NEET UG Physics · Chapter 3

Laws of Motion

Newton’s laws, impulse, connected bodies, friction and dynamics of circular motion.

NCERT Class 11NEET: 2 Qs per paper 11 min read

1. Chapter at a Glance

Kinematics told us how things move; the laws of motion tell us why. Newton's three laws connect force with the change in motion. Using free-body diagrams, they solve problems on blocks, strings, pulleys, lifts, friction and circular motion. NEET asks about two questions from this chapter every year and the ideas return in almost every mechanics chapter.

What NEET asks most
  • Acceleration and tension in connected bodies and pulleys.
  • Apparent weight in an accelerating lift.
  • Impulse and conservation of momentum (recoil).
  • Static and kinetic friction; angle of repose.
  • Maximum speed on flat and banked curves.

2. Newton's Laws

  • First law (law of inertia): a body stays at rest or keeps moving uniformly in a straight line unless an external force acts on it. Inertia depends on mass.
  • Second law: the rate of change of momentum equals the net external force, F = dp/dt. For constant mass, F = ma. Unit: newton (1 N = 1 kg m s⁻²).
  • Third law: to every action there is an equal and opposite reaction. The two forces act on different bodies, so they never cancel.
  • Momentum p = mv. Impulse J = F × Δt = Δp. A cricketer lowers the hands while catching to increase Δt and reduce the force.
  • Conservation of momentum: if no net external force acts, the total momentum of a system is constant. Examples: recoil of a gun, rocket propulsion, a boy jumping off a boat.

3. Free-Body Diagrams and Applications

For each body, draw all forces acting on it — weight mg, normal reaction N, tension T, friction f, applied force — and apply ΣF = ma along convenient axes.

3 kg 2 kg T f T mg
Connected bodies: the same tension T acts on both blocks; friction opposes the table block's motion.
SystemResultNEET Note
Atwood machine (m₁ > m₂)a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂)Light, frictionless pulley
Lift going up with aN = m(g + a)Person feels heavier
Lift going down with aN = m(g − a)Zero in free fall
Smooth inclinea = g sinθIndependent of mass
Bodies pushed together by Fa = F/(m₁ + m₂); contact force = m₂aPush applied on m₁

4. Friction

  • Friction opposes relative motion (or its tendency) between surfaces in contact.
  • Static friction is self-adjusting: it equals the applied force up to a limit fs,max = μsN.
  • Kinetic friction fk = μkN acts once sliding starts; μk < μs. Rolling friction is the smallest.
  • Angle of repose α: the maximum incline on which a body stays at rest; tan α = μs.
  • Sliding down a rough incline: a = g(sinθ − μkcosθ).
  • Friction is necessary for walking and for vehicles to move; it is reduced by lubricants and ball bearings.

5. Dynamics of Circular Motion

  • A body in circular motion needs a net force mv²/r towards the centre (centripetal force). It is provided by tension, friction, gravity or a component of the normal force — it is not a separate force.
  • Car on a flat curve: friction supplies it, so vmax = √(μsrg).
  • Banked road without friction: tanθ = v²/rg. With friction: vmax = √[rg(μ + tanθ)/(1 − μ tanθ)].
  • Vertical circle on a string: least speed at the top √(gr); at the bottom √(5gr); tension at bottom exceeds that at top by 6mg.

6. Formula Master Table

QuantityFormula
Second lawF = dp/dt = ma
ImpulseJ = FΔt = Δp
RecoilMV = mv
Force from a stream of bulletsF = n m v (n per second)
Frictionfs ≤ μsN; fk = μkN
Angle of reposetan α = μs
Flat curve / bankingv = √(μrg); tanθ = v²/rg

7. Master Summary Table

ConceptKey pointNEET Note
InertiaMeasured by massPassengers jerk forward when a bus stops
Action–reactionEqual, opposite, on different bodiesNever cancel each other
Momentum conservationHolds when Fext = 0Gun recoil, rockets, explosions
Apparent weightN = m(g ± a)Zero in free fall
Frictionμk < μsCheck limiting friction first
Circular motionNeeds mv²/r inwardBanking reduces reliance on friction

8. Mind Map

Laws of Motion
Newton's laws
  • Inertia
  • F = dp/dt
  • Action–reaction
Momentum
  • Impulse
  • Conservation
  • Recoil
Applications
  • Pulleys, lifts
  • Connected bodies
  • Inclines
Friction & circles
  • Static, kinetic
  • Angle of repose
  • Banking, vertical circle

9. Tips and Tricks

Shortcuts
  • For a system moving together: a = (net external force)/(total mass). Then isolate one body to find a tension or contact force.
  • Before using μk, compare the applied force with μsN; if smaller, the body stays still and friction equals the applied force.
  • A block will slide down an incline by itself only if tanθ > μs.
  • In a freely falling lift, a pendulum stops swinging and a spring balance reads zero.
Common NEET traps
  • Writing friction = μN for a body that is not sliding.
  • Drawing "centripetal force" as an extra force in a free-body diagram.
  • Thinking a horse cannot pull a cart because action and reaction cancel — they act on different bodies.

10. Important One-liners

  • Mass is the measure of inertia of a body.
  • Impulse equals the change in momentum.
  • A rocket moves forward by pushing gases backward (momentum conservation).
  • Kinetic friction is less than limiting static friction.
  • The coefficient of friction does not depend on the area of contact (approximately).
  • Roads are banked on curves so that the normal force provides part of the centripetal force.
  • A body in uniform circular motion is not in equilibrium.

11. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take g = 10 m s⁻². Click the button under a question to see the answer and working.

Average level

N1Average

What constant force is needed to take a 1500 kg car from rest to 20 m s⁻¹ in 10 s (ignore friction)?

Show answer & solution

Concept: F = ma with a = Δv/Δt.

  1. a = 20/10 = 2 m s⁻².
  2. F = 1500 × 2 = 3000 N.

Answer: 3000 N

N2Average

A fielder stops a 0.15 kg ball moving at 30 m s⁻¹ in 0.1 s. Find the average force on the hands. What would it be if the ball were stopped in 0.01 s?

Show answer & solution

Concept: F = Δp/Δt.

  1. Δp = 0.15 × 30 = 4.5 kg m s⁻¹.
  2. F = 4.5/0.1 = 45 N.
  3. In 0.01 s: F = 450 N — ten times larger, which is why fielders pull their hands back.

Answer: 45 N (450 N if stopped in 0.01 s)

N3Average

Blocks of 4 kg and 6 kg lie on a smooth floor, joined by a light string. A 20 N horizontal force pulls the 6 kg block. Find the acceleration and the string tension.

Show answer & solution

Concept: treat both as one system, then isolate the rear (4 kg) block.

  1. a = 20/(4 + 6) = 2 m s⁻².
  2. The string alone pulls the 4 kg block: T = 4 × 2 = 8 N.

Answer: 2 m s⁻²; 8 N

N4Average

A 50 kg girl stands on a weighing scale in a lift that accelerates downward at 3 m s⁻². What does the scale read?

Show answer & solution

Concept: the scale reads the normal force N = m(g − a) for downward acceleration.

  1. N = 50 × (10 − 3) = 350 N.
  2. That corresponds to a "reading" of 35 kg-wt instead of 50 kg-wt.

Answer: 350 N

N5Average

A 5 kg box is pushed across a floor (μk = 0.3) by a horizontal force of 25 N. Find its acceleration.

Show answer & solution

Concept: net force = applied − kinetic friction.

  1. fk = 0.3 × 5 × 10 = 15 N.
  2. a = (25 − 15)/5 = 2 m s⁻².

Answer: 2 m s⁻²

Advanced level

N6Advanced

In the figure above, a 3 kg block on a table (μk = 0.2) is tied over a light pulley to a hanging 2 kg block. Find the acceleration and the tension.

Show answer & solution

Concept: hanging block: 2g − T = 2a; table block: T − μ(3g) = 3a.

  1. Friction on the table block = 0.2 × 30 = 6 N.
  2. Add the equations: 20 − 6 = (2 + 3)a ⇒ a = 2.8 m s⁻².
  3. T = 2(10 − 2.8) = 14.4 N.

Answer: 2.8 m s⁻²; 14.4 N

N7Advanced

A 0.5 kg ball on a 2.5 m string is whirled in a vertical circle, just managing to complete it. Find its speed at the top and the string tension at the bottom.

Show answer & solution

Concept: just completing ⇒ T = 0 at the top, so mg = mvtop²/r; then energy conservation.

  1. vtop = √(gr) = √25 = 5 m s⁻¹.
  2. vbottom² = vtop² + 2g(2r) = 25 + 100 = 125.
  3. Tbottom = mg + mv²/r = 5 + 0.5 × 125/2.5 = 5 + 25 = 30 N (= 6mg).

Answer: 5 m s⁻¹; 30 N

N8Advanced

A block rests on an incline of 37° (sin 37° = 0.6, cos 37° = 0.8) with μs = 0.5 and μk = 0.5. Will it slide? If so, find its acceleration.

Show answer & solution

Concept: it slides if tanθ > μs.

  1. tan 37° = 0.75 > 0.5, so it slides.
  2. a = g(sinθ − μkcosθ) = 10(0.6 − 0.4).
  3. a = 2 m s⁻² down the slope.

Answer: yes; 2 m s⁻²

N9Advanced

A machine gun fires 10 bullets per second, each of mass 40 g, at 500 m s⁻¹. What average force must the gunner apply to hold it steady?

Show answer & solution

Concept: force = rate of change of momentum = n × m × v.

  1. Momentum per bullet = 0.04 × 500 = 20 kg m s⁻¹.
  2. Per second: 10 × 20 = 200 kg m s⁻².
  3. F = 200 N.

Answer: 200 N

N10Advanced

(a) Find the maximum safe speed of a car on an unbanked curve of radius 80 m if μs = 0.5. (b) At what angle should the road be banked so that a car at this speed needs no friction?

Show answer & solution

Concept: flat curve: μmg = mv²/r; banking: tanθ = v²/rg.

  1. (a) v = √(μrg) = √(0.5 × 80 × 10) = √400 = 20 m s⁻¹ (72 km h⁻¹).
  2. (b) tanθ = 400/(80 × 10) = 0.5.
  3. θ = tan⁻¹(0.5) ≈ 26.6°.

Answer: 20 m s⁻¹; about 26.6°

12. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

A passenger in a moving bus falls forward when the bus stops suddenly. This is due to:

  1. Inertia of rest
  2. Inertia of motion
  3. Newton's third law
  4. Friction
Show solution

The upper body tends to continue moving forward while the feet stop with the bus — inertia of motion. Inertia of rest explains falling backward when a bus starts suddenly.

Answer: (B)

Q2Easy

The area under a force–time graph gives:

  1. Work
  2. Power
  3. Impulse
  4. Acceleration
Show solution

∫F dt = impulse = change in momentum. Work is the area under a force–displacement graph.

Answer: (C)

Q3Medium

A 10 kg gun fires a 20 g bullet at 250 m s⁻¹. The recoil speed of the gun is:

  1. 0.25 m s⁻¹
  2. 0.5 m s⁻¹
  3. 2.5 m s⁻¹
  4. 5 m s⁻¹
Show solution

MV = mv ⇒ V = 0.02 × 250/10 = 0.5 m s⁻¹, opposite to the bullet.

Answer: (B)

Q4Medium

A block just begins to slide down an incline when the angle is 30°. The coefficient of static friction is:

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

At the angle of repose, μs = tan30° = 1/√3 ≈ 0.577.

Answer: (C)

Q5Medium

Masses of 3 kg and 5 kg hang over a light frictionless pulley. The acceleration of the system is (g = 10 m s⁻²):

  1. 1.25 m s⁻²
  2. 2.5 m s⁻²
  3. 5 m s⁻²
  4. 10 m s⁻²
Show solution

a = (5 − 3) × 10/(5 + 3) = 20/8 = 2.5 m s⁻².

Answer: (B)

Q6Medium

A 2 kg block lies on a floor with μs = 0.4. A horizontal force of 5 N is applied. The friction force is (g = 10 m s⁻²):

  1. 8 N
  2. 5 N
  3. 3 N
  4. 0
Show solution

Limiting friction = 0.4 × 20 = 8 N. Since 5 N < 8 N, the block stays at rest and static friction adjusts to exactly 5 N.

Answer: (B)

Q7Hard

A spring balance in a lift reads 60 N for a body when the lift is at rest. If the lift falls freely, the reading becomes:

  1. 60 N
  2. 120 N
  3. 30 N
  4. 0
Show solution

In free fall a = g, so the reading m(g − a) = 0. The body is weightless relative to the lift.

Answer: (D)

Q8Hard

A road of radius 100 m is banked for a speed of 20 m s⁻¹ with no friction. The banking angle θ satisfies (g = 10 m s⁻²):

  1. tanθ = 0.2
  2. tanθ = 0.4
  3. tanθ = 2
  4. sinθ = 0.4
Show solution

tanθ = v²/rg = 400/1000 = 0.4.

Answer: (B)

13. Quick Revision Summary

Remember before the exam
  • F = dp/dt = ma; impulse = FΔt = Δp.
  • Action and reaction act on different bodies; momentum is conserved when Fext = 0.
  • Lift: N = m(g ± a); free fall ⇒ weightless.
  • fs ≤ μsN (self-adjusting), fk = μkN; tan(repose) = μs.
  • Flat curve v = √(μrg); banking tanθ = v²/rg; vertical circle √(gr) at top.
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