Question

A particle of mass ${10}^{-2}\text{kg}$ is moving along the positive $\text{x-axis}$ under the influence of a force $F\left(x\right)=-\frac{K}{2x^2}$ where $K={10}^{-2}{\text{Nm}}^2$. At time $t=0$, particle is at $x=1\text{m}$ and its velocity is $v=0$. The time at which it reaches $x=0.25\text{m}$ is

• A. $1.7\text{s}$
• B. $1.2\text{s}$
• C. $1.9\text{s}$
• D. $1.5\text{s}$

Answer 1

The correct option is D $1.5\text{s}$

Given,
mass of the particle, $m={10}^{-2}\text{kg}$

Force acting on the particle,

$F=-\frac{K}{2x^2}...\left(1\right)$

We know that force acting on the moving particle,

$F=ma...\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$

$\Rightarrow ma=-\frac{K}{2x^2}$

Substituting the values of $m$ and $K$ in the above equation,

$\Rightarrow {10}^{-2}\times a=-\frac{{10}^{-2}}{2x^2}$

$\Rightarrow a=-\frac{1}{2x^2}$

$\Rightarrow v\frac{dv}{dx}=-\frac{1}{2x^2}$ $[\because a=v\frac{dv}{dx}]$

Rearranging and integrating the equation, as $x=1\text{m},v=0\text{ms}$,

$\Rightarrow \underset{0}{\overset{v}{\int }}vdv=-\frac{1}{2}\underset{1}{\overset{x}{\int }}-x^{-2}dx$

$\Rightarrow \frac{1}{2}{\left[v^2\right]}_0^v=+\frac{1}{2}{\left[x^{-1}\right]}_1^x$

$\Rightarrow v^2=[\frac{1}{x}-1]=\frac{1-x}{x}$

$\Rightarrow v=\pm \sqrt{\frac{1-x}{x}}$

Initially the particle was moving in $\text{+x-direction}$ at $x=1$. When the particle is at $x=0.5$, obviously its velocity will be in $\text{-x-direction}$. The force acting in $\text{-x-direction}$ first decreases the speed of the particle, bring it momentarily at rest and then changes the direction of motion of the particle.
Thus, we should consider the negative velocity.

$\Rightarrow v=\frac{dx}{dt}=-\sqrt{\frac{1-x}{x}}$

$\Rightarrow \underset{1}{\overset{0.25}{\int }}\sqrt{\frac{x}{1-x}}dx=-\underset{0}{\overset{t}{\int }}dt$

On taking $x={\sin }^2\theta$, we get $dx=2\sin \theta \cos \theta d\theta$.

And limits as when $x=1,\theta ={90}^{\circ }$
and when $x=0.25,\theta ={30}^{\circ }$.

$\underset{{90}^{\circ }}{\overset{{30}^{\circ }}{\int }}\sqrt{\frac{{\sin }^2\theta }{1-{\sin }^2\theta }}\times 2\sin \theta \cos \theta d\theta =-\underset{0}{\overset{t}{\int }}dt$

$\underset{{90}^{\circ }}{\overset{{30}^{\circ }}{\int }}2{\sin }^2\theta d\theta =-\underset{0}{\overset{t}{\int }}dt$

$\underset{{90}^{\circ }}{\overset{{30}^{\circ }}{\int }}(1-\cos 2\theta )=-\underset{0}{\overset{t}{\int }}dt$

$\Rightarrow {[\theta -\frac{\sin 2\theta }{2}]}_{{90}^{\circ }}^{{30}^{\circ }}=-{\left[t\right]}_0^t$

$\Rightarrow -t={30}^{\circ }-\frac{\sin {60}^{\circ }}{2}-{90}^{\circ }+\frac{\sin {180}^{\circ }}{2}$

$\Rightarrow -t=\frac{\pi }{6}-\frac{\sqrt{3}}{4}-\frac{\pi }{2}+0$

$\Rightarrow -t=-1.48$

$\therefore t=1.48\text{s}\approx 1.5\text{s}$