Question

A particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t=0$, a force $F=kt$ acts in the same direction on the moving particle during time interval $T$, so that its momentum changes from $p$ to $3p$. Here, $k$ is a constant. The value of $T$ is

• A. $\sqrt{\frac{2p}{k}}$
• B. $2\sqrt{\frac{p}{k}}$
• C. $\sqrt{\frac{2k}{p}}$
• D. $2\sqrt{\frac{k}{p}}$

Answer 1

The correct option is B $2\sqrt{\frac{p}{k}}$

Given, $F=kt$
When $t=0$, linear momentum $=p$
When $t=T$, linear momentum $=3p$
According to Newton's second law of motion,
Applied force, $F=\frac{dp}{dt}$
or $dp=F.dt$
or $dp=kt.dt$

Now, integrate both sides with proper limit.
$\underset{p}{\overset{3p}{\int }}dp=k\underset{0}{\overset{T}{\int }}tdt$
or $[p{]}_p^{3p}=k{\left[\frac{t^2}{2}\right]}_0^T$
or $(3p-p)=\frac{1}{2}k[T^2-0]$
or $T^2=\frac{4p}{k}$
or $T=2\sqrt{\frac{p}{k}}$