Impulse & MomentumJEE Main

Area under the F–t curve changes momentum — interactive Physics simulation for IIT-JEE.

Concept

Impulse is the time-integral of force — the area under the F–t graph — and it equals the change in momentum. A large force for a short time and a small force for a long time can deliver the same momentum change; that choice is the whole science of cushioning and impact.

Key formula

\vec{J} = \int \vec{F}\,dt = \Delta\vec{p} = m\vec{v} - m\vec{u}, \qquad F_{avg} = \frac{J}{\Delta t}

Derivation

Newton's second law in momentum form: $\vec F = d\vec p/dt$ . Integrating over the contact time,

$ $\int_0^{T} \vec F\,dt = \vec p(T) - \vec p(0)$ $

For the pulse shapes here the areas are $FT$ (rectangle), $\tfrac12 FT$ (triangle), and $\tfrac{2}{\pi}FT$ (half-sine) — the graph, not the peak, is what matters.

Scenarios to explore

Impulse & Momentum — Area under the F–t curve becomes momentum change.

Real-world applications

Airbags & crumple zones: stretch Δt to slash the peak force.
A cricketer pulling hands back while catching.
Rocket thrust integrated over burn time = momentum gained.

JEE exam tips

Bounce problems: |Δp| = m(v + u) when direction reverses; this is where most sign errors live.
Impulsive forces (huge F, tiny t) let you ignore gravity during contact.
For F–t graphs, just compute area — including negative lobes below the axis.

Common mistakes

Equating impulse to peak force × time regardless of pulse shape.
Dropping vector signs — a ball bouncing back has Δp = m(v+u), not m(v−u).
Confusing impulse (N·s) with work (J) — different integrals, different quantities.

Exam traps to avoid

N·s and kg·m/s are the same unit — expect either in options.
Same impulse ≠ same peak force. Cushioned landings survive because Δt grows.