Question

A small object is projected up along the incline from point $A$ with speed $u.$ The incline surface has a coefficient of friction equal to $\frac{1}{\sqrt{3}}$. The particle returns after some time passing through point $\text{A}$ with speed $v$ down the incline. What is the value of $\frac{v}{u}$?

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

Let, the object moves a distance $d$ before coming to rest.

While the object is moving up, friction acts down the incline.

$a_1=-g(\sin \theta +\mu \cos \theta )$

Using kinematic equations,

$v^2=u^2+2as$

Here, $v=0$

$\Rightarrow u^2=2dg(\sin \theta +\mu \cos \theta )....\left(1\right)$

While coming down, friction acts up the incline,

$a_2=g(\sin \theta -\mu \cos \theta )$

Using kinematic equations,

$v^2=u^2+2as$

Here, $u=0$

$v^2=2dg(\sin \theta -\mu \cos \theta )....\left(2\right)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\frac{v^2}{u^2}=\frac{\tan \theta -\mu }{\tan \theta +\mu }$

Given, $\mu =\frac{1}{\sqrt{3}}$ and $\theta ={60}^{\circ }$

$\Rightarrow \frac{v^2}{u^2}=\frac{\tan {60}^{\circ }-\frac{1}{\sqrt{3}}}{\tan {60}^{\circ }+\frac{1}{\sqrt{3}}}$

$\Rightarrow \frac{v^2}{u^2}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{\sqrt{3}+\frac{1}{\sqrt{3}}}$

$\Rightarrow \frac{v^2}{u^2}=\frac{1}{2}$

$\Rightarrow \frac{v}{u}=\frac{1}{\sqrt{2}}$

Hence, option $\left(d\right)$ is correct.