Almost every mechanics question in NEET — blocks on inclines, pulleys, cars on curves, rockets, collisions — is solved by applying Newton's three laws through a free-body diagram. Master the logic of this chapter and you effectively unlock work-energy, circular motion, and rotational dynamics as well.
Inertia and the First Law
Before Galileo, motion was thought to need a continuous "push". Galileo's inclined-plane experiments suggested the opposite: a ball rolling on a perfectly smooth horizontal surface would keep rolling forever. What stops real bodies is an external agent — friction or air resistance — not the absence of a pusher.
Newton's first law states: a body continues in its state of rest, or of uniform motion in a straight line, unless acted upon by a net external force. Two important readings of this statement:
- It defines force qualitatively — force is that which changes velocity (i.e. produces acceleration).
- It tells us that zero net force and zero acceleration are equivalent conditions. A parachutist falling at constant terminal velocity has zero net force even though gravity is acting.
The tendency of matter to resist a change in its state of motion is inertia, measured by mass. Three familiar faces of inertia:
- Inertia of rest — dust flies off when a carpet is beaten; the carpet moves, the dust tends to stay.
- Inertia of motion — a passenger in a bus that brakes suddenly lurches forward.
- Inertia of direction — mud from a spinning cycle tyre flies off along the tangent.
A subtle but exam-relevant point: the first law holds only in inertial frames — frames that are unaccelerated. Inside an accelerating lift or a turning car, a free object appears to accelerate with no real force on it; such frames are non-inertial, and to use Newton's laws there we must add a fictitious (pseudo) force of magnitude $ma$ opposite to the frame's acceleration.
Momentum, the Second Law and Impulse
Linear momentum is $\vec{p} = m\vec{v}$, a vector with SI unit kg m s⁻¹. Experience says that stopping difficulty depends on both mass and speed — a cricket ball at 100 km/h and a truck crawling at 2 km/h can be equally hard to stop — and momentum captures exactly that combination.
Newton's second law, in its most general form:
$$\vec{F}_{net} = \frac{d\vec{p}}{dt}$$
For constant mass this reduces to $\vec{F}_{net} = m\vec{a}$, which in components reads $F_x = ma_x$, $F_y = ma_y$, $F_z = ma_z$. Each component equation is independent — a force along $x$ can never change $v_y$. This is why projectile motion works: the vertical weight leaves the horizontal velocity untouched.
Features worth remembering:
- The law is local in time: acceleration at an instant is decided by the force at that same instant, not by the force's past history.
- 1 newton is the force that gives a 1 kg mass an acceleration of 1 m s⁻².
- Only the net external force enters; internal forces between parts of the system cancel in pairs.
Rewriting the second law as $\vec{F},dt = d\vec{p}$ and integrating over the duration of a force gives the impulse–momentum theorem:
$$\vec{J} = \int \vec{F},dt = \vec{F}_{avg},\Delta t = \Delta \vec{p}$$
Impulse is the area under the force–time graph. Its practical value lies in impulsive forces — collisions, kicks, hammer blows — where the force is huge, its detailed variation unknown, but the momentum change measurable. Since $\Delta p$ is fixed for a given landing, increasing $\Delta t$ reduces the peak force: hence padded dashboards, airbags, sand pits for long jumpers, and a cricketer drawing the hands back while catching.
The Third Law and Conservation of Linear Momentum
To every action there is an equal and opposite reaction, i.e. if body A exerts $\vec{F}{AB}$ on B, then B exerts $\vec{F}{BA} = -\vec{F}_{AB}$ on A. The crucial conditions: the two forces are of the same nature, act on different bodies, act simultaneously (neither is "caused" by the other), and therefore never cancel in a single body's free-body diagram.
Apply the second law to an isolated two-body system and the third law forces cancel internally, giving:
$$\vec{p}_1 + \vec{p}_2 = \text{constant}$$
When the net external force on a system is zero, its total linear momentum is conserved. This is a vector statement, so it may hold along one direction even if not along another (e.g. horizontal momentum is conserved for an exploding shell in flight, vertical momentum is not, because gravity acts).
Standard applications:
- Recoil of a gun: $M V = m v$, so the heavy gun recoils slowly compared with the bullet.
- Explosion of a stationary bomb: fragment momenta add vectorially to zero.
- Rocket propulsion: exhaust gases are pushed back, the rocket is pushed forward.
- Collisions: momentum is conserved in all collisions (elastic and inelastic); kinetic energy is conserved only in elastic ones.
During impulsive interactions, ordinary finite forces such as gravity contribute negligible impulse over the tiny collision time, so momentum conservation may be applied even to non-isolated systems for that brief interval.
Friction
Friction is the tangential contact force that opposes relative sliding (or the tendency to slide) between two surfaces. Along with the normal reaction $N$, it makes up the total contact force.
Static friction ($f_s$) is self-adjusting: it takes whatever value is needed to prevent slipping, up to a maximum, $$f_s \le f_s^{max} = \mu_s N$$ Kinetic friction acts once sliding begins and is essentially constant: $$f_k = \mu_k N, \qquad \mu_k < \mu_s$$ This is why a heavy crate is hardest to get moving and easier to keep moving. Both coefficients are dimensionless, roughly independent of contact area and (for $\mu_k$) of speed, and depend on the nature of the surfaces.
Key results to memorise:
- Angle of friction $\phi$: the angle between the resultant contact force and the normal at limiting equilibrium, $\tan\phi = \mu_s$.
- Angle of repose $\theta_r$: the incline angle at which a block just begins to slide, $\tan\theta_r = \mu_s$. Numerically $\theta_r = \phi$.
- On an incline of angle $\theta$, for a sliding block: $a = g(\sin\theta - \mu_k\cos\theta)$ down the slope.
- Rolling friction is far smaller than sliding friction, which is why wheels and ball bearings are used.
Friction is both a nuisance (wear, heat, energy loss — reduced by lubricants, polishing, bearings) and a necessity (walking, writing, braking, gripping, the very possibility of a car accelerating). Note that in walking and in rolling without slipping, static friction can point along the motion — friction opposes relative sliding at the contact, not the body's motion as a whole.
Circular Motion Dynamics
A body in uniform circular motion has constant speed but continuously changing velocity direction, hence a centripetal acceleration $a_c = v^2/r = \omega^2 r$ directed towards the centre. The second law then demands a net inward force:
$$F_c = \frac{mv^2}{r}$$
"Centripetal force" is not a new kind of force — it is the requirement met by some real agent: tension for a stone whirled on a string, gravity for a satellite, friction and/or normal force for a car on a bend, electrostatic attraction for an electron in Bohr's model.
Car on a level circular road: static friction supplies the centripetal force, so $\frac{mv^2}{r} \le \mu_s mg$, giving the safe speed limit $$v_{max} = \sqrt{\mu_s r g}$$ Note it is independent of the vehicle's mass.
Banked road: tilting the road by angle $\theta$ lets the horizontal component of the normal reaction help. Ignoring friction, $\tan\theta = v^2/rg$, i.e. the optimum speed is $v_0 = \sqrt{rg\tan\theta}$. With friction included, $$v_{max} = \sqrt{rg,\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}}$$
Vertical circle (string/rod, e.g. a bucket of water or a "death well"): at the topmost point, gravity and tension both point down, so $$T + mg = \frac{mv_{top}^2}{r} \Rightarrow v_{top} \ge \sqrt{gr} \text{ (for } T \ge 0)$$ and by energy conservation the minimum speed at the lowest point is $\sqrt{5gr}$.
Using Free-Body Diagrams: A Working Method
Almost all numericals yield to a fixed routine:
- Isolate each body and draw it separately.
- Mark only real external forces: weight $mg$ (always), normal reaction (perpendicular to contact), friction (along contact), tension (along the string, away from the body), applied forces. Never draw "$ma$" as a force.
- Choose axes cleverly — for an incline, take axes along and perpendicular to the surface.
- Write $\Sigma F = ma$ for each axis, for each body. For equilibrium of concurrent forces, $\Sigma F_x = 0$ and $\Sigma F_y = 0$ (three concurrent forces in equilibrium also satisfy Lami's theorem).
- Add constraint relations — bodies connected by an inextensible string over a pulley share the same magnitude of acceleration; a block on a moving surface may share its acceleration.
- Solve the simultaneous equations; check limiting cases and signs.
Two classic results: for two masses over a light frictionless pulley (Atwood machine), $a = \frac{(m_1-m_2)g}{m_1+m_2}$ and $T = \frac{2m_1m_2 g}{m_1+m_2}$. For a person of mass $m$ in a lift accelerating upward with $a$, the apparent weight (normal reaction) is $N = m(g+a)$; downward acceleration gives $m(g-a)$, and free fall $a=g$ gives weightlessness, $N = 0$.
Common Mistakes and Exam Traps
- Confusing the third-law pair with balanced forces. For a book on a table, $mg$ (by Earth) and $N$ (by table) act on the same body and are equal only because acceleration is zero — they are not action–reaction. The reaction to $N$ is the book pushing down on the table; the reaction to $mg$ is the book attracting the Earth.
- Assuming friction is always $\mu N$. Static friction equals $\mu_s N$ only at the verge of slipping. If a 10 N push acts on a block whose $\mu_s N = 15$ N, the actual static friction that acts is only 10 N (just enough to prevent motion), not 15 N — plugging in $\mu_s N$ regardless of the applied force is a very common error.
- Ignoring that the normal reaction is not always $mg$. On an accelerating lift, an incline, or under an additional applied force with a vertical component, $N$ must be found from $\Sigma F_y = ma_y$ for that specific situation; using $N = mg$ out of habit gives a wrong answer in any question involving vertical acceleration or a slanted force.
- Applying $v_{max} = \sqrt{\mu_s rg}$ to a banked road, or $\sqrt{rg\tan\theta}$ to a level road. The two formulas belong to different setups — flat roads rely entirely on friction, banked frictionless roads rely entirely on the horizontal component of the normal reaction — and mixing them up is a frequent slip when a question changes the geometry midway.