Question

A particle having a mass $m$ and velocity $V_m=3m/s$ in the y-direction is projected on to a horizontal belt that is moving with uniform velocity $V_b=4m/s$ in the $x$ direction as shown in figure. $\mu =0.5$ is the coefficient of friction between the particle and the belt. Assuming that the particle first touches the belt at the origin of the fixed xy coordinate system and remains on the belt, the coordinates of the point where the sliding stops is $(x,y)$. Then $x+y=$

Answer 1

Relative velocity of the particle w.r.t the belt after hitting the belt
$\overrightarrow{V_r}=V_m\hat{j}-V_b\hat{i}$
$V_r=\sqrt{V_m^2+V_b^2}$
Let the particle come to rest w.r.t the belt at time $t$. Then,
$0=V_r-\mu gt$
$\Rightarrow t=\frac{\sqrt{V_m^2+V_b^2}}{\mu g}$

After time $t$, particle starts to slide


$x=V_{b}t-\frac{1}{2}{a}_x{t}^2=V_b\frac{\sqrt{V_m^2+V_b^2}}{\mu g}-\frac{1}{2}\mu g\sin \theta {\left(\frac{\sqrt{V_m^2+V_b^2}}{\mu g}\right)}^2$
$=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2}$
($\because \tan \theta =\frac{V_b}{V_m}$, $\sin \theta =\frac{V_b}{\sqrt{V_m^2+V_b^2}}$)

Similarily,
$y=V_{m}t-\frac{1}{2}{a}_y{t}^2=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}$

Substituting the values of $V_b,V_m,\mu$,
$x=2,y=1.5$