Question

A particle is projected up on an incline at $t=0$ with a speed of $14m{s}^{-1}$ as shown in the figure. At the same moment, a box starts sliding down. The box has two small windows at a height of $2m$ on opposite walls. The particle enters from $1^{\text{st}}$ window and leaves through the other. The width of the box is $\frac{7\alpha }{\sqrt{3}}m$. The value of $\alpha$ is . (Assume friction to be absent everywhere and answer upto two digit after decimal point)

Answer 1

Solving this question by resolving acceleration due to gravity along and normal to the incline.

Magnitude of $g_{\text{along wedge}}=8m/s^2$

Magnitude of $g_{\text{normal to wedge}}=6m/s^2$

So, by writing second equation of motion along normal to the plane we get,
$2=7t+\frac{1}{2}(-6)t^2\Rightarrow t=(\frac{1}{3}\sec ,2\sec )$

$\Delta t=\frac{5}{3}sec$

$\because V_{\text{along wedge}}=14\times cos\left({30}^0\right)=7\sqrt{3}m/s$ and the accleration of the particle with respect to the wedge is zero,

Width of the box is $d=\left(\frac{5}{3}\right)\times 7\sqrt{3}=\frac{35}{\sqrt{3}}m=\frac{7\alpha }{\sqrt{3}}$

So, $\alpha =5$.