A force-time $(F-t)$ graph for a linear motionis shown in following figure. The segments shown are circular.…
A force-time $(F-t)$ graph for a linear motionis shown in following figure. The segments shown are circular. The linear momentum gained between 0 and $8 \mathrm{~s}$ is
$-2 \pi N s$
$0$
$4 \pi \mathrm{Ns}$
$6 \pi \mathrm{Ns}$
Solution
From graph, the force is varying sinusoidally with time by equation,
$F=-2 \cdot \cos \omega t$ ...(i)
where, $\omega$ be the angular frequency of variation of force.
Since, the graph is completing one cycle in $8 \mathrm{~s}$,
hence time period, $T=8 \mathrm{~s}$
Using expression of change of momentum,
$\Delta p=\int_0^8 F d t=\int_0^8-2 \cos \omega t d t$[from Eq. (i)]
$\begin{aligned} & =-\frac{2}{\omega}[\sin \omega t]_0^8 \\ & =\frac{-2}{\omega}[\sin \omega 8-\sin 0]\end{aligned}$
$=\frac{-2}{\omega}\left[\sin \frac{2 \pi}{8} 8-0\right]$ $\left(\because \omega=\frac{2 \pi}{T}\right)$
$=\frac{-2}{\omega}[\sin 2 \pi]=0$ $[\because \sin 0=\sin 2 \pi=0]$
Hence, the gain or change of linear momentum is zero.
