Q.
A body is projected up a smooth inclined plane with a velocity $v_{0}$ from the point $A$ as shown in figure. The angle of inclination is $45{^\circ}$ and top $B$ of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of $v_{0}$? Length of the inclined plane is $20\sqrt{2}m$, and $g = 10ms^{- 2}$
A) $20\ ms^{- 1}$
B) $20\sqrt{2}ms^{- 1}$
C) $40\ ms^{- 1}$
D) $40\sqrt{2}\ ms^{- 1}$
Explanation
(b) Let $v$ be the velocity acquired by the body at $B$ which will be moving making an angle $45{^\circ}$ with the horizontal direction. As the body just crosses the well so $\frac{v^{2}}{g} = 40$
or $v^{2} = 40g = 40 \times 10 = 400$
or $v = 20\ ms^{- 1}$
Taking motion of the body from $A$ to $B$ along the inclined plane we have
$u = v_{0},a = - g\sin{45{^\circ} = - \frac{10}{\sqrt{2}}}ms^{- 2}$
$s = 20m,v = 20ms^{- 1}$
As $v^{2} = u^{2} + 2as$
$\therefore\ 400 = v_{0}^{2} + 2\left( - \frac{10}{\sqrt{2}} \right) \times 20\sqrt{2}$
or $v_{0}^{2} = 400 + 400 = 800$ or $v = 20\sqrt{2}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.