Momentum: p = mv (vector)

Impulse: J = ∫ F dt = Δp = change in momentum.

Impulse = area under F-t graph.

Equilibrium: ΣF = 0 → a = 0. For non‑equilibrium, ΣF = ma.

Common forces: weight (mg), normal (N), tension (T), friction (f), applied force.

For ideal pulley (massless, frictionless), tension is same on both sides of a string.

Constraint relation: For a system of masses connected by strings, sum of accelerations weighted by string length changes = 0.

When lift accelerates upward: N = m(g + a) (apparent weight increases)

When lift accelerates downward: N = m(g – a) (apparent weight decreases)

Free fall: a = g → N = 0 (weightlessness)

💡 NEET TIPS & SHORTCUTS

• Always draw FBD separately for each body.
• For two bodies connected by a string over a pulley, tension same if pulley ideal.
• When a force is applied on a block kept on another block, check limiting friction condition.
• In spring problems, force is proportional to extension (F = kx).

⚠️ COMMON MISTAKES

• Forgetting to include pseudo force in accelerating frames.
• Assuming action-reaction forces cancel when they act on different bodies.
• Not resolving forces correctly along axes.
• Confusing mass and weight.

Newton's 2nd law: ΣF = ma

Impulse: J = Δp = mΔv

Atwood machine: a = (m₂ – m₁)g/(m₁ + m₂), T = 2m₁m₂g/(m₁ + m₂)

Lift upward acc: N = m(g + a)

Lift downward acc: N = m(g – a)

Spring force: F = –kx