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 $10$ m. 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 $t(\sqrt2 + 1)$ s. The value of $t$ is ______ (use $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 2
Take the planes as smooth.
$A \to B$: length $\dfrac{10}{\sin45^\circ} = 10\sqrt2$ m, retardation $g\sin45^\circ = 5\sqrt2\ \text{m s}^{-2}$. The block just stops at $B$, so (reversing the motion)
$$t_1 = \sqrt{\frac{2(10\sqrt2)}{5\sqrt2}} = 2\ \text{s}$$
$B \to C$: starts from rest, length $\dfrac{10}{\sin30^\circ} = 20$ m, acceleration $g\sin30^\circ = 5\ \text{m s}^{-2}$:
$$t_2 = \sqrt{\frac{2(20)}{5}} = 2\sqrt2\ \text{s}$$
Total $= 2 + 2\sqrt2 = 2(\sqrt2 + 1)$ s, so $t = 2$.
Solution by Sreeraj P, M.Sc Physics