Question

A particle is projected from the bottom of an inclined plane of inclination ${30}^{\circ }$ (with the horizontal) with speed $40\text{m/s}$ at an angle ${60}^{\circ }$ with the horizontal. The speed of the particle when its velocity vector is parallel to inclined plane is $(g=10{\text{m/s}}^2)$

• A. $\frac{40}{\sqrt{3}}\text{m/s}$
• B. $20\text{m/s}$
• C. $20\sqrt{3}\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is A $\frac{40}{\sqrt{3}}\text{m/s}$

Here, we choose our axes along and perpendicular to the inclined plane.



So, $u_x=u\cos {30}^{\text{o}}$ and $u_y=u\sin {30}^{\text{o}}$
Also the acceleration components will be $a_x=-g\sin {30}^{\text{o}}$ and $a_y=-g\cos {30}^{\text{o}}$

When velocity vector is along inclined plane we get $v_y=0$
By 1st law of motion, $v_y=u_y+a_{y}t$

i.e $v_y=40\sin {30}^{\circ }-g\cos {30}^{\circ }t=0$
i.e $t=\frac{40\sin {30}^{\circ }}{10\cos {30}^{\circ }}=\frac{4}{\sqrt{3}}\text{s}$

Hence the speed of the particle will be along $\text{x}-$axis only
$v_x=u_x+a_{x}t$

$v_x=40\cos {30}^{\circ }-g\sin {30}^{\circ }t$
$=40\frac{\sqrt{3}}{2}-10\times \frac{1}{2}\times \frac{4}{\sqrt{3}}=20\sqrt{3}-\frac{20}{\sqrt{3}}=\frac{40}{\sqrt{3}}\text{m/s}$

Hence option A is the correct answer.