Laws of Motion

• Sections 4.2 to 4.4 dismantle Aristotle's claim that a force is needed to keep a body moving, and rebuild motion on Galileo's law of inertia, arriving at Newton's first law of motion.
• Section 4.5 introduces momentum and derives the second law, $F = dp/dt = ma$, together with impulse for large forces that act for a short time.
• Section 4.6 states the third law and clears up the action-reaction misconceptions that cost the most marks in exams.
• Section 4.7 combines the second and third laws to derive the law of conservation of momentum, the key tool for collisions and recoils.
• Sections 4.8 and 4.9 cover the equilibrium of a particle under several concurrent forces, and the common forces of mechanics: normal reaction, tension, spring force, and static, kinetic and rolling friction.
• Section 4.10 applies everything so far to circular motion: the centripetal force, and cars cornering on flat and banked roads.
• Section 4.11 closes with Newton's own systematic method for solving any mechanics problem using free-body diagrams, which every worked example in this article follows.

From Aristotle's fallacy to the law of inertia

Aristotle held that a body in motion needs a continuous external push to keep moving; a moving toy car, left alone, coasts to a stop, so common experience seems to agree with him. Galileo saw the flaw. He rolled balls down one incline and up a second, smooth one, and noticed that the ball always climbed back to almost its original height, regardless of the second incline's slope. As the second slope was flattened toward the horizontal, the ball had to travel farther and farther to reach that same height, and in the limiting case of a perfectly horizontal, frictionless plane, it would never stop at all. Galileo's conclusion was that rest and uniform straight-line motion are physically equivalent states, and that a body needs no force whatsoever to sustain either one. What Aristotle missed was friction: the toy car stops because the floor drags on it, and the child must apply a force only to cancel that drag, not to sustain the motion itself.

Newton's first law and the meaning of inertia

Newton rephrased Galileo's insight as a precise law: every body continues in its state of rest or of uniform motion in a straight line unless compelled by an external force to change that state. Applied to acceleration, the law becomes a simple diagnostic you can use in either direction — from known forces to motion, or from observed motion back to the forces at play.

where: $\vec F_{\text{net}}$ is the vector sum of every external force acting on the body, and $\vec a$ is its acceleration. The physical reading runs both ways. If you already know the net force is zero — for instance, a spaceship coasting in deep space with its engines off — the first law tells you the acceleration must be zero, so it keeps whatever velocity it already had. More often on Earth you instead observe the motion first: a book lying at rest on a table must, by the first law, have zero net force on it, and since gravity pulls it down with weight $W$, the table's normal reaction $N$ must be exactly equal and opposite, $N = W$. Note carefully which way the logic runs — it is wrong to say "since $N = W$, the forces cancel and so the book is at rest"; the correct order is that the observed rest tells you the net force is zero, from which $N = W$ follows.

Momentum and Newton's second law

The first law only handles the special case of zero net force. To describe how a body speeds up, slows down, or changes direction, Newton needed a quantity that combines both mass and velocity, because everyday experience shows that a heavier body needs a larger force to produce the same change in speed in the same time, and a faster-moving body needs a larger force to stop it in a given time. That combined quantity is momentum.

where: $m$ is the mass of the body and $\vec v$ its velocity, so $\vec p$ is a vector pointing in the same direction as the velocity. Newton's second law states that the rate of change of momentum is directly proportional to the applied force and takes place in the direction the force acts; choosing the constant of proportionality to be one, in SI units, gives the working form of the law.

where: $\vec F$ is the net external force on the body, $d\vec p/dt$ is the instantaneous rate of change of momentum, and $\vec a = d\vec v/dt$ is the acceleration, valid here because the mass $m$ is constant. The physical reading is that force is not needed to maintain velocity, only to change momentum, exactly as the first law already said for the special case $\vec F = 0$. The law is also a vector equation, equivalent to three independent scalar equations along the $x$, $y$ and $z$ axes, so a force can change only the component of velocity along its own direction, leaving the perpendicular component untouched — this is why the horizontal velocity of a projectile stays constant even as gravity changes its vertical velocity.

Impulse: force acting briefly

Some interactions, a bat striking a ball, a wall stopping a billiard ball, involve a very large force acting for a very short, hard-to-measure time. What is easy to measure is the change in momentum the interaction produces, so Newton's second law is rewritten in a form that sidesteps the unknown force and duration separately.

where: $\vec J$ is the impulse, $\vec F$ is treated as an average force acting over the short interval $\Delta t$, and $\Delta \vec p$ is the resulting change in momentum. The physical reading explains a seasoned cricketer's soft catch: by drawing the hands backward, he stretches out the contact time $\Delta t$ for the same change in the ball's momentum, so the average force $\vec F = \Delta \vec p / \Delta t$ on his hands is much smaller than a novice who keeps his hands rigid and stops the ball almost instantly.

Newton's third law: action and reaction pairs

Where does the external force on a body come from? Newton's answer is that it always arises from some other body, and — crucially — the influence is never one-sided.

where: $\vec F_{AB}$ is the force exerted on body A by body B, and $\vec F_{BA}$ is the force exerted on body B by body A; the two are equal in magnitude, opposite in direction, and act simultaneously, with no cause-and-effect ordering between "action" and "reaction." The physical reading carries three traps students routinely fall into. First, action and reaction act on two different bodies, A and B, never on the same body, so they can never be added together to claim a net force of zero on either one alone. Second, the pair exists even without contact — the earth pulls a falling stone down, and by the third law the stone pulls the earth up with an equal force, simply too small to notice against the earth's enormous mass. Third, if you group A and B together as a single system, then $\vec F_{AB}$ and $\vec F_{BA}$ become internal forces of that system and do cancel in the sum, which is exactly the loophole that makes the second law applicable to an extended body or a system of particles.

Conservation of momentum

Combine the second and third laws for two bodies that interact only with each other, and a powerful conservation law falls out with no further assumptions.

where: $\vec p_A, \vec p_B$ are the momenta of the two bodies before the interaction and $\vec p_A{}', \vec p_B{}'$ their momenta afterward. The physical reading, worked out in full in the derivation below, is that the total momentum of an isolated system, one with no external force, stays constant however violently the bodies inside it interact, whether the collision is elastic or perfectly inelastic. A rifle recoiling as it fires a bullet, a nucleus splitting into fragments, and two skaters pushing off each other are all the same physics: momentum lost by one part is gained, in full, by the rest of the system.

Equilibrium of a particle

A particle is in equilibrium when the net external force on it is zero, so by the first law it is either at rest or moving with constant velocity. For two forces this simply means they are equal and opposite; for three or more concurrent forces, equilibrium requires their vector sum to vanish.

where: each $\vec F_i$ is one of the concurrent forces acting at the same point. The physical reading is geometric: if the force vectors are drawn head-to-tail, equilibrium under three forces means they close into a triangle, and under $n$ forces they close into an $n$-sided polygon. In components, this single vector equation splits into $F_{1x} + F_{2x} + F_{3x} + \cdots = 0$ along $x$ and the equivalent statement along $y$, which is exactly the algebra used to solve every free-body diagram in this chapter.

Common forces: normal reaction, tension and spring force

Besides gravity, which acts at a distance, mechanics deals almost entirely with contact forces. When two surfaces touch, the component of the contact force perpendicular to the surfaces is the normal reaction $N$, and the component along the surfaces, opposing relative or impending sliding, is friction. A string pulled taut develops an internal restoring force called tension, conventionally treated as uniform throughout a massless, inextensible string. A spring resists being stretched or compressed with a restoring force proportional to the displacement from its natural length, $F = -kx$, where $k$ is the spring's force constant and the negative sign shows the force always opposes the displacement. All of these contact forces, along with friction and the buoyant force a fluid exerts on an immersed solid, ultimately trace back to electrical forces between the charged particles that make up ordinary matter, but at the macroscopic scale of this chapter they are simplest to treat as empirical rules determined by experiment.

Friction: static, kinetic and rolling

Push gently on a heavy box resting on the floor and it does not move; push harder and, past some threshold, it suddenly starts to slide. Before sliding begins, an unseen force adjusts itself to exactly cancel your push, keeping the box at rest — this is static friction, and it can take any value up to a measured maximum.

where: $f_s$ is the static friction force, $N$ is the normal reaction between the surfaces, and $\mu_s$ is the coefficient of static friction, a number depending only on the nature of the two surfaces in contact, not on the area of contact. Once the applied force exceeds the limiting value $(f_s)_{\max} = \mu_s N$, the body begins to slide, and the resisting force switches to kinetic friction, which is found experimentally to be roughly independent of speed and given by a similar law.

where: $\mu_k$ is the coefficient of kinetic friction for the same pair of surfaces, and experiment consistently shows $\mu_k$ is a little less than $\mu_s$, which is why a body "breaks free" with a small jerk and then slides more easily. A third kind, rolling friction, opposes a wheel or ball rolling without slipping and is smaller than sliding friction by one or two orders of magnitude for the same load — precisely why the invention of the wheel, and later of ball bearings, mattered so much for reducing power lost to friction in machines.

Circular motion: centripetal force, flat and banked roads

A body moving with uniform speed $v$ around a circle of radius $R$ has acceleration $v^2/R$ toward the centre, so by the second law some real force must supply that inward pull.

where: $f_c$ is the centripetal force, $m$ the mass, $v$ its speed and $R$ the radius of the path. This reading is worth repeating: "centripetal force" is not a distinct kind of force alongside gravity or friction — it is the name for whichever real force points toward the centre. For a stone whirled on a string it is tension; for a planet it is gravity; for a car turning on a flat road it is tyre friction, which is why there is a strict maximum cornering speed on an unbanked curve.

where: $\mu_s$ is the coefficient of static friction between tyres and road, $R$ the turn's radius, and $g$ the acceleration due to gravity; notice the maximum speed does not depend on the car's mass at all. Banking the road at an angle lets the normal reaction itself contribute a horizontal, centre-pointing component, reducing how much the tyres must rely on friction — the full expression is worked out in the derivation below.

Solving problems in mechanics: the free-body diagram method

Most real mechanics problems involve not one isolated object but an assembly of connected bodies, each experiencing gravity plus forces from the others. Newton's own systematic approach is to sketch the full assembly first, then choose a convenient part of it as "the system," and draw a separate free-body diagram showing every force on that system from the rest of the assembly and from any other agency, but never the forces the system itself exerts on its surroundings. Known magnitudes and directions — the pull of gravity, the direction of a taut string — go directly on the diagram; everything else is treated as unknown and solved for using the first and second laws. If the problem needs it, the same procedure is repeated for a different choice of system, with the third law used to carry information between the two diagrams: a force shown as $\vec F$ acting on body A due to body B must appear as $-\vec F$ acting on body B due to A. This five-step recipe, repeated across every worked example below, is the single most transferable skill in this chapter.

Board level - stopping distance and average resistive force of a bullet

A bullet of mass $0.04\ \text{kg}$ moving at $90\ \text{m/s}$ enters a heavy wooden block and is brought to rest after penetrating $0.6\ \text{m}$. Find the average resistive force exerted by the block on the bullet.

Step 1. Find the average retardation using $v^2 = u^2 + 2as$ with $v = 0$, $u = 90\ \text{m/s}$, $s = 0.6\ \text{m}$: $0 = 90^2 + 2a(0.6)$, so $a = -8100/1.2 = -6750\ \text{m/s}^2$.

Step 2. Apply Newton's second law with the magnitude of the retardation: $F = 0.04 \times 6750 = 270\ \text{N}$. Answer: the average resistive force on the bullet is 270 N, opposite to its motion. Since the actual resistance inside the wood is unlikely to be perfectly uniform, this is only the average, exactly as NCERT's Example 4.2 notes.

Board level - tension and angle for a sideways-pulled rope

A mass of $6\ \text{kg}$ hangs from the ceiling by a rope of length $2\ \text{m}$. A horizontal force of $50\ \text{N}$ is applied at the midpoint $P$ of the rope. Taking $g = 10\ \text{m/s}^2$ and a massless rope, find the angle the upper segment makes with the vertical, and the tensions in both segments.

Step 1. In the free-body diagram of the hanging mass, only gravity ($60\ \text{N}$ down) and the lower tension $T_2$ (up) act, so by the first law $T_2 = mg = 60\ \text{N}$.

Step 2. At point $P$, three forces meet: $T_2$ pulling down, the upper tension $T_1$ at angle $\theta$ to the vertical, and the $50\ \text{N}$ horizontal force. Equilibrium along each axis gives $T_1\cos\theta = T_2 = 60\ \text{N}$ and $T_1\sin\theta = 50\ \text{N}$.

Step 3. Dividing eliminates $T_1$: $\tan\theta = 50/60 = 0.833$, so $\theta \approx 39.8^\circ$. Then $T_1 = \sqrt{60^2 + 50^2} \approx 78.1\ \text{N}$. Answer: the rope makes about 40 degrees with the vertical, with $T_1 \approx 78.1\ \text{N}$ and $T_2 = 60\ \text{N}$. As NCERT notes, this does not depend on the rope's length or where along it the force is applied.

Competitive level - connected block and trolley with kinetic friction (JEE and NEET pattern)

A 3 kg block hangs from a string over a smooth pulley and connects to a 20 kg trolley on a horizontal floor with $\mu_k = 0.04$ between trolley and floor. Taking $g = 10\ \text{m/s}^2$ and a massless string, find the acceleration and the tension.

Step 1. Since the string is inextensible and the pulley smooth, both bodies share the same acceleration $a$. For the hanging block: gravity ($30\ \text{N}$) pulls down, tension $T$ pulls up, so $30 - T = 3a$.

Step 2. For the trolley, the horizontal forces are the forward tension $T$ and the backward kinetic friction $f_k$. Its normal reaction equals only its own weight, $N = 20 \times 10 = 200\ \text{N}$ (the block hangs off the pulley, it does not press on the trolley), so $f_k = \mu_k N = 8\ \text{N}$, giving $T - 8 = 20a$.

Step 3. Add the two equations to eliminate $T$: $22 = 23a$, so $a = 22/23 \approx 0.96\ \text{m/s}^2$, and $T = 8 + 20(0.96) \approx 27.1\ \text{N}$. Answer: acceleration about 0.96 m/s², tension about 27.1 N. Timing note: under two minutes. Write both second-law equations immediately and add them so $T$ cancels — never guess $T$ first. The classic slip is computing the trolley's normal reaction as block-plus-trolley weight; since the block hangs from the pulley rather than resting on the trolley, only the trolley's own weight presses on the floor.

Competitive level - optimum and maximum speed on a banked racetrack (JEE and NEET pattern)

A circular racetrack of radius $300\ \text{m}$ is banked at $15^\circ$, tyre-road friction coefficient $0.2$, $g = 9.8\ \text{m/s}^2$. Find (a) the optimum speed needing no friction, and (b) the maximum permissible speed before skidding.

Step 1. For the optimum speed use $v_0 = \sqrt{Rg\tan\theta}$, valid when the horizontal component of $N$ alone supplies the centripetal force. With $\tan 15^\circ \approx 0.268$: $v_0 = \sqrt{300 \times 9.8 \times 0.268} = \sqrt{788} \approx 28.1\ \text{m/s}$.

Step 2. For the maximum speed use $v_{\max} = \left[Rg(\mu_s + \tan\theta)/(1 - \mu_s\tan\theta)\right]^{1/2}$, since friction now acts fully down the slope. Numerator factor $\mu_s + \tan\theta = 0.468$; denominator factor $1 - \mu_s\tan\theta = 0.946$.

Step 3. Combine: $v_{\max}^2 = 2940 \times (0.468/0.946) \approx 1454$, so $v_{\max} \approx 38.1\ \text{m/s}$. Answer: optimum speed about 28.1 m/s, maximum permissible speed about 38.1 m/s. Timing note: about ninety seconds. The common error is swapping the two formulas — optimum speed uses only $\tan\theta$, maximum speed uses the full expression with $\mu_s$. Sanity-check that $v_{\max} > v_0$, since friction only adds to the frictionless optimum here.

Reversing the logic of the first law is the single most common conceptual slip: the correct statement is "the book is at rest, therefore the net force is zero, therefore $N = W$," never "since $N = W$, the forces cancel and so the book is at rest." A related error is thinking constant velocity needs a continuous forward force — it needs only enough force to cancel friction, not to sustain the motion itself. Students routinely add the two forces of an action-reaction pair and claim they cancel; they never can, since they act on different bodies, and only forces on the same body may be summed to test equilibrium. Writing $f_s = \mu_s N$ unconditionally, instead of $f_s \le \mu_s N$, is another frequent slip: static friction is self-adjusting and only reaches its limiting value the instant sliding is about to begin. Some students add "$ma$" onto a free-body diagram as though it were an extra force; $ma$ is the effect of the net real force, never a force to be drawn or summed. Centripetal force is often treated as a distinct new force rather than the name given to whichever real force, tension, gravity, friction, or normal reaction, points toward the centre in that problem. A ball at the top of its throw has zero velocity but not zero force — gravity still acts, so "$v = 0$" must never be read as "$F = 0$." On banked-road problems, the direction friction points (down the slope above the optimum speed, up the slope below it) is the most common source of sign errors that flip the final answer.

• State Newton's three laws of motion precisely, and explain why a force is needed only to change, not to maintain, a body's velocity.
• Draw a correct free-body diagram for any single object in an assembly, showing every external force on that object and no forces it exerts on anything else.
• Apply $F = ma$ along independent directions to solve problems involving tension, normal reaction, gravity and friction together.
• Use impulse, $J = F\Delta t = \Delta p$, to relate a brief large force to the momentum change it produces, and explain why extending contact time reduces the force felt.
• Derive the law of conservation of momentum from the second and third laws, and apply it to collisions, recoils and explosions.
• Distinguish self-adjusting static friction ($f_s \le \mu_s N$) from kinetic friction ($f_k = \mu_k N$) and from rolling friction, and know why $\mu_k$ is smaller than $\mu_s$.
• Derive and apply the angle of repose, $\tan\theta_{\max} = \mu_s$, and explain why it does not depend on mass.
• Analyse circular motion on flat and banked roads, computing both the friction-free optimum speed and the maximum safe cornering speed with friction.
• Correctly identify genuine action-reaction pairs, and distinguish them from two forces on the same body that merely happen to be equal and opposite.
• Use Newton's five-step free-body diagram method to solve problems involving two or more connected bodies.

Ask the tutor: if you are stuck on a free-body diagram or want to check whether you have identified the correct action-reaction pair in a problem, ask the tutor to walk through it with you step by step.