Newton's Second Law and Impulse

Free notes · JEE Main Physics

Formulas

\vec F = \frac{d\vec p}{dt}, \qquad \vec F = m\vec a \; (\text{constant } m)

Newton's second law — general and constant-mass forms

\vec J = \int_{t_1}^{t_2} \vec F \, dt = \Delta \vec p

Impulse–momentum theorem

Key points

Impulse is the area under the F–t graph. JEE problems give you graphs or piecewise forces precisely so you integrate — never multiply F × t blindly when the force varies.

Worked example

A 150 g ball hits a wall at 20 m/s and rebounds at 15 m/s along the same line. Contact lasts 0.05 s. Find the impulse and the average force on the ball.

Take direction away from the wall as positive: u = −20 m/s, v = +15 m/s
J = Δp = m(v − u) = 0.15 × (15 − (−20)) = 0.15 × 35 = 5.25 N·s
F_avg = J / Δt = 5.25 / 0.05 = 105 N

Answer: J = 5.25 N·s (away from wall); F_avg = 105 N