Newton's Laws of Motion — Concepts, Numericals & Exam Tips

Newton's Laws of Motion carry serious weightage in every major exam — CBSE, JEE Main, JEE Advanced, and NEET. This is one topic where understanding the concept once means you can handle almost any numerical that comes your way.

We'll cover all three laws, free body diagrams, friction, tension in strings, and a range of solved numericals from NCERT level up to JEE Advanced.

Newton's First Law — The Law of Inertia

Every object stays at rest or moves in a straight line at constant velocity unless an external force acts on it. That's it. The law is deceptively simple but it hides a deep idea: inertia.

Inertia is the resistance of a body to any change in its state of motion. Heavier objects have more inertia — that's why it's harder to push a loaded truck than an empty one.

🎯 Exam Insider

First law gives us the definition of force — it's what causes a change in state of motion. It also defines inertial frames of reference, which JEE Advanced sometimes tests directly.

Real examples of First Law:

When a bus brakes suddenly, passengers lurch forward — their bodies want to keep moving.
Dust flies off a carpet when you beat it — the carpet moves but the dust stays behind momentarily.
A coin placed on a card over a glass drops into the glass when the card is flicked away.

Newton's Second Law — F = ma

The net force on an object equals its mass times its acceleration. Written properly:

Newton's Second Law

$F_{net} = ma$

Where:

$F_{net}$ = net (resultant) force in Newtons (N)
$m$ = mass in kg
$a$ = acceleration in m/s²

Also written as: $F = \dfrac{dp}{dt}$ (rate of change of momentum — preferred for JEE Advanced)

The second law is the workhorse of mechanics. Nearly every numerical in this chapter uses it. The key word is net force — you must account for all forces acting on the body before applying $F = ma$ .

⚠️ Common Mistake

Students often apply $F = ma$ using just one force instead of the net force. If a 10 N force acts on a body and friction is 4 N, the net force is 6 N — not 10 N. Always draw the FBD first.

Impulse and Momentum

When force acts for a short time, we use impulse:

Impulse-Momentum Theorem

$J = F \cdot \Delta t = \Delta p = m(v - u)$

This is why airbags save lives — they increase the time of impact, reducing the force on the passenger even though the change in momentum is the same.

Newton's Third Law — Action and Reaction

For every action, there is an equal and opposite reaction. Forces always come in pairs. If body A exerts force on body B, then body B exerts an equal and opposite force on body A.

📌 Note

Action-reaction pairs always act on different bodies. This is why they never cancel each other out. Students who think "equal and opposite forces cancel" are confusing action-reaction pairs with equilibrium.

The Horse and Cart Paradox

Classic confusion: if the cart pulls the horse backward with equal force, how does anything move?

The answer is that you're looking at the wrong system. The horse pushes the ground backward. The ground pushes the horse forward (reaction). This forward force on the horse exceeds the backward pull of the cart. Net force on the system is forward — so it accelerates.

Free Body Diagrams (FBDs)

Drawing FBDs is the single most important skill in this chapter. Get this right and half your work is done.

Steps to draw a correct FBD:

Isolate the body you're analyzing.
Draw all forces acting on that body (not forces it exerts).
Label each force clearly: weight ( $mg$ downward), normal force ( $N$ perpendicular to surface), tension ( $T$ along string), friction ( $f$ opposing motion).
Choose a coordinate system. Usually x along motion, y perpendicular.
Apply Newton's second law along each axis separately.

💡 Expert Tip

For inclined planes, tilt your coordinate axes along and perpendicular to the incline. This makes components much cleaner — you won't need to resolve normal force then.

Friction

Friction opposes relative motion between surfaces. It acts parallel to the surface of contact.

Laws of Friction

Static friction: $f_s \leq \mu_s N$

Kinetic (sliding) friction: $f_k = \mu_k N$

Rolling friction: $f_r = \mu_r N$ (much smaller than $\mu_k$ )

Always: $\mu_s > \mu_k > \mu_r$

Types of Friction

Static friction — acts when the body is at rest. It adjusts itself to match the applied force, up to a maximum value $\mu_s N$ . This maximum value is called limiting friction.

Kinetic friction — acts when the body is sliding. It has a fixed value $\mu_k N$ , independent of speed (for the range we deal with in exams).

Rolling friction — acts on rolling objects like wheels. Much smaller in magnitude, usually neglected unless specifically asked.

🎯 Exam Insider

JEE often asks: "Find whether the block moves." First calculate limiting friction $(\mu_s N)$ . If applied force $<$ limiting friction, the block doesn't move and static friction equals the applied force. If applied force $>$ limiting friction, the block moves and friction becomes kinetic $(\mu_k N)$ .

Friction on Inclined Planes

For a block on an incline of angle $\theta$ :

Inclined Plane — Key Equations

Forces along incline: $mg\sin\theta$ (down the incline)

Normal force: $N = mg\cos\theta$

Maximum static friction: $f_s = \mu_s \cdot mg\cos\theta$

Condition for sliding: $mg\sin\theta > \mu_s \cdot mg\cos\theta$ , i.e., $\tan\theta > \mu_s$

Acceleration when sliding: $a = g(\sin\theta - \mu_k\cos\theta)$

Tension in Strings

Tension is the force transmitted through a string or rope. For a massless, inextensible string, tension is the same throughout its length.

When two blocks are connected by a string and pulled by a force, treat the system first to find acceleration, then isolate one block to find tension.

💡 Expert Tip

For a string with mass $m_s$ : tension varies along the string. Tension at a point distance $x$ from the free end: $T = F \cdot \dfrac{M + m_s - m_x}{M + m_s}$ where $M$ is the block mass. JEE Advanced has asked this.

Solved Numericals

Numerical 1 — Basic F = ma (CBSE Level)

A body of mass 5 kg is moving at 10 m/s. A force is applied for 2 seconds and velocity becomes 20 m/s. Find the force.

$a = \frac{v - u}{t} = \frac{20 - 10}{2} = 5 \text{ m/s}^2$

$F = ma = 5 \times 5 = 25 \text{ N}$

Numerical 2 — Atwood Machine (JEE Main Level)

Two blocks of 3 kg and 5 kg connected by string over frictionless pulley. Find acceleration and tension.

Let $m_1 = 3$ kg, $m_2 = 5$ kg. Taking $m_2$ going down as positive:

$m_2 g - T = m_2 a \quad \cdots (i)$

$T - m_1 g = m_1 a \quad \cdots (ii)$

Adding: $(m_2 - m_1)g = (m_1 + m_2)a$

$a = \frac{(5 - 3) \times 10}{5 + 3} = \frac{20}{8} = 2.5 \text{ m/s}^2$

$T = m_1(g + a) = 3 \times (10 + 2.5) = 37.5 \text{ N}$

Numerical 3 — Block on Incline with Friction (JEE Advanced Level)

A 10 kg block on 30° incline, $\mu = 0.3$ . Does it slide? Find friction force.

$mg\sin 30° = 10 \times 10 \times 0.5 = 50 \text{ N (down the incline)}$

$N = mg\cos 30° = 10 \times 10 \times \frac{\sqrt{3}}{2} = 86.6 \text{ N}$

Maximum static friction: $f_{s,\max} = \mu N = 0.3 \times 86.6 = 26$ N

Since 50 N $>$ 26 N, the block slides.

Kinetic friction: $f_k = 0.3 \times 86.6 = 26$ N (opposing motion, so up the incline)

$a = g(\sin 30° - \mu\cos 30°) = 10(0.5 - 0.3 \times 0.866) = 10 \times 0.24 = 2.4 \text{ m/s}^2$

Numerical 4 — Three Blocks on a Surface

Three blocks of 2 kg, 3 kg, 5 kg are placed in contact. A force of 40 N pushes all three. Find contact force between first and second block. (No friction)

Total mass $= 10$ kg. Acceleration $= \dfrac{40}{10} = 4$ m/s²

Contact force between block 1 and block 2 acts on the system of blocks 2 and 3:

$F_{12} = (m_2 + m_3) \times a = (3 + 5) \times 4 = 32 \text{ N}$

5 Common Mistakes Students Make

⚠️ Common Mistake

Mistake 1 — Using total force instead of net force. Always find the resultant of all forces before writing $F = ma$ .

⚠️ Common Mistake

Mistake 2 — Confusing mass and weight. Mass is in kg, weight is in N. Weight $= mg$ . On moon, weight changes but mass doesn't.

⚠️ Common Mistake

Mistake 3 — Wrong direction for friction. Friction opposes relative motion or tendency of motion, not the applied force directly. On an incline where the block is being pushed up, friction acts downward.

⚠️ Common Mistake

Mistake 4 — Action-reaction pairs acting on same body. They always act on different bodies. Normal force on book from table and weight of book on table are action-reaction — they don't cancel because they act on different objects.

⚠️ Common Mistake

Mistake 5 — Forgetting pseudo force in non-inertial frames. In an accelerating lift or car, add a pseudo force $-ma_0$ in the direction opposite to acceleration of the frame.

Exam Tips

🎯 Exam Insider

JEE Main pattern: Expect 2-3 questions per paper. Most common: Atwood machine variants, inclined plane with friction, block-on-block problems. Weightage is consistent year after year.

🎯 Exam Insider

NEET pattern: Conceptual questions dominate. First Law (inertia examples), Second Law applications, and Third Law (identifying correct action-reaction pairs) are frequent. One or two straightforward numericals.

🎯 Exam Insider

CBSE Board: Definitions of all three laws, derivation of $F = ma$ from Second Law, numerical on friction and tension. 5-6 marks dedicated to this chapter reliably.

For FBD problems in JEE Advanced: Always check if the string goes slack. If the calculated tension comes out negative, the string is slack and you need to redo with $T = 0$ .

Pseudo force trick: In problems involving a block inside an accelerating vehicle, shift to the vehicle's frame and add pseudo force $ma_0$ on all objects (opposite to vehicle's acceleration). Converts a dynamic problem into a static one.

Practice Questions

A 2 kg block is pushed by a force of 10 N on a surface with $\mu = 0.3$ . Find acceleration. ( $g = 10$ m/s²)
Two blocks of 4 kg and 6 kg are on a frictionless surface connected by a string. A force of 20 N pulls the 6 kg block. Find tension in the string.
A block of 5 kg is on a 45° frictionless incline. Find its acceleration.
A lift of mass 800 kg carries a passenger of 60 kg. The lift accelerates upward at 2 m/s². Find normal force on passenger and tension in the lift cable.
A body of mass 10 kg is moving at 20 m/s. A braking force reduces velocity to 5 m/s in 3 seconds. Find the braking force.
Block A (3 kg) sits on Block B (7 kg) on a frictionless surface. Force of 40 N applied on B. $\mu$ between A and B is 0.2. Find acceleration of each block and friction force on A.
A bullet of 20 g is fired from a gun of 2 kg. Bullet velocity is 400 m/s. Find recoil velocity of gun.
A 5 kg block hangs from a spring balance inside a lift. If the balance reads 40 N, find the acceleration of the lift. ( $g = 10$ m/s²)

Newton's First Law — The Law of Inertia

Newton's Second Law — F = ma

Impulse and Momentum

Newton's Third Law — Action and Reaction

The Horse and Cart Paradox

Free Body Diagrams (FBDs)

Friction

Types of Friction

Friction on Inclined Planes

Tension in Strings

Solved Numericals

Numerical 1 — Basic F = ma (CBSE Level)

Numerical 2 — Atwood Machine (JEE Main Level)

Numerical 3 — Block on Incline with Friction (JEE Advanced Level)

Numerical 4 — Three Blocks on a Surface

5 Common Mistakes Students Make

Exam Tips

Practice Questions

Frequently Asked Questions

Practice Questions