Newton's laws of motion for JEE Main

Newton's laws are three short sentences, and almost every mechanics question in JEE Main rests on them. Candidates rarely lose marks here because they do not know F = ma. They lose marks because they draw the wrong forces, forget a constraint, or apply the law in an accelerating frame without correcting for it.

The fix is a method, used the same way every time. This guide sets out that method, the standard systems it applies to, and the traps that JEE questions like to set.

The three laws, stated usefully

1. First law (inertia). A body stays at rest or in uniform straight-line motion unless a net external force acts on it. It also defines an inertial frame: one in which this law holds. The ground is treated as inertial for JEE problems.
2. Second law. The net force equals the rate of change of momentum: F = dp/dt. For constant mass, F = ma. It applies separately along each axis: ΣFₓ = maₓ, ΣF_y = ma_y.
3. Third law. Forces come in pairs: if A pushes B, B pushes A with an equal and opposite force. The pair acts on different bodies, so the two forces never cancel each other in one free-body diagram.

Impulse. Integrating the second law, impulse = ∫F dt = change in momentum. For a large force over a short time, the average force is Δp / Δt.

The four-step method

1. Isolate each body and draw its free-body diagram (FBD): every force on that body, and nothing else.
2. Choose axes for each body, usually along and perpendicular to its motion.
3. Write F = ma along each axis.
4. Add constraint relations (string lengths, contact) and solve the equations together.

The forces that usually appear:

Constraint relations

A string of fixed length ties the motions of the bodies it connects.

• Over a fixed pulley: both ends move by the same distance, so the two blocks have equal and opposite accelerations.
• Movable pulley: if a block hangs from a pulley that is supported by two segments of one string, the block moves half as far as the free end of the string, so its acceleration is half.
• Bodies in contact: they share the same acceleration as long as they stay in contact.

When in doubt, write the length of the string in terms of the positions and differentiate twice.

Standard systems

Pseudo-force and apparent weight

Newton's laws hold in inertial frames. If you analyse motion from a frame accelerating with a, add a pseudo-force −ma on every body (opposite to the frame's acceleration). Then F = ma works in that frame.

Apparent weight in a lift. The scale reads the normal force N.

Note that the reading depends on the acceleration, not the direction of motion.

Worked problems

Problem 1: Atwood machine. Masses of 3 kg and 2 kg hang over a light frictionless pulley (g = 10 m/s²). Find the acceleration and tension.

a = (3 − 2) × 10 / 5 = 2 m/s².
T = 2 × 3 × 2 × 10 / 5 = 24 N. Check with the 2 kg block: T − 20 = 2 × 2, so T = 24 N.

Problem 2 (numerical answer): pulling a train of blocks. Blocks of 2 kg, 3 kg and 5 kg lie in a line on a smooth floor, joined by light strings. A 20 N force pulls the 5 kg block. Find the tension in the string between the 5 kg and 3 kg blocks.

a = 20 / 10 = 2 m/s².
That string pulls the 3 kg and 2 kg blocks together: T = (3 + 2) × 2 = 10 N.
(The string between the 3 kg and 2 kg blocks has T = 2 × 2 = 4 N.)

Problem 3: pendulum in an accelerating car. A bob of mass 0.1 kg hangs from the roof of a car accelerating at 5 m/s² (g = 10 m/s²). Find the angle of the string and its tension.

In the car's frame, a pseudo-force ma acts backward. The string lines up with the combined "effective gravity".
tan θ = a/g = 0.5, so θ ≈ 26.6° from the vertical, with the bob swinging backward.
T = m √(g² + a²) = 0.1 × √125 ≈ 1.12 N.

Problem 4: cutting a string below a spring. Block A (2 kg) hangs from a spring attached to the ceiling. Block B (1 kg) hangs from A by a string. The system is at rest. The string is cut. Find the accelerations of A and B just after.

Before cutting, the spring holds both: spring force = 3 × 10 = 30 N.
Just after, the spring force is still 30 N (the spring cannot change length instantly). On A: 30 − 20 = 2a, so a = 5 m/s² upward.
B is now falling freely: 10 m/s² downward.

Practice set

1. In an Atwood machine with 5 kg on each side, what is the acceleration?
2. A 1 kg body hangs from a spring balance in a lift accelerating down at 2 m/s² (g = 10 m/s²). What does the balance read?
3. A 60 kg person stands on a weighing scale in a lift moving up at a constant 3 m/s. What does it read (g = 10 m/s²)?
4. A gun fires 10 bullets a second, each of 20 g, at 400 m/s. What average force is needed to hold the gun steady?
5. Blocks of 4 kg and 6 kg joined by a string lie on a smooth floor. A 50 N force pulls the 6 kg block. Find the tension.
6. A force F = 6t newtons acts on a 2 kg body at rest from t = 0. Find its speed at t = 2 s.
7. A bob hangs in a car accelerating at g. What angle does the string make with the vertical?
8. A 2 kg block on a smooth table is joined by a string over a pulley at the edge to a hanging 3 kg block. Find the acceleration and tension (g = 10 m/s²).

Answers

1. 0; the masses balance.
2. m(g − a) = 1 × 8 = 8 N.
3. 600 N. Constant velocity means zero acceleration.
4. Momentum delivered per second = 10 × 0.02 × 400 = 80 N.
5. a = 50/10 = 5 m/s²; the string pulls only the 4 kg block, so T = 4 × 5 = 20 N.
6. Impulse = ∫6t dt from 0 to 2 = 3t² = 12 N s; v = 12/2 = 6 m/s.
7. tan θ = g/g = 1, so 45°.
8. a = 30/5 = 6 m/s²; T = 2 × 6 = 12 N. Check with the hanging block: 30 − 12 = 3 × 6.

What to do next

• Draw FBDs for ten systems before solving any of them, and check each against a friend's or a solution.
• Solve 30 connected-body and pulley problems, writing the constraint relation every time.
• Do ten lift and accelerating-car problems using pseudo-forces.
• Continue to friction, then circular motion, where the same FBD method applies.