Law of inertia

Sit in a car at a red light and everything feels calm — your coffee cup sits still in its holder, your body rests easy against the seat. The light turns green, the driver floors the accelerator, and your head snaps back into the headrest. Nobody pushed you forward; nothing pushed you back either. Your body simply tried to keep doing what it was already doing — staying still — while the car surged ahead beneath you. That stubbornness of matter, its refusal to change its own state of motion without being made to, is inertia, and the rule that describes it is Newton's first law of motion, the oldest and most quietly revolutionary idea in this chapter. For nearly two thousand years before Newton, the accepted wisdom (going back to Aristotle) was that objects need a continuous push to keep moving, and that rest is the "natural" state of everything. Galileo Galilei broke that spell with a thought experiment involving balls rolling down smooth inclines, and Newton turned Galileo's insight into a precise law: a body stays at rest, or moves in a straight line at constant speed, unless a net external force acts on it. That single sentence quietly redefines what a force actually is — not something needed to keep things moving, but something needed only to change how they move. This article builds outward from that first law into the machinery of the whole "Laws of Motion" chapter: linear momentum and the second law that ties force to its rate of change, the third law's action-reaction pairs, the conservation of momentum that follows from it, equilibrium and free-body diagrams, the everyday forces — gravity, normal reaction, tension, spring force, friction — and finally the dynamics of objects moving in circles, from a car rounding a bend to a pendulum swinging in a cone. Every idea here traces back to the same stubborn refusal of matter to change its motion on its own.

Key takeaways

Newton's first law (law of inertia): a body continues in its state of rest or uniform straight-line motion unless acted on by a net external force; inertia is measured by mass.
Momentum $\vec p = m\vec v$ is the true bookkeeping quantity of motion; Newton's second law says the net force equals its rate of change, $\vec F_{net} = d\vec p/dt = m\vec a$ for constant mass.
Impulse $\vec J = \vec F\,\Delta t$ equals the change in momentum — the impulse-momentum theorem — and explains why a longer contact time softens a given change in velocity.
Newton's third law: every action force has an equal and opposite reaction force, but the two act on two different bodies and never cancel each other on the same object.
With zero net external force, total linear momentum is conserved: $m_1u_1+m_2u_2 = m_1v_1+m_2v_2$ , which explains recoil of a gun and the thrust of a rocket.
Friction is self-adjusting up to a maximum ( $f_s \le \mu_s N$ ), then becomes kinetic ( $f_k = \mu_k N < \mu_s N$ ); it fixes both the maximum safe speed on a level curve and the optimum banking angle of a road.
Key idea: because a body resists any change to its own motion (inertia), a net external force is required to change momentum, $\vec F_{net}=d\vec p/dt$ , and this single idea — reapplied as action-reaction, conservation, equilibrium, friction and circular dynamics — is the engine of the entire Laws of Motion chapter.

The core insight

A body's tendency to resist any change in its state of rest or motion is inertia, quantified by mass, and Newton's laws simply describe how a net external force overcomes that inertia to change momentum, $\vec F_{net} = d\vec p/dt = m\vec a$ , while an equal and opposite reaction acts back on whatever applied the force.

Where it fits in the syllabus

Laws of Motion is a full, board-tested unit of CBSE Class 11 Physics (Chapter 5, NCERT), and every idea named above — inertia, momentum, the three laws, conservation of momentum, equilibrium, common forces, friction, and circular dynamics — sits squarely inside the current 2025-26 prescription with no rationalisation cuts. Expect direct numerical problems (blocks on inclines, collisions, banked curves) as well as conceptual and derivation-based questions in board exams, and heavy reliance on this chapter's ideas — momentum, friction, circular motion — throughout JEE and NEET mechanics. It builds directly on kinematics (position, velocity, acceleration) from the previous chapter and feeds forward into work-energy-power and rotational dynamics, both of which are deliberately kept out of this article.

The physical picture

Newton's first law states that a body remains at rest or in uniform motion along a straight line unless a net external force acts on it. This is not really a separate law from the second — it is the special case of the second law when the net force is zero — but it earns its own name because it defines what a force is (something that changes motion) and asserts the existence of frames in which the law holds (inertial frames). Momentum makes the idea of "state of motion" precise:

\vec p = m\vec v

where: $m$ is the body's mass (kg) and $\vec v$ is its velocity (m/s), so $\vec p$ is a vector in kg·m/s pointing along the direction of motion. Physically, momentum bundles "how much mass" and "how fast and in what direction" into one quantity, and it is momentum — not velocity alone — that a net force actually changes. Newton's second law makes that change quantitative:

\vec F_{net} = \frac{d\vec p}{dt}

where: $\vec F_{net}$ is the vector sum of all external forces on the body and $d\vec p/dt$ is the rate at which its momentum changes with time. For a body of constant mass this reduces to the familiar $\vec F_{net} = m\vec a$ , and the SI unit of force, the newton, is defined exactly so that $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ produces an acceleration of $1\ \text{m/s}^2$ in a $1\ \text{kg}$ mass. Closely related is impulse, $\vec J = \vec F\,\Delta t$ for a force applied over a short interval $\Delta t$ ; the impulse-momentum theorem says this impulse exactly equals the change it produces in momentum, $\vec J = \Delta \vec p$ — which is why a cricketer "gives with the catch," stretching $\Delta t$ to reduce the peak force for a given $\Delta p$ . Newton's third law completes the picture: whenever body A exerts a force on body B, body B exerts an equal and opposite force back on body A, and these two forces act on different bodies, at the same instant, along the same line. Because they act on different bodies they never add up to zero on a single object — they only cancel if you (wrongly) draw both on the same body.

Figure

A block resting on a rough incline with four labelled force arrows, weight acting straight down, normal reaction acting perpendicular to the incline surface, tension acting up the slope from a string, and friction acting along the surface opposing sliding, illustrating a free body diagram

Newton's second law from momentum, and force as its rate of change

Derivation AFrom momentum to F = ma, and where the newton comes from

The goal is to show precisely how Newton's second law, stated in terms of the rate of change of momentum, reduces to the familiar $F=ma$ for a body of constant mass, and how that reduction fixes the size of the unit we call the newton. This matters because board answers that jump straight to $F=ma$ without showing the momentum step lose marks for an incomplete derivation, and because the momentum form is the one that survives into variable-mass problems like rockets, which $F=ma$ alone cannot handle.

Step 1. Start from Newton's second law in its general momentum form, $\vec F_{net} = \dfrac{d\vec p}{dt}$ , where $\vec p = m\vec v$ is the body's momentum. This form says the net external force equals however fast momentum is changing, regardless of whether that change comes from a changing velocity, a changing mass, or both.

Step 2. Expand the derivative using the product rule on $\vec p = m\vec v$ : $\dfrac{d\vec p}{dt} = m\dfrac{d\vec v}{dt} + \vec v\dfrac{dm}{dt}$ . This is the fully general result, and it is exactly this second term, $\vec v\,dm/dt$ , that a rocket's thrust equation needs, because a rocket's mass keeps falling as it burns fuel.

Step 3. Apply the restriction that matters for this chapter: constant mass, $dm/dt = 0$ . This is true for every situation in the CBSE syllabus except the qualitative mention of rockets. The second term in Step 2 vanishes, leaving $\vec F_{net} = m\dfrac{d\vec v}{dt}$ .

Step 4. Recognise $d\vec v/dt$ as the acceleration $\vec a$ , giving the familiar $\vec F_{net} = m\vec a$ . This is not a separate law; it is the constant-mass special case of the momentum form from Step 1. Setting the size of the newton so that a $1\ \text{kg}$ mass under a $1\ \text{N}$ net force accelerates at exactly $1\ \text{m/s}^2$ removes any proportionality constant from the equation, which is precisely why $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ .

Worked examples

Board level - force to stop a moving trolley

A trolley of mass $20\ \text{kg}$ moving at $6\ \text{m/s}$ is brought to rest in $3\ \text{s}$ by a constant braking force. Find (a) the trolley's initial momentum, (b) the impulse needed to stop it, and (c) the braking force.

Step 1. Initial momentum: $p_i = mv = 20 \times 6 = 120\ \text{kg}\,\cdot\,\text{m/s}$ . The final momentum is $p_f = 0$ since the trolley is brought to rest.

Step 2. Impulse needed equals the change in momentum (impulse-momentum theorem): $J = \Delta p = p_f - p_i = 0 - 120 = -120\ \text{kg}\,\cdot\,\text{m/s}$ ; the magnitude of impulse required is $120\ \text{kg}\,\cdot\,\text{m/s}$ , directed opposite to the motion.

Step 3. Braking force from $J = F\Delta t$ : $F = \dfrac{J}{\Delta t} = \dfrac{-120}{3} = -40\ \text{N}$ , the minus sign showing the force acts opposite to the trolley's motion. Answer: initial momentum $120\ \text{kg}\,\cdot\,\text{m/s}$ , impulse $120\ \text{kg}\,\cdot\,\text{m/s}$ (backward), braking force $40\ \text{N}$ .

Competitive level - maximum safe speed on a banked curve with friction

A car of mass $900\ \text{kg}$ negotiates a curve of radius $r = 100\ \text{m}$ banked at $\theta = 15^\circ$ , where the coefficient of static friction between tyres and road is $\mu_s = 0.25$ . Take $g = 9.8\ \text{m/s}^2$ and $\tan 15^\circ \approx 0.268$ . Find the maximum speed at which the car can round the curve without skidding.

Step 1. Use the banked-road formula with friction at the point of slipping outward, $v_{max} = \sqrt{\dfrac{rg(\mu_s+\tan\theta)}{1-\mu_s\tan\theta}}$ . This formula reduces to the frictionless optimum $\tan\theta = v^2/(rg)$ when $\mu_s = 0$ , and it is the correct one to use whenever friction is given alongside a banking angle.

Step 2. Compute the numerator factor: $\mu_s + \tan\theta = 0.25 + 0.268 = 0.518$ , so $rg(\mu_s+\tan\theta) = 100 \times 9.8 \times 0.518 = 507.6$ .

Step 3. Compute the denominator factor: $1 - \mu_s\tan\theta = 1 - (0.25)(0.268) = 1 - 0.067 = 0.933$ .

Step 4. Divide and take the square root: $v_{max}^2 = \dfrac{507.6}{0.933} = 544.0$ , so $v_{max} = \sqrt{544.0} \approx 23.3\ \text{m/s}$ , which is about $84\ \text{km/h}$ . Answer: $v_{max} \approx 23.3\ \text{m/s} \approx 84\ \text{km/h}$ ; note the car's mass of $900\ \text{kg}$ never entered the calculation, because mass cancels out of the circular-motion force balance exactly as it does in free fall. Timing note: under two minutes once the formula is memorised. The one trap is picking the wrong banked-road formula — check whether friction is mentioned; if not, use the simpler frictionless $\tan\theta = v^2/(rg)$ , and always sanity-check that mass has cancelled out of the final expression.

Common traps

Drawing an action and its reaction on the same body. Newton's third law pairs always act on two different bodies (the block pushes the table down, the table pushes the block up); they never both appear in one object's free-body diagram, and they cannot be used to argue that a body's own weight is "cancelled" by its reaction to the ground.
Treating $f_s = \mu_s N$ as an equation rather than a limit. Static friction is self-adjusting and satisfies $f_s \le \mu_s N$ ; it equals whatever value is needed to prevent sliding, up to that maximum. Only at the point of slipping does $f_s = \mu_s N$ exactly.
Inventing a centrifugal force in an inertial (ground) frame. In the frame of the road or the ground, the only real forces on a car going round a curve are gravity, the normal reaction and friction; their net inward component is the centripetal force, $F_c = mv^2/r$ , not a separate outward force to be added and cancelled.
Confusing mass with weight when discussing inertia. Inertia is measured by mass (kg), which does not change with location; weight $W=mg$ does change from planet to planet, but the body's resistance to a change in motion does not.
Forgetting the newton's definition when checking units. $1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2$ ; any force answer that does not reduce cleanly to this combination signals an arithmetic slip upstream.

Common trapThe third law pair never acts on the same body, and there is no real outward force in an inertial frame

The single costliest mistake in this chapter is pairing an action force with its reaction on the same object, then "cancelling" them to argue nothing can ever accelerate — this is wrong because the two forces of a third-law pair act on two different bodies and can never appear together in one body's equation of motion. The mirror-image mistake is adding an outward "centrifugal force" to a free-body diagram drawn in the ground (inertial) frame when analysing circular motion; in that frame there is no such force, only gravity, normal reaction and friction, whose net inward component supplies the real centripetal force $mv^2/r$ . Keep every free-body diagram tied to one body and one frame, and both errors disappear.

You should be able to…

State Newton's first, second and third laws precisely, including that a third-law pair acts on two different bodies.
Define momentum and impulse, and apply the impulse-momentum theorem to find an average force or a change in velocity.
Apply conservation of linear momentum to simple recoil and collision problems, $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ .
Draw a correct free-body diagram for a body in equilibrium under gravity, normal reaction, tension and friction.
Distinguish static friction (an inequality, self-adjusting) from kinetic friction (an equation, $f_k=\mu_k N$ ), and state $\mu_k<\mu_s$ .
Apply the centripetal force idea and the banked-road formulas (with and without friction) to circular motion problems.

Ask the tutor: Want to see how the same free-body-diagram method handles a block on a frictionless incline connected over a pulley to a hanging mass — ask the AI tutor to walk through that classic two-body problem step by step.

If every action has an equal and opposite reaction, how does anything ever move?

Because the two forces of a third-law pair act on two different bodies, they never cancel for either body individually. When you push against the ground while walking, you exert a backward force on the ground and the ground exerts an equal forward force on you — that forward force on your body is what accelerates you, and it is not cancelled by anything, because the reaction to it (your push) acts on the ground, not on you. Motion happens because we track the forces on one body at a time, not both bodies at once.

Why is static friction written as an inequality, f_s less than or equal to mu_s N, instead of a plain equation like kinetic friction?

Because static friction's whole job is to prevent sliding, and it does this by automatically supplying however much force is needed, up to a ceiling. If you push a heavy cupboard gently, static friction matches your push exactly and the cupboard does not move — friction here is far below its maximum. Push harder and friction keeps matching you, right up until you exceed the maximum value $\mu_s N$ ; beyond that point the cupboard starts to slide and the (usually smaller) kinetic friction $f_k=\mu_k N$ takes over. Writing $f_s = \mu_s N$ always would incorrectly claim friction is at its maximum even when nothing is trying to move the object.

Why does banking a road at the correct angle let a car go around a curve without needing any friction at all?

Because on a frictionless banked curve, the normal reaction from the road is tilted inward (toward the centre of the curve) rather than straight up. That tilt gives the normal force a horizontal component pointing toward the centre, and at exactly the right speed this horizontal component alone supplies the centripetal force needed to keep the car turning, with the vertical component of the normal force balancing the car's weight. The angle that makes this work out exactly is $\tan\theta = v^2/(rg)$ ; only speeds other than this optimum need friction to make up the difference.

Does mass really not matter for how fast a car can go around a banked curve?

For the idealised formulas given here, correct — mass cancels out of both the frictionless banking condition $\tan\theta=v^2/(rg)$ and the with-friction maximum speed formula, exactly as it does in the derivation of $g=GM/R^2$ for free fall. Physically this is because both the centripetal force required ( $mv^2/r$ ) and the forces that supply it (gravity $mg$ , friction $\mu_s N$ where $N$ itself scales with $m$ ) scale with the same mass, so it cancels from the final speed condition. A heavier car and a lighter car, tyres and road surface being identical, can safely take the same banked curve at the same maximum speed.