Question

Show that impulse is equal to the product of average force and the time interval for which the force acts.

Answer 1

$$\begin{gathered} \mathrm{According}\mathrm{to}\mathrm{newton}\text{'}\mathrm{s}\mathrm{second}\mathrm{law}\mathrm{of}\mathrm{motion} \\ \overrightarrow{F}=\frac{d\overrightarrow{P}}{dt} \\ \therefore \overrightarrow{\mathrm{F}}\mathrm{dt}=\mathrm{d}\overrightarrow{\mathrm{P}} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{within}\mathrm{the}\mathrm{limits},\mathrm{we}\mathrm{get} \\ {\int }_0^{\mathrm{t}}\overrightarrow{F}dt={\int }_{\overrightarrow{p_{\mathit{1}}}}^{\overrightarrow{p_{\mathit{2}}}}\mathrm{d}\overrightarrow{P} \\ \mathrm{Where},{\overrightarrow{P}}_{\mathit{1}}\mathit{}\mathrm{is}\mathrm{initial}\mathrm{linear}\mathrm{momentum}\mathrm{at}t\mathit{=}0\mathrm{and}\overrightarrow{P_{\mathit{2}}}\mathrm{is}\mathrm{final}\mathrm{momentum}\mathrm{at}\mathrm{time}t\mathit{.} \\ \mathrm{If}\mathit{}\stackrel{\mathit{\rightharpoonup }}{F_{\mathit{av}}}\mathit{}\mathrm{is}\mathrm{the}\mathrm{constant}\mathrm{force}\mathrm{during}\mathrm{this}\mathrm{time}.\mathrm{We}\mathrm{may}\mathrm{write} \\ \overrightarrow{F_{\mathit{av}}}{\left[t\right]}_0^{\mathrm{t}}={\left[\overrightarrow{p}\right]}_{\overrightarrow{{\mathrm{p}}_1}}^{\overrightarrow{{\mathrm{p}}_2}} \\ \Rightarrow \mathrm{Impulse},\overrightarrow{I}={\overrightarrow{F}}_{\mathrm{av}}\times t={\overrightarrow{p}}_2-{\overrightarrow{p}}_1 \\ \mathrm{Thus},\mathrm{impulse}\mathrm{is}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{average}\mathrm{force}\mathrm{and}\mathrm{time}\mathrm{interval}\mathrm{for}\mathrm{which}\mathrm{the}\mathrm{force}\mathrm{act}. \end{gathered}$$