Question

A particle is projected at angle ${60}^{\circ }$ with speed $10\sqrt{3}\text{m/s}$ , from the point $‘A^{\prime }$ as shown in the figure. At the same time the wedge is made to move with speed $10\sqrt{3}\text{m/s}$ towards right as shown in the figure.Then, the time(in $s$) after which particle will strike the wedge is

Answer 1

Velocity of particle
$V_P=10\sqrt{3}\cos {60}^{\circ }\hat{i}+10\sqrt{3}\sin {60}^{\circ }\hat{j}$
$V_P=5\sqrt{3}\hat{i}+15\hat{j}$
Velocity of wedge
$V_W=10\sqrt{3}\hat{i}$
Velocity of particle wrt wedge
$V_{PW}=V_P-V_W=-5\sqrt{3}\hat{i}+15\hat{j}$
$|V_{PW}|=\sqrt{25\times 3+225}=\sqrt{300}=10\sqrt{3}\text{m/s}$
$\Rightarrow \tan \theta =\frac{15}{5\sqrt{3}}=\sqrt{3},\theta ={60}^{\circ }$
Time after which particle will strike the wedge is
$T=\frac{2u_{\perp }}{g_{\perp }}=2\times \frac{\left(10\sqrt{3}\sin {30}^{\circ }\right)}{\left(g\cos {30}^{\circ }\right)}$
$T=\frac{2\times 10\sqrt{3}\times \frac{1}{2}}{10\times \frac{\sqrt{3}}{2}}=2\text{sec}$