Question

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 sides down on inclined plane BC. Time it takes to reach to the point C from point A is t(√2 + 1) s. The value of t is ______.
(Use g = 10 m/s2)

Fig.

Answer 1

The correct answer is 2
$AB = 10\sqrt2m$
$v_A = \sqrt{2×10×10}$
$= 10\sqrt2$ m/s
$v_C = 10\sqrt2$ m/s
$a_{BC} = g\sin(30°)$
= 5 m/s2
$t_{BC} = 2\sqrt2s(\frac{v_c}{a_{BC}})$
$t_{AB} = \frac{v_A}{5\sqrt2} = 2s$
$t_{AB} + t_{BC} = 2(\sqrt2 + 1)$
⇒ t = 2
Therefore , value of t is 2