📚 StudyHub

⚛️ Physics  ·  Class 11  ·  NEET & JEE

Laws of Motion

Newton's three laws, friction, circular motion, and free body diagrams.

Start Laws of Motion Chapterwise Test - 100% Free →
Reading time~9 min
Revision time~3 min
Last updated2026-08-18
1Read the chapter~9 min

🎯 Key Points

  • F=ma (Second Law); action-reaction pairs act on DIFFERENT bodies, never cancel each other on the same body (Third Law)
  • Static friction (fs ≤ μ_s N) adjusts to prevent sliding; kinetic friction (fk = μ_k N) is constant once sliding, and μ_k < μ_s
  • Centripetal force F=mv²/r always points toward the centre - it's not a new force, just the net result of existing forces (tension, gravity, normal, friction)
  • Banking angle: tan θ = v²/rg - designed so friction isn't needed at the design speed
  • Impulse J = FΔt = Δp; momentum is conserved when net external force is zero

Newton's Laws

  • First Law: An object remains at rest or in uniform motion unless acted on by an external force (inertia)
  • Second Law: F = ma (net force = mass × acceleration)
  • Third Law: Every action has an equal and opposite reaction

Friction

incline surfaceθmgNfmg sinθmg cosθ

Free-body diagram of a block on an incline: weight (mg) resolves into mg sinθ along the slope and mg cosθ into the slope, balanced by the normal force N and friction f.

  • Static friction: fs ≤ μ_s·N (maximum just before sliding)
  • Kinetic friction: fk = μ_k·N (constant during sliding, μ_k < μ_s)
  • Angle of friction: tan(λ) = μ

Circular Motion

  • Centripetal acceleration: a = v²/r = ω²r
  • Centripetal force: F = mv²/r (directed toward centre)
  • Banking angle: tan(θ) = v²/rg
  • Critical speed at top of loop: v = √(rg)

Impulse and Momentum

  • Impulse J = F·Δt = Δp (change in momentum)
  • Conservation of momentum: if ΣFext = 0, then Σp = constant

Free Body Diagrams (FBD) - The Method

  • Isolate ONE body and draw only the forces acting ON it: weight mg (always downward), normal reaction N (perpendicular to the contact surface), tension T (along the string, away from the body), applied force, and friction f (along the surface, opposing relative motion/tendency).
  • Choose convenient axes (for inclines, take axes ALONG and PERPENDICULAR to the slope), then write ΣF = ma separately along each axis.
  • Never include forces the body exerts on others - an action-reaction pair acts on two DIFFERENT bodies, so the two never appear in the same FBD.
  • Solve the resulting simultaneous equations for the unknowns (acceleration, tension, normal force, etc.).

Connected Bodies: Pulley, Lift, and Chain

  • Bodies on a table joined over a pulley: for a hanging mass m₁ pulling a mass m₂ on a smooth horizontal surface, a = m₁g/(m₁ + m₂) and tension T = m₁m₂g/(m₁ + m₂).
  • Atwood machine (two masses over a pulley): a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂).
  • Apparent weight in a lift: N = m(g + a) when accelerating up, N = m(g − a) when accelerating down; N = mg at rest or constant velocity; N = 0 (weightlessness) in free fall (a = g).
  • Inextensible string: all connected bodies share the same magnitude of acceleration, and tension is uniform in a massless string over a frictionless pulley.

Friction on an Inclined Plane

  • On an incline of angle θ, the weight resolves into mg sinθ (down the slope) and mg cosθ (into the surface), so N = mg cosθ.
  • Angle of repose (α): the maximum incline angle at which a block just stays at rest → tan α = μ_s. If θ ≤ angle of repose, the block does not slide.
  • Block sliding DOWN a rough incline: acceleration a = g(sinθ − μ_k cosθ).
  • Block pushed UP a rough incline (friction acts down-slope): retardation a = g(sinθ + μ_k cosθ).

Banking of Roads (With Friction)

  • For a frictionless banked road, the safe speed is fixed: v = √(rg tanθ), i.e. tanθ = v²/(rg).
  • With friction (coefficient μ), a RANGE of safe speeds is allowed: vmax = √[rg(tanθ + μ)/(1 − μ tanθ)] and vmin = √[rg(tanθ − μ)/(1 + μ tanθ)].
  • On a flat (unbanked) road, the turn relies on friction alone: vmax = √(μ rg).

Pseudo Forces in Non-Inertial Frames

  • Newton's laws hold directly only in inertial frames (non-accelerating). In an accelerating (non-inertial) frame, they can still be applied by adding a pseudo force = −m·aframe on every body, opposite to the frame's acceleration.
  • Example: inside a car accelerating forward with acceleration a, a passenger feels a backward pseudo force ma; a hanging pendulum settles at tanφ = a/g from the vertical.
  • Pseudo force explains the "apparent weight" in a lift and the outward "centrifugal force" felt in the rotating frame of a turning vehicle - it is a bookkeeping force, not a real interaction.

🚀 JEE Advanced Edge

Pseudo force in non-inertial frames: When analysing motion from inside an accelerating frame (like a lift or accelerating car), add a pseudo force = −ma_frame (opposite to the frame's acceleration) to make Newton's laws apply in that frame. This is the fastest way to solve "apparent weight in a lift" or "pendulum hanging in an accelerating car" problems without switching to the ground frame.

Connected bodies / Atwood machine with pulleys: For two masses connected over a pulley, write F=ma for EACH mass separately (using the SAME tension T and SAME magnitude of acceleration a, since the string is inextensible), then solve the two equations simultaneously: a = (m₁−m₂)g/(m₁+m₂) for a simple Atwood machine.

Worked problem: A block of mass 2 kg on a rough incline (θ=30°, μ=0.5) - will it slide? Approach: Compare tan θ to μ: tan(30°)≈0.577 > μ=0.5? No, actually tan(30°)=0.577 > 0.5, so since tan θ > μ, gravity's component down the slope EXCEEDS the maximum available static friction, and the block DOES slide. (If tan θ ≤ μ, it would stay put.)

Conservation of Linear Momentum

  • When the net external force on a system is zero, its total linear momentum remains constant: this follows directly from Newton's second law.
  • For an isolated two-body system, m1u1 + m2u2 = m1v1 + m2v2, so any momentum lost by one body is gained by the other.
  • Recoil of a gun: the forward momentum of the bullet equals the backward momentum of the gun, so the gun recoils.
  • Explosion of a stationary bomb: the fragments fly off with momenta that add vectorially to zero.
  • Momentum conservation holds during collisions and explosions even when kinetic energy is not conserved.

Equilibrium of Concurrent Forces

  • A particle is in equilibrium when the vector sum of all forces acting on it is zero, so its acceleration is zero.
  • For coplanar forces, equilibrium requires the sum of components along both the x and y axes to be zero separately.
  • Lami's theorem: for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two.
  • The three forces in equilibrium can be represented in magnitude and direction by the three sides of a triangle taken in order.
  • Equilibrium may be static (body at rest) or dynamic (body moving with constant velocity).

Inertia and Its Types

  • Inertia is the inherent property of a body to resist any change in its state of rest or of uniform motion; it is measured by mass.
  • Inertia of rest: a body at rest tends to stay at rest, e.g. dust flies off when a carpet is beaten.
  • Inertia of motion: a moving body tends to keep moving, e.g. a passenger lurches forward when a bus stops suddenly.
  • Inertia of direction: a body resists change in its direction of motion, e.g. mud flies off tangentially from a spinning wheel.
  • Greater mass means greater inertia; this is the basis of Newton's first law, also called the law of inertia.

Common Forces in Mechanics

  • Weight is the gravitational force mg acting vertically downward on a body.
  • The normal reaction N is the contact force exerted by a surface perpendicular to it.
  • Tension is the pulling force transmitted along a string or rope; for an ideal massless string it is the same throughout.
  • The spring force obeys Hooke's law, F = -kx, directed so as to restore the spring to its natural length.
  • Contact forces (normal and friction) arise from electromagnetic interactions between surfaces, while weight is a non-contact (field) force.
2Revise~3 min before the exam

📐 Formula Sheet

  • Newton's second law: Fnet = ma = dp/dt
  • Momentum: p = mv  |  Impulse: J = FΔt = Δp
  • Friction: fs(max) = μsN (limiting)  |  fk = μkN (kinetic, μk < μs)
  • Inclined plane: along the slope mg·sinθ; perpendicular N = mg·cosθ; slides when tanθ > μs
  • Circular motion: Fc = mv²/r = mω²r
  • Banked road: vmax = √(rg·tanθ) (frictionless)  |  ideal banking angle tanθ = v²/rg
  • Conservation of momentum: total p is constant when no external force acts
3Practiceapply it

✍️ Worked Examples

Example 1 - Two blocks in contact
Q: A 3 kg and a 2 kg block sit touching on a frictionless surface. A 10 N force pushes the 3 kg block. Find the acceleration and the contact force between them.
Step 1 - Treat both as one system: a = F/(m₁ + m₂) = 10/5 = 2 m/s².
Step 2 - Isolate the 2 kg block: the only horizontal force on it is the contact force N.
Step 3 - Apply F = ma to it alone: N = 2 × 2 = 4 N.
Answer: a = 2 m/s², contact force = 4 N. Note: checking with the 3 kg block: 10 − 4 = 6 = 3 × 2 ✓.

Example 2 - Block on a rough incline
Q: A 10 kg block rests on a 30° incline with μs = 0.5. Does it slide? (g = 10 m/s²)
Step 1 - Driving force down the slope: mg·sinθ = 10 × 10 × 0.5 = 50 N.
Step 2 - Maximum static friction: N = mg·cosθ = 10 × 10 × 0.866 = 86.6 N, so fmax = 0.5 × 86.6 = 43.3 N.
Step 3 - Compare: 50 N > 43.3 N, so friction cannot hold it.
Answer: yes, it slides. Shortcut: tan30° = 0.577 > μs = 0.5 gives the same verdict instantly.

Example 3 - Recoil of a gun
Q: A 4 kg gun fires a 20 g bullet at 200 m/s. Find the recoil velocity of the gun.
Step 1 - Momentum before firing is zero (everything at rest).
Step 2 - Conservation: 0 = mbvb + mgvg.
Step 3 - Substitute: 0 = (0.02)(200) + 4vg ⇒ 4vg = −4 ⇒ vg = −1 m/s.
Answer: 1 m/s backwards. Trap: leaving the bullet mass in grams makes the recoil 1000× too large.

Start Laws of Motion Chapterwise Test - 100% Free →

Frequently Asked Questions - Laws of Motion

What are the key concepts in Laws of Motion?
Newton's three laws, friction, circular motion, and free body diagrams.
Is Laws of Motion important for NEET & JEE?
Yes. Laws of Motion is part of the Physics Class 11 NCERT syllabus and is directly tested in NEET and JEE examinations. StudyHub provides structured notes, diagrams, and practice questions covering all exam-level subtopics.
How can I practice Laws of Motion questions on StudyHub?
Open StudyHub and select Physics → Laws of Motion. Choose Easy, Medium, or Hard difficulty. Hard-tier questions are at NEET & JEE level with full step-by-step explanations.

References

  1. NCERT Class 11 Physics Textbook - Chapter: Laws of Motion
  2. CBSE Curriculum - Physics (Class 11)
  3. NTA NEET UG Official Syllabus - subject-wise topic list
  4. NTA JEE Main Official Syllabus - subject-wise topic list