Laws of Motion Connects Everything
This chapter is not standalone. JEE Advanced exclusively tests Newton's Laws combined with other chapters. Master these connections to unlock the hardest problems.
Chapter Connections
Newton's 2nd Law gives us acceleration. Work-Energy theorem gives us velocity. In multi-step problems, use both.
Step 1: Energy: v = √(2gh - 2μgh/tanθ) ... using work-energy
Step 2: Force: F = mv²/(2d) ... using Newton's 2nd law
Newton's Law gives acceleration 'a'. Kinematics equations (v=u+at, s=ut+½at²) use 'a'. These two are always combined.
Step 1: Newton's Law: a = (F-f)/m = (30-10)/5 = 4 m/s²
Step 2: Kinematics: s = ut + ½at² = 0 + ½×4×16 = 32 m
Newton's Laws have rotational analogs: τ = Iα (torque = moment of inertia × angular acceleration). Rolling bodies involve both.
• F = ma (linear) ↔ τ = Iα (rotational)
• p = mv (linear momentum) ↔ L = Iω (angular momentum)
• Impulse-momentum ↔ Impulsive torque-angular impulse
Rolling without slipping: friction provides torque for rotation AND translational force.
Gravity is a force. Newton's Laws govern its effects. Satellite motion = centripetal force = gravitational force.
• Satellite orbit: GMm/r² = mv²/r → orbital velocity
• Apparent weightlessness in satellite: centripetal = gravity
• Weight variation: W = mg at surface, W' = mg(R/(R+h))²
• Free fall: a = g (all objects, irrespective of mass — Newton's 2nd + Universal Gravity)
SHM is defined by F = -kx (restoring force). This is Newton's 2nd Law applied: F = ma → -kx = ma → a = -(k/m)x → ω² = k/m.
• Spring-mass system: Newton's Law gives F = -kx → SHM
• Block on spring in lift: effective g changes
• Conical pendulum: centripetal force from tension component
• Spring with friction: SHM equation modified with friction term
Buoyancy, viscosity, terminal velocity — all explained through Newton's Laws applied to fluid scenarios.
• Terminal velocity: Weight = Buoyancy + Viscous drag (Net F = 0)
• Object in fluid: Apparent weight = Weight - Buoyancy
• Fluid pressure: F = PA (pressure × area = force)
• Pascal's Law: Pressure transmitted equally → multiplied force
Newton's Laws + Energy + Circular Motion
A block of mass m is attached to a string of length L. The other end is fixed at a point O on the ceiling. The block is given a horizontal velocity at the lowest point. Find the minimum velocity at the lowest point for the block to complete a vertical circle. Then find the tension at the top and bottom.
At top: T + mg = mv_top²/L (both tension and weight provide centripetal force)
Minimum condition: T = 0 → v_top(min) = √(gL)
Concept 2 — Energy Conservation (bottom to top):
½mv_bottom² = ½mv_top² + mg(2L)
v_bottom² = v_top² + 4gL = gL + 4gL = 5gL
v_bottom(min) = √(5gL)
Concept 3 — Newton's Law at bottom:
T_bottom - mg = mv_bottom²/L = m(5gL)/L = 5mg
T_bottom = 6mg