Impulse

Consider the motion of a single particle (i.e., a body of negligible spatial extent). Newton's second law of motion, (1.19), can be written

$${\bf f} = m\,\frac{d{\bf v}}{dt},$$ (1.22)

which implies that

$${\bf f}\,dt = m\,d{\bf v}.$$ (1.23)

Suppose that the particle in question has an instantaneous velocity ${\bf v}_1$ at an initial time $t_1$, and an instantaneous velocity ${\bf v}_2$ at a final time $t_2$. Integrating the previous equation between the initial and the final time, we obtain

$$\int_{t_1}^{t_2} {\bf f}(t)\,dt = m\int_{{\bf v}_1}^{{\bf v}_2} d{\bf v} = m\,({\bf v}_2-{\bf v}_1),$$ (1.24)

where we have taken into account the fact that the force ${\bf f}$ is, in general, a function of time. The quantity

$${\bf I} = \int_{t_1}^{t_2}\,{\bf f}(t)\,dt$$ (1.25)

is known as impulse, and is essentially the `area' under the ${\bf f}(t)$ curve between times $t_1$ and $t_2$. It is clear from the previous two equations that

$${\bf I} = m\,{\mit\Delta}{\bf v},$$ (1.26)

where ${\mit\Delta}{\bf v}= {\bf v}_2-{\bf v}_1$ is the change in the particle's velocity between the initial and the final times. Equation (1.18) can be combined with the previous two equations to give

$${\bf I} = {\mit\Delta}{\bf p}.$$ (1.27)

In other words, the net impulse acting on a particle between an initial and a final time is equal to the change in the momentum of the particle between the same two times.