Question

A particle having a mass $m$ and velocity $V_m$ in the y-direction is projected on to a horizontal belt that is moving with uniform velocity $V_b$ in the $x-direction$ as shown in figure. $\mu$ is the coefficient of friction between particle and belt. Assuming that the particle first touches the belt at the origin at the fixed $x-y$ coordinate system and remains on the belt, find the coordinates $(x,y)$ of the point where the sliding stops.

• A. $x=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2},y=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}$
• B. $x=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2},y=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2}$
• C. $x=\frac{V_b}{\mu g}\sqrt{V_m^2+V_b^2},y=\frac{V_m}{\mu g}\sqrt{V_m^2+V_b^2}$
• D. $x=\frac{V_b}{\mu g}\sqrt{V_m^2+V_b^2},y=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}$

Answer 1

The correct option is A $x=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2},y=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}$

Horizontal direction:

$Initialvelocity=0$
$Finalvelocity=V_b$
Applying eqn. $v=u+at$,
$\Rightarrow V_b=a_{1}t$
$\Rightarrow t=\frac{V_b}{a_1}..............\left(i\right)$
Vertical direction:

$Initialvelocity=V_m$
$Finalvelocity=0$
Using same eqn. as above
$V_m=a_{2}t$
$\Rightarrow t=\frac{V_m}{a_2}.......\left(ii\right)$
From eq. (i) and (ii) $\frac{V_b}{a_1}=\frac{V_m}{a_2}.......\left(iii\right)$
Net friction force acting $=\mu mg$
$\Rightarrow m\sqrt{a_1^2+a_2^2}=\mu mg$
$or\sqrt{a_1^2+a_2^2}=\mu g$
$\Rightarrow V_b\sqrt{\frac{a_1^2}{V_b^2}+\frac{a_2^2}{V_b^2}}=\mu g$
Substituting $\frac{a_1^2}{V_b^2}$ from eq. (iii)
$\Rightarrow \sqrt{V_b^2+V_m^2}{a}_2=\mu g{V}_m$
$\Rightarrow a_2=\frac{\mu g}{\sqrt{V_b^2+V_m^2}}{V}_b$
Similarly, $a_1=\frac{\mu g}{\sqrt{V_b^2+V_m^2}}{V}_m$
Now using eq. $v^2=u^2+2as$
Horizontal direction: $V_b^2=2a_{1}x$
$\Rightarrow x=\frac{V_b}{2a_1}$
$\Rightarrow x=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2}$
Similarly $y=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}$
Thus correct answer is (a)