Laws of motion for NEET

Laws of motion is where kinematics meets cause. Almost every mechanics question after this, in work and energy, rotation or fluids, begins with the same step: draw the forces on one body and apply Newton's second law. Get that habit right here and the rest of mechanics becomes easier.

NEET questions from this chapter are usually one- or two-step numericals: a block on an incline, two blocks joined by a string, a person in a lift, a car on a curve. The traps are almost always in the normal force and the direction of friction.

Newton's three laws, properly stated

First law (inertia). A body stays at rest or in uniform straight-line motion unless a net external force acts on it. Mass measures inertia. The law also defines an inertial frame: one in which the first law holds.

Second law. The rate of change of momentum equals the net external force: F = dp/dt. For constant mass this becomes F = ma. It is a vector law, so apply it separately along each axis. Only external forces count; internal forces cancel in pairs.

Third law. Forces come in pairs: if A pushes B, B pushes A with an equal and opposite force. The two forces act on different bodies, so they never cancel each other.

Impulse and momentum

Impulse J = F × Δt = Δp. For a varying force, impulse is the area under the force–time graph.

This explains everyday cases. A cricketer draws the hands back while catching, which increases the time of contact and so reduces the average force for the same change in momentum.

Conservation of momentum follows from the second and third laws. If no net external force acts on a system, its total momentum stays constant. Gun recoil, explosions and collisions all use this: in a recoil, m_bullet × v_bullet = M_gun × V_gun.

Equilibrium of concurrent forces

A body is in equilibrium when the vector sum of forces on it is zero: ΣFₓ = 0 and ΣF_y = 0. Three forces in equilibrium form a closed triangle when drawn head to tail. Lami's theorem gives a shortcut: each force is proportional to the sine of the angle between the other two.

Friction

• Static friction adjusts itself to match the applied force, up to a maximum f_s(max) = μ_s N.
• Kinetic friction acts once sliding starts: f_k = μ_k N, with μ_k < μ_s.
• Friction does not depend on the area of contact, only on the nature of the surfaces and N.
• Rolling friction is much smaller than sliding friction, which is why wheels and ball bearings help.

The friction graph in words: as the applied force increases from zero, friction rises equally (a 45° line), reaches the limiting value μ_s N, then drops a little to the constant kinetic value once the block moves.

On an incline of angle θ:

• Normal force N = mg cos θ (only if no other force presses the block).
• Angle of repose: the block just begins to slide when tan θ = μ_s.
• Sliding down with friction: a = g(sin θ − μ_k cos θ).

Connected bodies and lifts

For two masses m₁ > m₂ over a light frictionless pulley (Atwood machine):

• a = (m₁ − m₂)g / (m₁ + m₂)
• T = 2m₁m₂g / (m₁ + m₂)

For a person of mass m in a lift, the scale reading (apparent weight) is:

Circular motion: where the centripetal force comes from

Centripetal force mv²/r is not a new force. It is the net inward force, supplied by something real: tension, friction, gravity or a component of the normal force.

• Level road: friction supplies it. The maximum safe speed is v_max = √(μ_s r g).
• Banked road, no friction: the horizontal component of N supplies it. The ideal speed is v₀ = √(r g tan θ), so tan θ = v²/(rg).
• Banked road with friction: v_max = √[r g (μ_s + tan θ) / (1 − μ_s tan θ)].

Worked numericals

Take g = 10 m s⁻², sin 37° = 0.6 and cos 37° = 0.8.

Example 1: sliding down a rough incline

A 10 kg block slides down a 37° incline with μ_k = 0.25. Find its acceleration.

• Along the incline: mg sin θ − μ_k mg cos θ = ma.
• a = 10(0.6 − 0.25 × 0.8) = 10(0.6 − 0.2) = 4 m s⁻².
• The mass cancels, so the answer is the same for any block on this surface.

Example 2: pulling at an angle

A 10 kg block on a floor (μ_k = 0.2) is pulled by 50 N at 37° above the horizontal. Find the acceleration.

• Vertical: N = mg − F sin θ = 100 − 50 × 0.6 = 70 N.
• Friction = 0.2 × 70 = 14 N.
• Horizontal: 50 × 0.8 − 14 = 40 − 14 = 26 N, so a = 26/10 = 2.6 m s⁻².
• Using N = 100 N by mistake gives friction 20 N and a = 2 m s⁻², which is likely to be one of the options.

Example 3: Atwood machine

Masses of 5 kg and 3 kg hang over a light frictionless pulley.

• a = (5 − 3) × 10 / 8 = 2.5 m s⁻².
• T = 2 × 5 × 3 × 10 / 8 = 37.5 N.
• Check on the 5 kg mass: 50 − T = 5 × 2.5, so T = 50 − 12.5 = 37.5 N ✓.

Example 4: impulse on a cricket ball

A 0.15 kg ball arrives at 20 m s⁻¹ and is hit straight back at 30 m s⁻¹. Contact lasts 0.01 s.

• Taking the return direction as positive, Δp = 0.15 × (30 − (−20)) = 0.15 × 50 = 7.5 N s.
• Average force = 7.5 / 0.01 = 750 N.
• Forgetting the reversal (using 30 − 20) gives only 1.5 N s, a common wrong option.

Practice MCQs

1. A 2 kg gun fires a 20 g bullet at 400 m s⁻¹. The recoil speed of the gun is: (a) 2 m s⁻¹ (b) 4 m s⁻¹ (c) 8 m s⁻¹ (d) 40 m s⁻¹
2. The maximum safe speed on a flat curve of radius 40 m with μ_s = 0.4 is about: (a) 4 m s⁻¹ (b) 8 m s⁻¹ (c) 12.6 m s⁻¹ (d) 16 m s⁻¹
3. A road of radius 100 m is banked with tan θ = 0.1. The speed at which no friction is needed is: (a) 10 m s⁻¹ (b) 31.6 m s⁻¹ (c) 100 m s⁻¹ (d) 1 m s⁻¹
4. A 50 kg person stands in a lift moving upward at a constant 3 m s⁻¹. The scale reads: (a) 0 (b) 350 N (c) 500 N (d) 650 N
5. A block just begins to slide when an incline reaches 30°. The coefficient of static friction is: (a) 0.5 (b) 0.87 (c) √3 (d) 1/√3
6. Blocks of 1 kg, 2 kg and 3 kg are placed in contact on a smooth floor. A 12 N force pushes the 1 kg block. The force between the 2 kg and 3 kg blocks is: (a) 2 N (b) 6 N (c) 10 N (d) 12 N
7. A force on a 2 kg body at rest rises from 0 to 100 N and falls back to 0 over 0.2 s (a triangle on the F–t graph). The final speed is: (a) 5 m s⁻¹ (b) 10 m s⁻¹ (c) 20 m s⁻¹ (d) 2.5 m s⁻¹
8. For a book resting on a table, an action–reaction pair is: (a) the book's weight and the table's normal force on it (b) the normal force and friction (c) the book's weight and the table's weight (d) the table's push on the book and the book's push on the table

Answers

1. (b) 0.02 × 400 = 2 × V, so V = 4 m s⁻¹.
2. (c) v = √(0.4 × 40 × 10) = √160 ≈ 12.6 m s⁻¹.
3. (a) v = √(100 × 10 × 0.1) = √100 = 10 m s⁻¹.
4. (c) Constant velocity means a = 0, so the reading is mg = 500 N.
5. (d) μ_s = tan 30° = 1/√3 ≈ 0.58.
6. (b) a = 12/6 = 2 m s⁻²; the 3 kg block needs 3 × 2 = 6 N.
7. (a) Impulse = ½ × 0.2 × 100 = 10 N s; v = 10/2 = 5 m s⁻¹.
8. (d) The pair acts on different bodies; option (a) acts on the same body.

What to do next

• For the next 20 questions you solve, draw a free-body diagram first, even when it feels obvious.
• Write N for five situations (flat floor, incline, pull at an angle, push at an angle, lift) and check each against the examples above.
• Solve previous-year NEET questions on friction, connected bodies and banking, timed.
• Move on to work, energy and power, which reuses every force you drew here.

For the motion side of these problems, revise kinematics for NEET.