JEE Main Physics · Chapter 3

Laws of Motion

Newton’s laws, impulse and momentum, free-body diagrams, pulleys, lifts, friction and banking of roads.

NCERT Class 11JEE Main: 1–2 Qs per shift 11 min read

1. Chapter at a Glance

Newton's laws of motion explain what causes motion to change. With free-body diagrams, these three laws solve problems on blocks, pulleys, lifts, inclines and friction. The chapter also covers momentum conservation, impulse, and the dynamics of circular motion (banked roads, conical pendulum). JEE Main asks one or two questions every year, and the ideas are needed in almost every mechanics chapter.

What JEE asks most
  • Connected bodies: pulleys, strings, contact forces.
  • Friction: static vs kinetic, blocks on inclines, stacked blocks.
  • Apparent weight in a lift; pseudo force in accelerated frames.
  • Impulse and momentum conservation; banked roads.

2. Newton's Three Laws

  • First law (inertia): a body stays at rest or in uniform motion unless a net external force acts. Mass is the measure of inertia.
  • Second law: Fnet = dp/dt = ma (for constant mass). The force and acceleration are in the same direction.
  • Third law: forces come in equal and opposite pairs acting on different bodies, so they never cancel each other.
  • Impulse J = FavgΔt = Δp. Catching a ball with moving hands increases Δt and reduces the force.
  • Conservation of linear momentum: if the net external force on a system is zero, its total momentum stays constant (gun recoil, explosions, rockets).

3. Free-Body Diagrams and Connected Bodies

Draw each body separately, show every force on it (weight, normal reaction, tension, friction, applied force), choose axes, and write ΣF = ma for each body. Add a constraint relation (for example, both blocks on one string have the same acceleration).

mg N f (friction) mg sinθ (down slope) θ
Block on a rough incline: N = mg cosθ; along the slope, mg sinθ acts downward and friction opposes the (tendency of) motion.
SystemResult
Atwood machine (m₁ > m₂)a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂)
Lift accelerating up (a)apparent weight = m(g + a)
Lift accelerating down (a)apparent weight = m(g − a); zero in free fall
Smooth inclinea = g sinθ
Two blocks pushed by F (contact)a = F/(m₁ + m₂); contact force = m₂a

Pseudo force: in a frame accelerating with a, add a force −ma on every body to use Newton's laws in that frame.

4. Friction

  • Static friction adjusts itself up to a maximum fs,max = μsN. It equals the applied force until motion starts.
  • Kinetic friction fk = μkN, roughly constant; μk < μs. Rolling friction is smaller still.
  • Angle of friction λ: tan λ = μs. Angle of repose: the steepest incline on which a block stays at rest; tan α = μs.
  • Rough incline: sliding down a = g(sinθ − μcosθ); force to push up at constant speed F = mg(sinθ + μcosθ).

5. Dynamics of Circular Motion

  • Net force towards the centre = mv²/r (the centripetal force is not a new force — it is provided by tension, friction, gravity or normal reaction).
  • Flat curve: vmax = √(μrg).
  • Banked road (no friction): tanθ = v²/rg. With friction: vmax² = rg(μ + tanθ)/(1 − μ tanθ).
  • Conical pendulum: T cosθ = mg, T sinθ = mω²r ⇒ ω² = g/(l cosθ).
  • Vertical circle (string): minimum speed at top √(gr), at bottom √(5gr).

6. Formula Master Table

QuantityFormula
Second lawF = dp/dt = ma
ImpulseJ = FΔt = Δp
Recoilm₁v₁ + m₂v₂ = 0
Frictionfs ≤ μsN; fk = μkN
Angle of reposetan α = μs
Rough incline (sliding down)a = g(sinθ − μcosθ)
Bankingtanθ = v²/rg
Flat curvevmax = √(μrg)

7. Mind Map

Laws of Motion
Newton's laws
  • Inertia
  • F = dp/dt
  • Action–reaction
Momentum
  • Impulse
  • Conservation
  • Recoil
Applications
  • Pulleys, lifts
  • Pseudo force
  • Contact force
Friction & circles
  • Static, kinetic
  • Inclines
  • Banking, conical pendulum

8. Tips and Tricks

Shortcuts
  • For a system moving together, first find a = Fnet,external/Mtotal, then isolate one body to get internal forces.
  • Before using μk, check whether the applied force exceeds μsN; if not, friction just equals the applied force.
  • For stacked blocks, the top block can accelerate at most μg (friction is the only horizontal force on it).
  • In a freely falling lift, everything is "weightless" and a pendulum stops oscillating.
Mistakes students make
  • Writing friction = μN even when the block is at rest.
  • Adding centripetal force as an extra force in the free-body diagram.
  • Cancelling action and reaction forces — they act on different bodies.

9. Important One-liners

  • Mass is a measure of inertia.
  • Impulse equals change in momentum.
  • Static friction is self-adjusting; kinetic friction is nearly constant.
  • μs is usually greater than μk.
  • Roads are banked so that the normal reaction supplies part of the centripetal force.
  • A rocket works on momentum conservation, not by pushing against air.

10. Solved Numericals: Average to Advanced

Five average-level and five advanced-level problems. Click to reveal each answer and its full working. Take g = 10 m s⁻².

Average level

N1Average

Masses of 5 kg and 3 kg hang on either side of a light frictionless pulley (Atwood machine). Find the acceleration and the string tension.

Show answer & solution

Concept: the heavier mass goes down; both have the same acceleration.

  1. 5 kg: 50 − T = 5a. 3 kg: T − 30 = 3a.
  2. Add: 20 = 8a ⇒ a = 2.5 m s⁻².
  3. T = 30 + 3 × 2.5 = 37.5 N (check: 2m₁m₂g/(m₁ + m₂) = 300/8 = 37.5 N).

Answer: 2.5 m s⁻²; 37.5 N

N2Average

A 60 kg person stands on a weighing machine in a lift. What does the machine read (in newtons) when the lift accelerates (a) upward at 2 m s⁻², (b) downward at 2 m s⁻²?

Show answer & solution

Concept: the machine reads the normal reaction N.

  1. (a) N − mg = ma ⇒ N = 60(10 + 2) = 720 N.
  2. (b) mg − N = ma ⇒ N = 60(10 − 2) = 480 N.

Answer: 720 N; 480 N

N3Average

A 5 kg gun fires a 50 g bullet at 400 m s⁻¹. Find the recoil speed of the gun.

Show answer & solution

Concept: total momentum is zero before and after firing.

  1. 0 = 0.05 × 400 + 5 × V.
  2. V = −20/5 = −4 m s⁻¹ (opposite to the bullet).

Answer: 4 m s⁻¹ backward

N4Average

A 0.2 kg ball hits a wall at 20 m s⁻¹ and rebounds along the same line at 15 m s⁻¹. The contact lasts 0.05 s. Find the impulse and the average force.

Show answer & solution

Concept: impulse = change in momentum; take the rebound direction as positive.

  1. Δp = 0.2 × 15 − 0.2 × (−20) = 3 + 4 = 7 N s.
  2. Favg = Δp/Δt = 7/0.05 = 140 N.

Answer: 7 N s; 140 N

N5Average

A 2 kg block rests on a floor with μs = 0.5 and μk = 0.4. Find the friction force and acceleration when it is pushed horizontally with (a) 8 N, (b) 15 N.

Show answer & solution

Concept: compare the push with the limiting friction μsN.

  1. N = 20 N, so fs,max = 0.5 × 20 = 10 N; fk = 0.4 × 20 = 8 N.
  2. (a) 8 N < 10 N ⇒ no motion; static friction = 8 N; a = 0.
  3. (b) 15 N > 10 N ⇒ it slides; friction = 8 N.
  4. a = (15 − 8)/2 = 3.5 m s⁻².

Answer: (a) 8 N, 0; (b) 8 N, 3.5 m s⁻²

Advanced level

N6Advanced

A 10 kg block slides down a 30° incline with μk = 0.2. Find its acceleration.

Show answer & solution

Concept: along the slope, mg sinθ − μmg cosθ = ma.

  1. a = g(sinθ − μ cosθ).
  2. = 10(0.5 − 0.2 × 0.866) = 10(0.5 − 0.173).
  3. ≈ 3.27 m s⁻².

Answer: about 3.27 m s⁻²

N7Advanced

A curve of radius 100 m is banked so that tanθ = 0.25, and μs = 0.2 between tyres and road. Find the maximum safe speed.

Show answer & solution

Concept: at maximum speed friction acts down the slope; resolving gives v² = rg(μ + tanθ)/(1 − μ tanθ).

  1. Numerator: μ + tanθ = 0.2 + 0.25 = 0.45.
  2. Denominator: 1 − 0.2 × 0.25 = 0.95.
  3. v² = 100 × 10 × 0.45/0.95 ≈ 473.7 m² s⁻².
  4. v ≈ 21.8 m s⁻¹ (about 78 km h⁻¹).

Answer: about 21.8 m s⁻¹

N8Advanced

Block A (4 kg) rests on block B (6 kg); μ between them is 0.3 and the floor is smooth. A horizontal force F pulls B. (a) Find the largest F for which they move together. (b) Find the accelerations of A and B when F = 50 N.

Show answer & solution

Concept: only friction accelerates A, so aA ≤ μg.

  1. (a) amax for A = μg = 0.3 × 10 = 3 m s⁻².
  2. Moving together: F = (4 + 6) × 3 = 30 N.
  3. (b) 50 N > 30 N, so A slips; kinetic friction = 0.3 × 40 = 12 N.
  4. aA = 12/4 = 3 m s⁻².
  5. aB = (50 − 12)/6 ≈ 6.33 m s⁻².

Answer: (a) 30 N; (b) 3 m s⁻² and 6.33 m s⁻²

N9Advanced

A 0.5 kg bob on a 1 m string moves as a conical pendulum with the string at 60° to the vertical. Find the tension, the angular speed and the time for one revolution.

Show answer & solution

Concept: T cosθ = mg (vertical); T sinθ = mω²r with r = l sinθ.

  1. T = mg/cos60° = 5/0.5 = 10 N.
  2. T sinθ = mω² l sinθ ⇒ ω² = T/(ml) = 10/0.5 = 20 ⇒ ω ≈ 4.47 rad s⁻¹ (same as g/(l cosθ)).
  3. Period = 2π/ω ≈ 1.40 s.

Answer: 10 N; 4.47 rad s⁻¹; about 1.40 s

N10Advanced

A 5 kg block is on an incline with sinθ = 0.6, cosθ = 0.8 and μ = 0.5. Find (a) the force parallel to the incline needed to push it up at constant speed, (b) the least such force that stops it sliding down.

Show answer & solution

Concept: friction always opposes the (tendency of) motion; N = mg cosθ = 40 N, so μN = 20 N; mg sinθ = 30 N.

  1. (a) Moving up: friction acts down. F = mg sinθ + μN = 30 + 20 = 50 N.
  2. (b) Tending to slide down: friction acts up. F + μN = mg sinθ ⇒ F = 30 − 20 = 10 N.
  3. So any force between 10 N and 50 N keeps the block at rest.

Answer: (a) 50 N; (b) 10 N

11. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Easy

A person in a lift that is falling freely feels:

  1. Heavier
  2. Lighter but not weightless
  3. Weightless
  4. No change
Show solution

With a = g downward, N = m(g − g) = 0, so the apparent weight is zero.

Answer: (C)

Q2Medium

Two blocks of 2 kg and 3 kg on a smooth floor are pushed by 20 N applied to the 2 kg block. The force between them is:

  1. 8 N
  2. 12 N
  3. 20 N
  4. 10 N
Show solution

a = 20/5 = 4 m s⁻². The 3 kg block is pushed only by the contact force: 3 × 4 = 12 N.

Answer: (B)

Q3Easy

A block stays at rest on an incline for angles up to 30°. The coefficient of static friction is:

  1. 0.5
  2. 1/√3
  3. √3
  4. 0.866
Show solution

Angle of repose: μs = tan30° = 1/√3.

Answer: (B)

Q4Hard

A car takes a flat curve of radius 40 m. If μs = 0.4, the maximum speed without skidding is (g = 10 m s⁻²):

  1. 4 m s⁻¹
  2. 8 m s⁻¹
  3. 12.6 m s⁻¹
  4. 16 m s⁻¹
Show solution

vmax = √(μrg) = √(0.4 × 40 × 10) = √160 ≈ 12.6 m s⁻¹.

Answer: (C)

Q5Numerical

A force–time graph for a 2 kg body at rest is a triangle rising from 0 to 10 N in 2 s and falling back to 0 at 4 s. Find the final speed in m s⁻¹.

Show solution

Impulse = area = ½ × 4 × 10 = 20 N s. v = 20/2 = 10 m s⁻¹.

Answer: 10

Q6Numerical

A curve of radius 90 m is banked for a speed of 30 m s⁻¹ with no friction. Find tanθ (g = 10 m s⁻²).

Show solution

tanθ = v²/rg = 900/900 = 1 (θ = 45°).

Answer: 1

12. Quick Revision Summary

Remember before the exam
  • F = dp/dt; impulse = Δp; momentum is conserved when Fext = 0.
  • Always draw a free-body diagram for each body.
  • Static friction ≤ μsN; kinetic friction = μkN; tan(angle of repose) = μs.
  • Lift: N = m(g ± a); free fall ⇒ weightless.
  • Banking: tanθ = v²/rg; flat curve vmax = √(μrg).
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