Laws of Motion

Covers NCERT Class 11 Physics Part I, Chapter "Laws of Motion". Section numbers below follow the standard chapter sequence — the chapter number itself can shift by ±1 between print editions, the internal section order does not.

Foundation Advanced High-Yield Revision
Advanced NTA favourite High weightage

⭐ NTA-favourite subtopics in this chapter

Friction (static vs kinetic, angle of friction, minimum force to move a block), circular motion applications (banking of roads with/without friction, conical pendulum), pulley & connected-block systems (including pseudo force in accelerating lifts), Lami's theorem for concurrent-force equilibrium, and momentum-conservation collision problems.

Exam-oriented notes

NCERT-tagged theory, simplified

NCERT §1–2

Aristotle's fallacy: force is not needed to keep a body moving at constant velocity — Galileo/Newton's law of inertia corrects this. A body continues in its state of rest or uniform motion unless acted on by a net external force.

NCERT §4

Newton's Second Law: F = dp/dt = ma (for constant mass). This is the working equation for nearly every numerical in this chapter — always define a consistent positive direction before writing it.

NCERT §5

Third Law: action and reaction act on different bodies and are simultaneous — they never cancel each other for the same object. A very common assertion-reason trap confuses this with equilibrium of a single body.

NCERT §6

Conservation of linear momentum follows directly from the third law for an isolated system: total momentum before = total momentum after, applied component-wise for 2D collisions.

NCERT §7

Equilibrium of a particle under three concurrent, coplanar forces: Lami's theorem, F₁/sinα = F₂/sinβ = F₃/sinγ, where each angle is opposite to the corresponding force.

NCERT §8

Static friction is self-adjusting up to a maximum f_s(max) = μₛN; once motion starts, kinetic friction f_k = μₖN applies and is very slightly less than the maximum static value (μₖ < μₛ generally).

NCERT §9

Rolling friction is much smaller than sliding friction — this is why wheels are more efficient than dragging, and is a frequent one-line factual MCQ.

NCERT §10

Circular motion needs a net centripetal (centre-seeking) force, F = mv²/r. Banking of roads reduces reliance on friction: for a frictionless banked curve, tanθ = v²/(rg); with friction, the safe speed range widens on both ends.

NCERT §10

Conical pendulum: bob moves in a horizontal circle, string sweeps a cone; tanθ = v²/(rg) again — recognise this as structurally the same equilibrium as banking, just with tension replacing normal reaction.

Problem-solving §11

Pseudo force (= −ma_frame) is added only in a non-inertial (accelerating) reference frame, e.g. inside a lift or accelerating truck — direction is always opposite to the frame's acceleration.

Printable

Formula sheet

Laws of Motion

Newton's second law
F = ma = dp/dt
Impulse
J = FΔt = Δp
Momentum conservation
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Lami's theorem
F₁/sinα = F₂/sinβ = F₃/sinγ
Max static friction
f_s(max) = μₛN
Kinetic friction
f_k = μₖN
Angle of friction
tanλ = μₛ
Centripetal force
F = mv² / r
Banking (frictionless)
tanθ = v² / (rg)
Banking (with friction, max speed)
v²_max = rg(μ+tanθ)/(1−μtanθ)
Atwood machine
a = (m₁−m₂)g / (m₁+m₂)
Atwood string tension
T = 2m₁m₂g / (m₁+m₂)
Pseudo force
F_pseudo = −ma_frame
Apparent weight in lift (acc. a upward)
N = m(g+a)
PYQ-style practice

Practice set, NEET pattern

Original questions modelled on recurring NEET question types for this chapter — not verbatim reproductions of any official paper.

Q1. A block of mass 5 kg rests on a rough horizontal surface (μₛ = 0.4). The minimum horizontal force needed to just move the block is: (g = 10 m/s²)
  • 10 N
  • 20 N
  • 40 N
  • 50 N
Show solution
f_s(max) = μₛN = 0.4×5×10 = 20 N. Answer: B
Q2. A car takes a turn on a frictionless banked road of radius 100 m banked at θ where tanθ = 0.5. The maximum safe speed is nearest to: (g = 10 m/s²)
  • 10 m/s
  • 22 m/s
  • 50 m/s
  • 70 m/s
Show solution
v² = rg tanθ = 100×10×0.5 = 500 ⇒ v ≈ 22.4 m/s. Answer: B
Q3. Two blocks of masses 3 kg and 2 kg connected by a string pass over a frictionless pulley (Atwood machine). The acceleration of the system is: (g = 10 m/s²)
  • 1 m/s²
  • 2 m/s²
  • 5 m/s²
  • 10 m/s²
Show solution
a = (m₁−m₂)g/(m₁+m₂) = (3−2)×10/5 = 2 m/s². Answer: B
Q4. A person stands on a weighing scale inside a lift accelerating upward at 2 m/s². If their true weight is 600 N, the scale reads: (g = 10 m/s²)
  • 480 N
  • 600 N
  • 620 N
  • 720 N
Show solution
m = 60 kg. N = m(g+a) = 60×12 = 720 N. Answer: D
Q5. A bullet of mass 20 g moving at 300 m/s embeds in a stationary block of mass 2 kg on a frictionless surface. The common velocity after collision is:
  • 1.5 m/s
  • 3 m/s
  • 6 m/s
  • 15 m/s
Show solution
Momentum conservation: 0.02×300 = (2.02)v ⇒ v ≈ 2.97 ≈ 3 m/s. Answer: B
Q6. Assertion: A rocket accelerates forward by ejecting mass backward, even in the vacuum of space. Reason: Newton's third law does not require a medium to act through.
  • Both true, reason explains assertion
  • Both true, reason does not explain assertion
  • Assertion true, reason false
  • Both false
Show solution
Rocket propulsion is a direct momentum-conservation/third-law consequence and needs no external medium — unlike friction or sound. Answer: A
Daily Practice Problems

DPP — Laws of Motion (15 questions)

Scaled from direct NCERT application (Q1–Q6) to mixed NEET-level difficulty (Q7–Q15). Attempt in 18 minutes, then check the answer key.

1.State Newton's first law of motion in your own words.
2.A force of 10 N acts on a 2 kg mass at rest. Find the acceleration and velocity after 4 s.
3.Define coefficient of static friction and state its typical range of values.
4.Why does a gun recoil when fired? Name the law involved.
5.A 1000 kg car moving at 20 m/s is brought to rest in 5 s. Find the average retarding force.
6.State the condition for equilibrium of a particle under three concurrent forces.
7.A block on a 30° incline (μ = 0.2) — will it slide on its own? Justify with the angle of friction.
8.A conical pendulum of length 1 m makes an angle of 30° with the vertical. Find the speed of the bob. (g = 10 m/s²)
9.Two masses 4 kg and 6 kg are connected via a pulley on a frictionless table with 6 kg hanging. Find the system's acceleration.
10.A lift accelerates downward at 3 m/s². Find the apparent weight of a 50 kg person. (g = 10 m/s²)
11.A ball of mass 0.5 kg strikes a wall with 10 m/s and rebounds with 8 m/s. Find the impulse imparted by the wall.
12.Explain why it is easier to pull a lawn roller than to push it, using force components.
13.A car of mass 1200 kg negotiates a curve of radius 90 m banked at tanθ = 0.25 with μ = 0.1. Find the maximum safe speed.
14.Two bodies of masses m and 2m moving in opposite directions with speed v collide perfectly inelastically. Find the velocity of the combined mass.
15.Assertion: A cyclist leans while turning. Reason: leaning provides the necessary centripetal force component. Judge the pair.
Show answer key
1. Body stays at rest/uniform velocity unless acted on by net external force  |  2. a = 5 m/s², v = 20 m/s  |  3. μₛ = f_s(max)/N, typically 0.1–1.0  |  4. Newton's third law (momentum conservation)  |  5. F = 4000 N  |  6. Vector sum of forces = 0 (Lami's theorem applies)  |  7. tan30° ≈ 0.577 > μ = 0.2, so it slides  |  8. v ≈ 2.4 m/s  |  9. a = 6 m/s²  |  10. N = 350 N  |  11. J = 9 kg·m/s (direction reversed)  |  12. Pushing adds a downward force component increasing normal force and friction; pulling reduces it  |  13. ≈17 m/s  |  14. v/3 in direction of 2m's motion  |  15. Both true, reason correctly explains assertion
Printable mnemonic chart

Quick-recall chart for Laws of Motion

🧠 "Same String, Same Tension"

For an ideal string (massless, inextensible) over a frictionless pulley, tension is identical on both sides — only changes if the pulley itself has mass/friction, or the string has mass. Say it as you set up every pulley problem.

🧠 "Push Up, Weigh More · Fall Down, Weigh Less"

Lift accelerating upward → apparent weight N = m(g+a) → feels heavier. Lift accelerating downward → N = m(g−a) → feels lighter. Free fall (a=g) → N = 0 → weightlessness.

🧠 Friction direction rule

Friction always opposes relative motion (or attempted motion) between the two surfaces in contact — never the motion of the body "in general." Always draw the relative-slip direction first, then friction points opposite to it.