Class 11 Physics · NCERT notes for NEET

Laws of Motion: NCERT Notes, Formulas and Tricks for NEET

Newton's three laws, momentum and impulse, friction and the dynamics of circular motion. It is the base of all mechanics; NEET regularly asks on lifts, connected blocks, friction on inclines and banked roads, and the same ideas return in work-energy and rotation.

• 1. Newton's laws and momentum
• 2. Free-body diagrams, connected bodies, pulleys
• 3. Friction
• 4. Dynamics of circular motion
• 5. Impulse and conservation of momentum
• Quick revision

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Open the interactive version (simulations and tap-to-answer quick checks).

Topic 1

Newton's laws and momentum

A bus brakes suddenly and you lurch forward. Nothing pushed you: your body simply tried to keep moving. That is inertia.

• Galileo (inclined planes) showed that a body keeps moving with constant velocity if no net force acts. Friction, not a lack of force, stops a rolling ball. Aristotle's idea that motion needs a force is wrong.
• Inertia is the resistance of a body to a change in its state of rest or uniform straight-line motion. Mass is the measure of inertia.
• First law: zero net external force means zero acceleration. Rest and uniform velocity are physically equivalent.
• Momentum $\vec{p}=m\vec{v}$ is a vector, unit ${\text{kg m s}}^{-1}$. The same force for the same time gives the same $\Delta p$ to a light or a heavy body.
• Second law: $\vec{F}=d\vec{p}/dt$, which is $m\vec{a}$ for constant mass. It is a vector law, so it splits into $x$, $y$, $z$ equations.
• For a system, $\vec{F}$ is the total external force only; internal forces cancel. The law is local: $\vec{a}$ at an instant depends on $\vec{F}$ at that instant, not on past motion.
• Third law: ${\vec{F}}_{AB}=-{\vec{F}}_{BA}$. The two forces act on different bodies, at the same instant, with no cause and effect order.
• Equilibrium of a particle: $\sum \vec{F}=0$. Three concurrent forces in equilibrium form a closed triangle head to tail.
• Gravity acts at a distance. Normal reaction, tension, friction, spring force $F=-kx$ and fluid forces are contact forces.
• Newton's laws hold in inertial (non-accelerating) frames.

Formulas

• $\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}$
  Second law; $m\vec{a}$ form only for constant mass
• $F_x=m{a}_x,F_y=m{a}_y,F_z=m{a}_z$
  Second law in components
• ${\vec{F}}_{AB}=-{\vec{F}}_{BA}$
  Third law; the pair acts on two different bodies
• ${\vec{F}}_1+{\vec{F}}_2+{\vec{F}}_3=\vec{0}$
  Equilibrium under three concurrent forces
• $1\text{N}=1{\text{kg m s}}^{-2}=10^5\text{dyne}$
  Unit of force

Tips, tricks and hacks

Common mistakes

NEET traps

NEET option: 'the reaction acts after the action'. False: both act at the same instant, and either one can be called the action.

A stone whirled at constant speed in a circle, or a satellite in orbit, is NOT in equilibrium: its velocity direction changes, so there is a net force.

'Internal forces can accelerate a system' is false: you cannot move a car by pushing its dashboard from inside.

Topic 2

Free-body diagrams, connected bodies, pulleys

In a lift that starts going up, a bathroom scale under a 60 kg person can read about 72 kg. Your mass did not change; the push of the floor did.

• Free-body diagram (FBD): isolate one body, draw every external force ON it (weight, normal, tension, friction, applied), choose axes, then write $\sum F=ma$ along each axis.
• Never draw forces the body exerts on others. For a chosen system, internal forces are not drawn.
• Normal reaction is perpendicular to the contact surface. It equals $mg$ only on a level surface with no other vertical forces and no vertical acceleration. On an incline $N=mg\cos \theta$.
• A light, inextensible string over a light, frictionless pulley has the same tension everywhere.
• Connected bodies share one acceleration. Treat them as one system to get $a$, then isolate one body to get the internal force (tension or contact force).
• Two blocks in contact pushed by $F$ on a smooth floor: $a=F/(m_1+m_2)$, and the contact force equals the mass ahead of the contact times $a$.
• Lift: apparent weight is the normal reaction $R$. $R=m(g+a)$ for upward acceleration (or slowing while going down), $R=m(g-a)$ for downward acceleration, $R=mg$ at constant velocity, $R=0$ in free fall.
• Atwood machine ($m_1>m_2$): $a=(m_1-m_2)g/(m_1+m_2)$ and $T=2m_1{m}_{2}g/(m_1+m_2)$, with $m_{2}g<T<m_{1}g$.
• A heavy rope hanging vertically has the largest tension at the top, since that point supports the whole rope below it.
• NCERT example: a 0.05 kg pebble falling freely has net force $mg=0.5$ N downward, whether it is going up, at the top, or coming down.

Formulas

• $a=\frac{(m_1-m_2)g}{m_1+m_2}$
  Atwood machine, $m_1>m_2$
• $T=\frac{2m_1{m}_{2}g}{m_1+m_2}$
  Atwood string tension
• $R=m(g+a)$
  Lift accelerating up, or moving down and slowing
• $R=m(g-a)$
  Lift accelerating down, or moving up and slowing
• $a=g\sin \theta ,N=mg\cos \theta$
  Smooth incline of angle $\theta$
• $a=\frac{F}{m_1+m_2+m_3},T_{\text{front}}=(m_2+m_3)a$
  Three blocks pulled at the front by $F$; tension just behind the front block pulls the two rear blocks
• $a=\frac{m_{2}g}{m_1+m_2}$
  Block $m_1$ on a smooth table pulled by hanging block $m_2$

Tips, tricks and hacks

Common mistakes

NEET traps

'A man in a freely falling lift has zero weight.' His true weight $mg$ is unchanged; only his apparent weight (scale reading) is zero.

In a lift moving up and slowing down, acceleration is downward, so $R=m(g-a)$, not $m(g+a)$.

For blocks in contact, the contact force changes if you push from the other side: it equals the mass ahead of the contact times $a$.

Topic 3

Friction

Friction is the force that both wastes fuel in your car's engine and lets the tyres grip the road. Without it you could not even walk.

• The contact force between surfaces has two parts: normal reaction $N$ (perpendicular) and friction (parallel to the surfaces).
• Friction opposes relative motion, or impending relative motion, between the surfaces in contact. It does not always oppose the motion of the body.
• Static friction is self-adjusting: it equals the applied force, up to the limiting value $f_s^{\max }={\mu }_{s}N$.
• Limiting friction is (nearly) independent of the area of contact and proportional to $N$. ${\mu }_s$ depends only on the pair of surfaces.
• Once sliding starts, kinetic friction $f_k={\mu }_{k}N$ acts. It is nearly independent of speed and area, and ${\mu }_k<{\mu }_s$.
• Angle of friction $\lambda$: the angle between the resultant of $N$ and limiting friction and the normal; $\tan \lambda ={\mu }_s$. Angle of repose ${\theta }_r$ (steepest incline on which a body stays at rest) has $\tan {\theta }_r={\mu }_s$, so $\lambda ={\theta }_r$.
• Friction is useful (walking, brakes, turning cars) and harmful (heat and wear in machines). Brakes use kinetic friction to stop motion quickly.
• On a rough floor $a=(F-{\mu }_{k}N)/m$. After $F$ is removed, the body decelerates at ${\mu }_{k}g$.
• NCERT examples: a box on a train floor with ${\mu }_s=0.15$ stays at rest for train acceleration up to ${\mu }_{s}g=1.5{\text{m s}}^{-2}$; a block that just starts sliding at $15^{\circ }$ has ${\mu }_s=\tan 15^{\circ }\approx 0.27$.

Formulas

• $f_s\le {\mu }_{s}N$
  Static friction; equal sign only at the point of slipping
• $f_k={\mu }_{k}N$
  Kinetic friction
• $\tan {\theta }_r=\tan \lambda ={\mu }_s$
  Angle of repose = angle of friction
• $a=g(\sin \theta -{\mu }_k\cos \theta )$
  Sliding down a rough incline
• $a=g(\sin \theta +{\mu }_k\cos \theta )$
  Deceleration while sliding up a rough incline
• $d=\frac{v^2}{2{\mu }_{k}g}$
  Stopping distance on a level rough road
• $a_{\max }={\mu }_{s}g$
  Largest acceleration of a truck so that a box on it does not slip

Tips, tricks and hacks

Common mistakes

NEET traps

A block at rest on an incline: friction is $mg\sin \theta$, NOT ${\mu }_{s}mg\cos \theta$, unless it is just about to slide.

'Friction is self-adjusting' is true for static friction only; kinetic friction has a fixed value ${\mu }_{k}N$.

Topic 4

Dynamics of circular motion

Race tracks and highway curves are tilted inward on purpose. The tilt lets the road itself push the car around the bend.

• A body moving on a circle of radius $r$ at speed $v$ needs a net force $m{v}^2/r$ toward the centre (centripetal force).
• Centripetal force is not a new kind of force. Tension, gravity, friction, normal reaction or their components supply it.
• Stone on a string: tension. Moon around Earth: gravity. Car on a flat curve: static friction from the road.
• Flat curve: no skidding if $m{v}^2/r\le {\mu }_{s}mg$, so $v_{\max }=\sqrt{{\mu }_{s}rg}$, independent of the car's mass.
• Banked road: the horizontal component $N\sin \theta$ of the normal reaction supplies part (or all) of $m{v}^2/r$. This allows higher safe speeds and reduces tyre wear.
• At the optimum speed $v_0=\sqrt{rg\tan \theta }$ no friction is needed. Above $v_0$ friction acts down the slope; below $v_0$ it acts up the slope.
• Maximum speed on a banked road: $v_{\max }=\sqrt{rg({\mu }_s+\tan \theta )/(1-{\mu }_s\tan \theta )}$. With ${\mu }_s=0$ it reduces to $v_0$.
• A cyclist on a turn leans inward by angle $\theta$ from the vertical, with $\tan \theta =v^2/(rg)$.
• NCERT example: a cyclist at 18 km/h ($v=5$ m/s) on a flat turn of radius 3 m with ${\mu }_s=0.1$ slips, since $v^2=25$ but ${\mu }_{s}rg=2.94{\text{m}}^2{\text{s}}^{-2}$.
• NCERT example: radius 300 m, banking $15^{\circ }$, ${\mu }_s=0.2$: optimum speed about 28.1 m/s, maximum speed about 38.1 m/s.

Formulas

• $F_c=\frac{m{v}^2}{r}=m{\omega }^{2}r$
  Centripetal force, the net force toward the centre
• $v_{\max }=\sqrt{{\mu }_{s}rg}$
  Flat curve
• $v_0=\sqrt{rg\tan \theta }$
  Banked road, no friction needed
• $v_{\max }=\sqrt{rg\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }}$
  Banked road with friction
• $\tan \theta =\frac{v^2}{rg}$
  Banking angle for speed $v$; also the lean angle of a cyclist

Tips, tricks and hacks

Common mistakes

NEET traps

'Uniform circular motion has constant velocity' is false: speed is constant, but velocity direction changes, so the net force is not zero.

In $v_{\max }=\sqrt{{\mu }_{s}rg}$ use static friction, not kinetic, because the tyres do not slip sideways while turning safely.

For a banked road question that says 'no friction', the answer is $v=\sqrt{rg\tan \theta }$; the term ${\mu }_s$ should not appear.

Topic 5

Impulse and conservation of momentum

A cricketer catching a fast ball pulls the hands back. The ball still stops, but the force on the palms drops sharply.

• Impulse $J=F_{\text{av}}\Delta t=\Delta p$. It is a vector, unit N s (same as ${\text{kg m s}}^{-1}$). For a varying force, impulse is the area under the $F$-$t$ graph.
• Impulsive force: a large force for a very short time (bat on ball, hammer on nail). We judge it by the change in momentum it produces.
• For a fixed $\Delta p$, a longer contact time means a smaller force: pulling hands back while catching, airbags, sand pits for long jump, packing straw for glassware.
• Conservation of momentum: if the net external force on a system is zero, its total momentum stays constant. It follows from the second and third laws.
• In any collision or explosion, internal forces are equal and opposite and act for the same time, so the momentum lost by one body is gained by the other.
• Gun recoil: total momentum is zero before firing, so $mv+MV=0$; the heavy gun moves back slowly.
• Momentum along one direction is conserved if no external force acts along that direction, even if other components change.
• A shell exploding in flight: just after the burst, the fragments' total momentum equals the shell's momentum just before.
• NCERT example: a ball turned back by a bat through $45^{\circ }$ with the same speed gets an impulse along the normal to the bat, of size $2mu\cos 22.5^{\circ }$, because only the normal component of velocity reverses.
• Kinetic energy need not be conserved in a collision even when momentum is.

Formulas

• $J=F_{\text{av}}\Delta t=mv-mu$
  Impulse-momentum theorem
• $J=\int Fdt$
  Impulse of a variable force = area under the $F$-$t$ graph
• ${\vec{p}}_A+{\vec{p}}_B={\vec{p}}_A^{\prime }+{\vec{p}}_B^{\prime }$
  Two-body collision with no external force
• $V=-\frac{mv}{M}$
  Recoil of a gun of mass $M$ firing a bullet $m$ at speed $v$
• $\Delta p=2mv\cos \theta$
  Ball at angle $\theta$ to the normal rebounding with the same speed
• $F=v\frac{dm}{dt}$
  Force for a mass flow rate $dm/dt$ at speed $v$ (machine gun, water jet, conveyor)

Tips, tricks and hacks

Common mistakes

NEET traps

Angled impact: NEET gives the angle with the wall instead of with the normal. If $\alpha$ is with the wall, $\Delta p=2mv\sin \alpha$.

The area under the $F$-$t$ graph gives $\Delta p$, not the final momentum; add the initial momentum to get $p_{\text{final}}$.

Two bodies released from rest pushing each other apart (a spring between them): total momentum stays zero, so their speeds are in the inverse ratio of their masses.

Quick revision: Laws of Motion the night before NEET

1. No force is needed for uniform motion; mass is the measure of inertia.
2. $\vec{F}=d\vec{p}/dt$; for constant mass $\vec{F}=m\vec{a}$; acceleration is always along net force.
3. Action and reaction act on different bodies, at the same instant, and never cancel.
4. Weight and normal reaction on a book are NOT a third-law pair.
5. Lift: $R=m(g+a)$ for upward $a$, $R=m(g-a)$ for downward $a$, $R=0$ in free fall.
6. Atwood: $a=(m_1-m_2)g/(m_1+m_2)$, $T=2m_1{m}_{2}g/(m_1+m_2)$.
7. Smooth incline: $a=g\sin \theta$, $N=mg\cos \theta$.
8. Static friction is self-adjusting up to ${\mu }_{s}N$; $f_k={\mu }_{k}N$; ${\mu }_k<{\mu }_s$.
9. Angle of repose = angle of friction: $\tan \theta ={\mu }_s$.
10. Rolling friction is much smaller than sliding friction (ball bearings).
11. Flat curve: $v_{\max }=\sqrt{{\mu }_{s}rg}$; banked, no friction: $v_0=\sqrt{rg\tan \theta }$; no mass in either.
12. Centripetal force is the net inward force supplied by real forces, never an extra force.
13. Impulse $=F\Delta t=\Delta p$ = area under $F$-$t$ graph, unit N s.
14. Rebound with the same speed: $\Delta p=2mv$ (head-on) or $2mv\cos \theta$ (angle with normal).
15. No external force: total momentum is conserved; gun recoil $V=-mv/M$.

Practise Laws of Motion questions

Topic-wise practice from Level 1 to Level 3, NCERT-based questions and NEET PYQs, with a full solution after every answer.