $μ_{s} = 0.5$ . Does it depend on the car's mass? (g = 10 m s $^{- 2}$ )
Hint
$v_{ma x} = μ_{s} r g$ .
Answer
$v_{ma x} = 0.5 \times 45 \times 10 = 225 = 15$ m s $^{- 1}$ (54 km h $^{- 1}$ ). No: m cancels.
Q8 · 2 marks · Very short answerHarder exam level
A cyclist goes round a level circular track of radius 20 m at 10 m s $^{- 1}$ . At what angle to the vertical must he lean? (g = 10 m s $^{- 2}$ )
Hint
$tan θ = v^{2} / r g$ .
Answer
$tan θ = \frac{100}{200} = 0.5$ , θ ≈ 26.6°.
Q9 · 3 marks · Short answerHarder exam level
A 4 kg block on a horizontal table is joined by a string over a smooth pulley at the edge to a 1 kg block hanging freely. The coefficient of kinetic friction between the 4 kg block and the table is 0.1. Find the acceleration and the tension. (g = 10 m s $^{- 2}$ )
Hint
Friction on the table block is $0.1 \times 40$ N.
Answer
Hanging: $10 - T = 1 \times a$ ; table: $T - 4 = 4 a$ . Adding: $6 = 5 a$ , a = 1.2 m s $^{- 2}$ ; T = 10 − 1.2 = 8.8 N.
Q10 · 3 marks · Short answerHOTS / competency
Explain, using Newton's laws, why a person standing in a bus is thrown outwards when the bus takes a sharp turn, and what supplies the centripetal force to the bus itself.
Hint
Inertia of direction; friction at the tyres.
Answer
The person's body tends to keep moving in a straight line (inertia of direction), while the bus turns towards the centre; relative to the bus the person seems pushed outwards until friction at the feet or a handrail provides the inward force. The bus's centripetal force is supplied by sideways static friction between the tyres and the road (and by banking, if any).
Q11 · 2 marks · Very short answerHOTS / competency
A machine gun fires 10 bullets per second, each of mass 40 g, at 500 m s $^{- 1}$ . What average force must the gunner apply to hold the gun steady?
Hint
Force = momentum given per second.
Answer
Momentum per bullet = 0.04 × 500 = 20 N s; per second = 10 × 20 = 200 N s, so the average force is 200 N.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Newton's first law.Show answer
A body stays at rest or in uniform motion in a straight line unless a net external force acts on it.
2.State Newton's second law in terms of momentum.Show answer
The rate of change of momentum equals the net external force and is in its direction: $F = d p / d t$ .
3.Define impulse and give its unit.Show answer
Impulse = force × time = change in momentum; unit N s.
4.State the law of conservation of linear momentum.Show answer
The total momentum of an isolated system remains constant.
5.What is the angle of repose?Show answer
The angle of an incline at which a body just starts to slide; $tan θ = μ_{s}$ .
6.Why is rolling easier than sliding?Show answer
Rolling friction is much smaller than sliding friction.
7.Write the optimum speed on a banked road.Show answer
$v_{0} = r g tan θ$ .
8.What does a weighing scale read in a freely falling lift?Show answer
Zero.

Best order to study it

Read Aristotle's fallacy and Galileo's experiments; learn inertia and the first law.
Learn momentum and the second law, with its vector form and units.
Learn impulse and its everyday uses.
Learn the third law and practise free-body diagrams for blocks, strings and lifts.
Prove conservation of momentum and solve recoil and explosion problems.
Practise equilibrium of concurrent forces by resolving.
Learn static, kinetic and rolling friction and solve incline problems.
Finish with level and banked roads, deriving the speed limits.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state all three laws and give an example of each?
Can you explain why a cricketer moves the hands back while catching?
Can you find impulse from an F-t graph?
Can you derive conservation of momentum from the second and third laws?
Can you draw free-body diagrams for an Atwood machine and find a and T?
Can you find the reading of a scale in a lift for all cases?
Can you decide whether a box moves for a given push, and find its acceleration?
Can you find the acceleration of a block on a rough incline?
Can you derive $v_{ma x}$ for a level road and for a banked road with friction?
Can you explain why it is easier to pull a lawn roller than to push it?

Laws of Motion: frequently asked questions

Which parts of this chapter were removed for 2026-27?

The main text is intact. The rationalisation removed the old additional exercises (5.24 to 5.40 of the old book). Everything in sections 4.1 to 4.11 of the current NCERT book is in scope, including rolling friction and lubrication.

Is centripetal force a real force?

It is the name for the net force towards the centre. It is always supplied by a real force such as tension, friction, gravity or a component of the normal reaction.

Why does a horse manage to pull a cart if the cart pulls back equally?

The equal and opposite forces act on different bodies. The horse moves forward because the ground pushes it forward (reaction to its push on the ground) more than the cart pulls it back; the cart moves because the horse's pull exceeds friction on the cart.

Is static friction always equal to $\mu_sN$?

No. \mu_sN is only its maximum (limiting) value. Below that, static friction is exactly as large as needed to prevent sliding.

How is this chapter tested in Class 11?

In the school annual exam through MCQs and assertion-reason on the laws, short numericals on impulse, recoil, lifts and friction, and long answers on banked roads or connected bodies. It is also among the most important mechanics chapters for JEE and NEET.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part I, Textbook for Class XI, Chapter 4 (Reprint 2026-27)