Question

A body is projected up a smooth inclined plane of length $20\sqrt{2}m$ from point $A$ as shown in the figure. The top $B$ of the inclined plane is connected to a well of diameter $40m$. If the body just manages to cross the well then the velocity of projection is
(Acceleration due to gravity, $g=10m{s}^{-2}$ )

• A. $40m{s}^{-1}$
• B. $40\sqrt{2}m{s}^{-1}$
• C. $20m{s}^{-1}$
• D. $20\sqrt{2}m{s}^{-1}$

Answer 1

Answer: (D)

According to the question,

In $\Delta ABC$,
$\frac{h}{20\sqrt{2}}=\cos 45^{\circ }$
$\Rightarrow h=20\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)$ or $h=20m$
From third equation of the motion,
$v^2=u^2+2gh$
$\Rightarrow v=\sqrt{2gh}(\because u=0)$
$\Rightarrow v=\sqrt{2\times 10\times 20}=20m/s$
Net energy at a point $B$,
$\Rightarrow \frac{1}{2}m{v}^2+mgh=\frac{1}{2}m(20{)}^2+m\times 10\times 20$
$=m(200+200)=400m$
Net energy at point $D$ to cross the well,
$E_D=\frac{1}{2}m{v}_D^2$
So, $\frac{1}{2}m{v}_D^2=400m$
$\Rightarrow v_D=20\sqrt{2}m{s}^{-1}$