Question

A particle is projected with a velocity $u$ in horizontal direction as shown in the figure. Find $u$(approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

• A. $10m/s$
• B. $20m/s$
• C. $10\sqrt{2}m/s$
• D. None of these

Answer 1

Answer: (C)

$10\sqrt{2}m/s$

Considering plane along Wedge axis,
when particle collides with the plane, its component of velocity parallel to inclined plane should be zero.
$\Rightarrow u\cos \theta -g\sin \theta \times t=0$ {Equation along the plane}
$t=\frac{u}{g\tan \theta }$
$u=gt\times \tan \theta =\frac{gt}{\sqrt{3}}$
Now,
Consider the plane along original x-y axis,
Thus,
Horizontal displacement before it hits the plane $=ut$
vertical displacement during the time t $=\frac{1}{2}g{t}^2$
Now , $\tan {30}^{\circ }=\frac{h-\frac{1}{2}g{t}^2}{ut}$
$\frac{1}{\sqrt{3}}=\frac{h-\frac{1}{2}g{t}^2}{ut}$
Hence , $\frac{1}{3}=(\frac{h}{g{t}^2}-\frac{1}{2})$

$t=\sqrt{\frac{6h}{5g}}$
Hence , $t=2.5938$ s
So,
$u=gt\tan {30}^{\circ }=\frac{9.8\times 2.5938}{\sqrt{3}}=14.69\simeq 10\sqrt{2}$ m/s