Question

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

• A. $2s$
• B. $2\sqrt{3}s$
• C. $\frac{4}{\sqrt{3}}s$
• D. $noneofthese$

Answer 1

The correct option is A $2s$

velocity of particle is given as $V_P=5\sqrt{3}\hat{i}+15\hat{j}$

velocity of wedge is given as $V_{wedge}=10\sqrt{3}\hat{i}$

The relative velocity of particle w.r.t. wedge is given as

$V_{P/wedge}=V_P-V_{wedge}$

$V_{P/wedge}=5\sqrt{3}\hat{i}+15\hat{j}-10\sqrt{3}\hat{i}$

$V_{P/wedge}=-5\sqrt{3}\hat{i}+15\hat{j}$

Now it can be analysed as projectile motion on a stationary inclined plane, where the time period of its flight is given as :

$T=\frac{2u\sin (\alpha -\beta )}{g\cos \beta }$

where $\alpha ={60}^{\circ },\beta ={30}^{\circ },u=10\sqrt{3}$

$T=\frac{2\left(10\sqrt{3}\right)\sin ({60}^{\circ }-{30}^{\circ })}{10\cos {30}^{\circ }}=2s$