Question

A force $F$ acts on an object (mass$=1kg$) which is initially at rest as shown in the figure. Draw the graph showing the momentum of the object varying during the time for which the force acts.

Answer 1

Graph of momentum (N ms) vs time (ms)

From the graph of force (N) vs time (ms), equations of F(t) can be computed as

$F\left(t\right)=\frac{t}{10}$ , $0\le t\le 50$

$F\left(t\right)=\frac{-t}{10}+10$ , $50\le t\le 100$

From Newton's second law,

$F\left(t\right)=m\times a\left(t\right)$

$\Rightarrow F\left(t\right)=m\times \frac{dv\left(t\right)}{dt}$

$\Rightarrow F\left(t\right)dt=mdv\left(t\right)$

$\Rightarrow \int F\left(t\right)dt=\int mdv\left(t\right)$

$\Rightarrow \int F\left(t\right)dt=\int dp\left(t\right)$

$\Rightarrow \int F\left(t\right)dt=p\left(t\right)$

$\therefore p\left(t\right)=\int \frac{t}{10}dt=\frac{t^2}{20}+c_1$ , $0\le t\le 50$ ...................(1)

$p\left(t\right)=\int (\frac{-t}{10}+10)dt=\frac{-t^2}{20}+10t+c_2$ , $50\le t\le 100$...................(2)

here $c_1$ and $c_2$ are constants of integration.

If $t=10ms$, per eqn(1) $p=10Nms\Rightarrow c_1=5Nms$.

And if $t=60ms$, per eqn(2) $p=240Nms\Rightarrow c_2=-180Nms$.

Hence, $p\left(t\right)=\int \frac{t}{10}dt=\frac{t^2}{20}+5$ , $0\le t\le 50$

$p\left(t\right)=\int \frac{-t}{10}dt=\frac{-t^2}{20}+10t-180$ , $50\le t\le 100$