Q.Define momentum and impulse. Starting from Newton's second law, $F = \dfrac{dp}{dt}$, show that for a force acting over a short interval $\Delta t$, the impulse delivered equals the change produced in the body's momentum.

Momentum of a body is defined as the product of its mass and velocity, $p = mv$, a vector quantity in the same direction as the body's velocity. Impulse is defined as the product of a force and the (typically short) time interval for which it acts, $J = F\,\Delta t$.

Starting from Newton's second law in its general momentum form, $F = \frac{dp}{dt}$ and considering a force that acts from time $t_1$ to time $t_2$, integrating both sides with respect to time gives $\int_{t_1}^{t_2} F\,dt = \int_{p_1}^{p_2} dp = p_2 - p_1 = \Delta p$ The left-hand side, $\int F\,dt$, is exactly the impulse delivered by the force over that interval (reducing to the simple product $F\,\Delta t$ when the force is constant, or is treated as an average force $F_{\text{avg}}$ over the interval). So $J = \Delta p$ -- the impulse delivered to a body during any interval of time is always exactly equal to the change produced in its momentum during that same interval, whatever the detailed way the force varies over that time.

Integrate Newton's second law, $F = dp/dt$, over the time interval during which the force acts; the left side becomes the impulse, the right side becomes the change in momentum.

• Treating impulse and momentum as the same quantity, rather than impulse being what changes momentum.
• Forgetting that this result holds even when the force is not constant, provided $F_{\text{avg}}\,\Delta t$ or the full integral is used.