Question

Two inclined planes $OA$ and $OB$ having inclination (with horizontal) ${30}^{\circ }$ and ${60}^{\circ }$, respectively, intersect each other at $O$ as shown in figure. A particle is projected from point $P$ with velocity $u=10\left(\sqrt{3}\right){\text{ms}}^{-1}$ along a direction perpendicular to the plane $OA$. If the particle strikes the plane $OB$ perpendicular at $Q$, calculate the displacement $PQ$.

• A. $2.5\text{m}$
• B. $20\text{m}$
• C. $10\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is B $20\text{m}$

Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane $OA$. It means $\overrightarrow{u}$ is parallel to plane $OB$.
At the instant of collision of the particle with $OB$, its velocity is perpendicular to $OB$ or the velocity component parallel to $OB$ is zero.

First considering motion of particle parallel to plane $OB$,
Initial velocity along $OB,u=10\left(\sqrt{3}\right){\text{ms}}^{-1}$
Component of acceleration due to gravity along $OB,g_{OB}=-g\sin {60}^{\circ }$
$\Rightarrow g_{OB}=-5\sqrt{3}{\text{ms}}^{-2}$
The final velocity of the particle along $OB$ is
$v_{OB}=0$
The time of flight ($t$) of particle is
$v_{OB}=u-g_{OB}t$
$\Rightarrow 0=10\sqrt{3}-5\left(\sqrt{3}\right)t$
$\Rightarrow t=2\text{s}$

The distance between $OQ$ is
$s=ut-\frac{1}{2}{g}_{OB}{t}^2$
$\Rightarrow OQ=10\left(\sqrt{3}\right)\times 2-\frac{1}{2}5\left(\sqrt{3}\right)(2{)}^2$
$\Rightarrow OQ=10\left(\sqrt{3}\right)\text{m}$

Now considering motion of the particle normal to the plane $OB$.
Initial velocity along $OA=0$
Acceleration along $OA,g_{OA}=g\cos {60}^{\circ }=5{\text{ms}}^{-2}$
The time of flight will remain same,
$t=2\text{s}$
The distance between $PO$ is
$s=\frac{1}{2}{g}_{OA}{t}^2$
$\Rightarrow s=\frac{1}{2}\times 5\times 2^2$
$\Rightarrow s=PO=10\text{m}$
Distance between $PQ$ is
$PQ=\sqrt{(PO{)}^2+(OQ{)}^2}$
$\Rightarrow PQ=\sqrt{\left({10}^2\right)+(10\sqrt{3}{)}^2}$
$PQ=20\text{m}$