Laws of Motion Formula Sheet — JEE Main Physics

Every key Laws of Motion formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 15 sub-topics

Force and inertia
Newton's first law of motion
Momentum
Newton's second law of motion
Impulse
Newton's third law of motion
Law of conservation of linear momentum and its applications
Equilibrium of concurrent forces
Static and kinetic friction
Laws of friction
Rolling friction
Lubrication
Dynamics of uniform circular motion
Centripetal force
Examples of circular motion

Newton's Three Laws & Inertia

First law (law of inertia): If the net external force is zero, a body stays at rest or in uniform motion:

F_{n e t} = 0 \Rightarrow a = 0

Inertia =

resistance to change of state; mass measures it.

Frames of reference

▸ $Inertial$ frame: at rest or uniform velocity — Newton's laws hold.
▸ $Non-inertial$ (accelerating) frame: laws need a pseudo-force (page 4).

Second law

F = \frac{d p}{d t} = m a, p = m v

F =

net external force (N),

p =

momentum,

m =

mass (constant),

a =

acceleration;

1 N = 1 kg m s^{- 2}

About the 2nd law

▸Vector law — one equation per axis: $F_{x} = m a_{x}$ , $F_{y} = m a_{y}$ .
▸Only $external$ forces count; $a$ is that of the centre of mass for a system.
▸ $Local$ : $a$ now depends on $F$ now, not on history.

Comparative: the three laws

Law	States	Gives
1st (inertia)	$F_{n e t} = 0 \Rightarrow a = 0$	defines inertial frame & force
2nd	$F = d p / d t = m a$	magnitude of force / accn
3rd	$F_{A B} = - F_{B A}$	forces in equal-opposite pairs

Third law

F_{A B} = - F_{B A}

F_{A B} =

force on A by B; to every action an equal & opposite reaction

About the 3rd law

▸Action & reaction act on $different$ bodies — never cancel on one body.
▸They act at the same instant (no cause–effect order).
▸Internal forces of a system cancel in pairs $\Rightarrow$ sum to zero.

★ Remember · Aristotle's fallacy

A force is

not

needed to keep a body in uniform motion — only to overcome friction. Rest and uniform motion are equivalent states (both

a = 0

🚫 Examiner Trap · Newton's laws & inertia

(1) Action–reaction act on

two different

bodies, so they can NEVER cancel each other — don't pair N with mg on the same block (those are a 2nd-law balance, not a 3rd-law pair). (2)

F = m a

uses the

net

force; sum all forces first. (3)

F = d p / d t = m a

only when m is constant — for variable mass use

d p / d t

. (4) Newton's laws fail in an accelerating frame unless you add a pseudo-force.

⚡ Shortcut · Spotting an action–reaction pair

Swap the two bodies in the phrase 'force on A by B'

\to

'force on B by A'; same type, same line, equal magnitude, opposite sense,

different

body. If both forces act on the same body, it is NOT a 3rd-law pair.

Impulse, Momentum & Conservation

Linear momentum:

p = m v

— a vector, the 'quantity of motion'. Same force for the same time gives the same

Δ p

to any mass.

Impulse–momentum theorem

J = \int F d t = F_{a v} Δ t = Δ p

J =

impulse (N s)

=

area under the F-t curve,

F_{a v} =

average force,

Δ p =

change in momentum

★ Remember · Impulsive force

A very large force for a very short time (bat–ball, collision). Increase

Δ t

(bend knees, draw hands back) to cut the force for a given

Δ p

Conservation of momentum

F_{e x t} = 0 \Rightarrow p = const; p_{A} + p_{B} = p_{A}^{'} + p_{B}^{'}

F_{e x t} =

net external force; primes

=

after interaction; follows from 2nd

+

3rd laws, holds for elastic or inelastic

Comparative: elastic vs inelastic (1-D)

	Elastic	Perfectly inelastic
$e$	$1$	$0$
$p$	conserved	conserved
KE	conserved	max loss (stick together)
After	separate	common velocity $v$

Applications

▸ $Recoil$ (gun–bullet): $m_{g} v_{g} = - m_{b} v_{b}$ .
▸ $Machine gun$ firing n bullets/s of mass m at speed v: force on the gunner $F = nmv$ (rate of momentum given out).
▸ $Explosion$ : fragments' momenta add to the initial (often $0$ ).

Variable mass (rocket)

F_{t h r u s t} = v_{re l} \frac{d m}{d t}; v = u + v_{re l} ln \frac{m _{0}}{m}

v_{re l} =

exhaust speed relative to rocket,

m_{0} =

initial mass,

m =

instantaneous mass,

u =

initial speed; the

ln

result neglects gravity

Centre of mass (two particles)

x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}; v_{c m} = \frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}}

F_{e x t} = M a_{c m}

with

M = m_{1} + m_{2}

; if

F_{e x t} = 0

v_{c m}

const

🎯 Exam · Coefficient of restitution

e = \frac{speed of separation}{speed of approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

(

e = 1

elastic,

0

perfectly inelastic). Energy analysis of collisions

\to

Work-Energy unit.

🚫 Examiner Trap · Momentum & impulse

(1) Momentum is a

vector

— conserve it component-wise; in 2-D explosions use

p_{x}

and

p_{y}

separately. (2) Momentum is conserved in

every

collision (elastic OR inelastic); only KE is lost in inelastic ones — don't say

p

is lost. (3) Impulse

= Δ p

, so a ball bouncing back has

Δ p = m (v + u)

(reversal), not

m (v - u)

. (4)

F_{e x t} = 0

is required for conservation — gravity/normal during a long collision are usually negligible only over the SHORT impact time.

⚡ Shortcut · Perfectly inelastic speed

Two bodies stick:

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

(

= v_{c m}

). KE lost

= \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}

(reduced mass

μ

) — the maximum loss among all e.

Force Analysis: FBD & Equilibrium

Common forces

▸ $Weight$ mg — vertically down.
▸ $Normal$ N — $⊥$ to the contact surface (a push, never a pull).
▸ $Tension$ T — along the string; same throughout a $massless$ string over a smooth pulley.
▸ $Spring$ $F = - k x$ (restoring, $\propto$ deformation).

Comparative: contact-force facts

Force	Direction	Key fact
Normal $N$	$⊥$ surface	push only; $N \neq = m g$ in general
Tension $T$	along string	pull only; uniform if massless
Spring $k x$	along spring	$F = - k x$ , restoring
Friction $f$	along surface	opposes relative slip

Equilibrium of a particle

Concurrent forces

\sum F = 0 \Rightarrow \sum F_{x} = 0, \sum F_{y} = 0

\sum F =

vector sum of all forces; the forces form a closed polygon (head-to-tail)

Lami's theorem (3 concurrent forces)

\frac{F _{1}}{sin α} = \frac{F _{2}}{sin β} = \frac{F _{3}}{sin γ}

α, β, γ

are the angles

opposite

F_{1}, F_{2}, F_{3}

; valid only for 3 concurrent coplanar forces in equilibrium

🎯 Exam · FBD method

(1) Isolate one body. (2) Draw

only

the external forces on it (weight, normal, tension, friction, applied). (3) Choose axes (along the motion / incline). (4) Apply

\sum F = ma

per axis.

Tips

▸On a $smooth$ incline of angle $θ$ : $a = g sin θ$ , $N = m g cos θ$ .
▸Massless string $\Rightarrow$ same tension at both ends; massless rod can also push.
▸Two blocks in contact (force F on $M_{1}$ touching $M_{2}$ ): contact force $N = \frac{M _{2} F}{M _{1} + M _{2}}$ .

★ Remember · Reading apparent forces

A weighing machine reads the

normal reaction

N; a spring balance reads the

tension

T — not necessarily mg.

🚫 Examiner Trap · FBD & equilibrium

(1) Never put a force on a body that another body exerts on a THIRD body — draw only what touches THIS body. (2)

N = m g

only on a horizontal surface with no vertical applied force; on an incline

N = m g cos θ

, and a downward push raises N. (3) Lami's angles are the ones

opposite

each force, not between them. (4) Tension is uniform only for a

massless

string over a

smooth

pulley — a heavy or accelerating string has different tensions.

⚡ Shortcut · 3-force equilibrium

For exactly 3 concurrent forces, either close the triangle (sides

∥

forces) or use Lami. If two forces are equal and the third bisects their angle, the system is symmetric — exploit it.

Connected Bodies, Pulleys, Pseudo-force & Lift

Atwood machine (smooth pulley)

Atwood machine: masses m1 and m2 over a smooth pulley, tension T up and weight down on each — Two masses over a smooth, massless pulley.

Acceleration & tension

a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}; T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}

m_{2} > m_{1}

assumed;

a =

accn of each block,

T =

string tension,

g =

gravity

Block M on a rough table connected over a pulley to a hanging block m — Block on table $+$ hanging block.

Block on table

+

hanging block

a = \frac{m g - μ M g}{M + m}; T = \frac{M m g ( 1 + μ )}{M + m}

M =

table block,

m =

hanging block,

μ =

table friction coeff (

μ = 0

if smooth)

Constraint relations

▸Inextensible string: components of acceleration along the string are equal.
▸Movable (light) pulley: $a_{p u ll ey} = \frac{a _{1} + a _{2}}{2}$ ; tension on the support $= 2 T$ .
▸Differentiate the length equation twice to get the acceleration constraint.

More set-ups (frequently asked)

▸ $Heavy rope$ (mass M, length L) pulled by F on a smooth floor: tension at distance x from the pulled end $T = F (1 - \frac{x}{L})$ . Hanging rope: T at height y above the free end $= \frac{M}{L} g y$ .
▸ $Chain on a rough table$ : the largest fraction that can hang over the edge without sliding is $\frac{μ}{1 + μ}$ .
▸ $Block on a smooth wedge$ (angle $θ$ ): it stays at rest on the wedge if the wedge accelerates horizontally at $a = g tan θ$ (then $N = \frac{m g}{cos θ}$ ).

Non-inertial frame & pseudo-force

Pseudo-force

F_{p se u d o} = - m a_{0}

m =

body's mass,

a_{0} =

acceleration of the (non-inertial) frame; add it on every body in that frame

Mass in a lift: normal reaction N, weight mg, acceleration a; apparent weight — Apparent weight $N = m (g \pm a)$ .

Apparent weight in a lift

N = m (g + a)

up,

m (g - a)

down,

0

in free fall

N =

reading of the floor/balance,

a =

lift acceleration magnitude; free fall (

a = g

)

\Rightarrow

weightlessness

Comparative: apparent weight in a lift

Lift motion	$N$ reads	Felt
At rest / uniform $v$	$m g$	normal
Accel. up (or decel. down)	$m (g + a)$	heavier
Accel. down (or decel. up)	$m (g - a)$	lighter
Free fall $a = g$	$0$	weightless

🚫 Examiner Trap · Connected bodies & frames

(1) On a frictionless pulley a

single

string has ONE tension throughout — but the hanging and table blocks have the SAME acceleration magnitude, not the same forces. (2) A

movable

pulley splits/halves:

a_{p u ll ey} = (a_{1} + a_{2}) /2

and the supporting tension is

2 T

. (3) Pseudo-force

- m a_{0}

is added ONLY in the accelerating frame; never add it in the ground frame. (4) Apparent weight

\neq =

true weight:

N = m (g \pm a)

; deceleration flips the sign.

⚡ Shortcut · System acceleration

For bodies moving together,

a = \frac{net driving force}{total mass}

(treat as one block). Then isolate one body to get the internal tension/contact force. Two-block contact:

N = \frac{m _{ahead}}{M _{total}} F

, where

m_{ahead}

is the mass beyond the contact (the part being pushed).

Friction

Friction: Contact force parallel to the surface, opposing relative motion (or its tendency).

Static (self-adjusting)

f_{s} \leq μ_{s} N

f_{s} =

static friction,

μ_{s} =

static coeff,

N =

normal;

f_{s}

adjusts to balance the applied force up to

μ_{s} N

Kinetic

f_{k} = μ_{k} N, μ_{k} < μ_{s}

f_{k} =

kinetic friction,

μ_{k} =

kinetic coeff; constant once sliding, nearly speed-independent

FBD of a block resting on a rough incline: N perpendicular, friction f up the slope, mg down with components mg sin theta and mg cos theta — Block on a rough incline (angle $θ$ ), tending to slide down: friction acts $up$ the slope.

Laws of friction

▸ $f \propto N$ ; independent of the apparent area of contact and of speed.
▸ $Rolling$ friction $≪$ sliding $<$ limiting static — smallest of the three.

Comparative: static vs kinetic

	Static $f_{s}$	Kinetic $f_{k}$
Value	$\leq μ_{s} N$ (variable)	$μ_{k} N$ (fixed)
When	no slipping (yet)	sliding
Max	$μ_{s} N$ (limiting)	—
Order	larger ( $μ_{s} > μ_{k}$ )	smaller

Angle of friction / repose

tan λ = μ_{s}; tan θ_{re p} = μ_{s}

λ =

angle of friction,

θ_{re p} =

angle of repose

=

max incline before sliding; both equal

tan^{- 1} μ_{s}

Rough incline (angle

θ

)

a_{d o w n} = g (sin θ - μ cos θ); a_{u p} = - g (sin θ + μ cos θ)

a_{d o w n} =

accn sliding down,

a_{u p} =

retardation of a block sent up;

μ = μ_{k}

while sliding

Minimum pull on a horizontal surface

F_{m i n} = \frac{μ m g}{1 + μ ^{2}}

applied at angle

θ = tan^{- 1} μ

above the horizontal (this angle minimises the pull)

🎯 Exam · Block on block

For a block sliding on another, friction acts on

both

surfaces (equal & opposite); check whether they move together (compare required friction with

μ_{s} N

🚫 Examiner Trap · Friction

(1) Static friction is

\leq μ_{s} N

, NOT

= μ_{s} N

— it equals only the force needed, up to the limit. Using

μ_{s} N

when the body isn't about to slip is the #1 error. (2) Friction is independent of contact AREA. (3) On an incline, friction enters

N = m g cos θ

via

μ cos θ

, not

μ sin θ

. (4) Friction opposes

relative

slipping, so it can act

forward

(e.g. it drives a car / a block carried by a sheet). (5) Once moving, use

μ_{k}

(smaller), not

μ_{s}

⚡ Shortcut · Will it slide?

On an incline a block stays put iff

θ \leq θ_{re p} = tan^{- 1} μ_{s}

(i.e.

tan θ \leq μ_{s}

). To just hold/start it, the required friction

= m g sin θ

; compare with

μ_{s} m g cos θ

Circular Dynamics: Level, Banking & Conical

Centripetal force

F_{c} = \frac{m v ^{2}}{r} = m ω^{2} r

F_{c} =

net inward force,

v =

speed,

r =

radius,

ω =

angular speed; not a new force — provided by tension/gravity/friction/normal

Vehicle on a level road

Friction provides

F_{c}

\frac{m v ^{2}}{r} \leq μ_{s} m g \Rightarrow v_{m a x} = μ_{s} r g

v_{m a x} =

max safe speed on a flat road; independent of mass m

Banked road (angle

θ

)

Car resting on a banked road: normal N perpendicular to the bank, weight mg, horizontal direction towards the centre of the turn — Banking: $N sin θ$ points to the centre; $tan θ = \frac{v ^{2}}{r g}$ (no friction).

Optimum / maximum / minimum speed

v_{0} = r g tan θ;

v_{m a x} = r g \frac{μ _{s} + tan θ}{1 - μ _{s} tan θ};

v_{m i n} = r g \frac{tan θ - μ _{s}}{1 + μ _{s} tan θ}

v_{0}

: no friction needed;

v_{m a x}

: friction acts

down

the bank;

v_{m i n}

: friction acts

up

the bank (

v_{m i n} = 0

μ_{s} \geq tan θ

);

θ =

bank angle

Comparative: providing the centripetal force

Setup	Provided by	Limit / key
Level road	friction	$v_{m a x} = μ_{s} r g$
Banked (no $μ$ )	$N sin θ$	$tan θ = v^{2} / r g$
Conical pend.	$T sin θ$	$tan θ = v^{2} / r g$

Conical pendulum

Equations

T cos θ = m g; T sin θ = \frac{m v ^{2}}{r}

T =

string tension,

θ =

angle from vertical,

r =

circle radius; dividing gives

tan θ = v^{2} / r g

Angular speed & period

ω = \frac{g}{L cos θ}; T_{p} = 2 π \frac{L cos θ}{g}

L =

string length,

T_{p} =

period,

r = L sin θ

★ Remember · Key idea

Always resolve forces toward the centre (

= \frac{m v ^{2}}{r}

) and perpendicular to it (

= 0

). The centripetal direction is the acceleration direction.

🚫 Examiner Trap · Circular dynamics

(1) Centripetal force is NOT a new/extra force — never add

m v^{2} / r

as a separate arrow; it is the

net inward

component of real forces. (2) 'Centrifugal force' exists

only

in the rotating (non-inertial) frame as a pseudo-force; never use it in the ground frame. (3) Banking

tan θ = v^{2} / r g

assumes NO friction — it gives

v_{0}

, not

v_{m a x}

. (4)

v_{m a x}

and

v_{m i n}

on a banked road have friction acting down/up the slope respectively; sign of the

μ

term flips.

⚡ Shortcut · Banking & conical share

tan θ

Frictionless banked road and conical pendulum give the

same

relation

tan θ = \frac{v ^{2}}{r g} = \frac{ω ^{2} r}{g}

— both have a tilted normal/tension whose horizontal part is

m v^{2} / r

and vertical part is mg.

Vertical Circle

At angle

θ

from the lowest point

v^{2} = u^{2} - 2 g l (1 - cos θ); T = \frac{m v ^{2}}{l} + m g cos θ

u =

speed at the bottom,

v =

speed at angle

θ

l =

radius,

T =

string tension there

Vertical circle of radius l at the critical speed: at the top T = 0 and mg alone gives v = sqrt(gl); at the bottom T = 6mg and v = sqrt(5gl) — Just completing the loop (string): at the top $T = 0$ and gravity alone supplies $m v^{2} / l$ .

Just completes the loop (string)

v_{t o p} = g l, v_{s i d e} = 3 g l, v_{b o t} = 5 g l

v_{t o p}, v_{s i d e}, v_{b o t} =

minimum speeds at top/side/bottom; from energy conservation between points

Tensions at critical speed

▸ $T_{t o p} = 0$ , $T_{s i d e} = 3 m g$ , $T_{b o t} = 6 m g$ .
▸Condition to complete the loop: $u \geq 5 g l$ at the bottom.

Comparative: string vs rod (vertical circle)

	String / track	Light rod
Top condition	$v_{t o p} \geq g l$	$v_{t o p} \geq 0$
Min $u$ (bottom)	$5 g l$	$4 g l$
Support can	pull only	push or pull
If too slow	goes slack / leaves	still rigid

🎯 Exam · Rod vs string

light rod

can push, so the bob just reaches the top with

v_{t o p} = 0

(

u \geq 4 g l

). A

string

needs

v_{t o p} \geq g l

(else it goes slack and the bob leaves the circle).

2 g l < u < 5 g l

(string)

▸The string goes slack between the horizontal and the top — the bob becomes a projectile.
▸For $u \leq 2 g l$ : the bob oscillates (never crosses the horizontal).

🚫 Examiner Trap · Vertical circle

(1) At the top the minimum speed is

g l

for a STRING/track (where

T = 0

and gravity alone supplies

m v^{2} / l

), but

0

for a RIGID rod — don't use

g l

blindly. (2)

T_{b o t} - T_{t o p} = 6 m g

for any speed (energy + dynamics), a favourite result. (3) Speed at the top is NOT zero for a string — it must be

\geq g l

. (4) For

2 g l < u < 5 g l

on a string the bob doesn't oscillate AND doesn't loop — it leaves the circle as a projectile.

⚡ Shortcut · Top-to-bottom relations

At the critical (just-completes) case:

v_{b o t}^{2} = 5 g l

v_{s i d e}^{2} = 3 g l

v_{t o p}^{2} = g l

(differ by

2 g l

each, from

Δ h

). Tension always satisfies

T_{b o t} - T_{t o p} = 6 m g

regardless of u.

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