Question 82 of 183

Two inclined planes are placed as shown in figure.
A block is projected from the Point $A$ of inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10m$. After reaching the Point $B$ the block slides down on inclined plane $BC$. Time it takes to reach to the point $C$ from point $A$ is $t\left(\sqrt{2}+1\right)s$. The value of t is _____ (use $g=10m{s}^{-2}$)

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Answer: 2

From energy conservation at point $A$ and $B$,
$\dfrac{1}{2}m{v}_{0}^{2}=mgh$
$\Rightarrow {v}_{0}=\sqrt{gh}=\sqrt{10\times 10}$
$\Rightarrow {v}_{0}=10\sqrt{2}m{s}^{-1}$
For $A\rightarrow B$
At $B,v=0$
Acceleration of a particle moving on a smooth incline is $g\sin \theta$. Therefore, along $AB$, $a=-g\sin 45^{\circ}=\dfrac{-10}{\sqrt{2}}m{s}^{-2}$.
Using equation of motion,
$v=u+a{t}_{1}$
$0=10\sqrt{2}-\dfrac{10}{\sqrt{2}}{t}_{1}\Rightarrow {t}_{1}=2s$For $B\rightarrow C$
Using second equation of motion,
$s=u{t}_{2}+\dfrac{1}{2}a{t}_{2}^{2}$
$\Rightarrow \dfrac{10}{\sin 30^{\circ}}=\dfrac{1}{2}\left(10\sin 30^{\circ}\right){t}_{2}^{2}$
${t}_{2}=2\sqrt{2}s$
So total time $T={t}_{1}+{t}_{2}$ $=2\sqrt{2}+2$
$=2\left(\sqrt{2}+1\right)s$