Two inclined planes are placed as shown in figure.
A block is projected from the Point $A$ of inclined plane $A B$ 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 $B C$ . Time it takes to reach to the point $C$ from point $A$ is $t (\sqrt{2} + 1) s$ . The value of t is _____ (use $g = 10 m s^{- 2}$ )

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Aicharya · Accepted Answer

From energy conservation at point A and B , 1 2 m v 0 2 = m g h &#8658; v 0 = g h = 10 &#215; 10 &#8658; v 0 = 10 2 &#160; m &#160; s - 1 For A &#8594; B At B , &#160;&#160; v = 0 Acceleration of a particle moving on a smooth incline is g sin &#952; . Therefore, along A B , a = - g sin 45 &#176; = - 10 2 &#160; m &#160; s - 2 . Using equation of motion, v = u + a t 1 0 = 10 2 - 10 2 t 1 &#8658; t 1 = 2 &#160; s For B &#8594; C Using second equation of motion, s = u t 2 + 1 2 a t 2 2 &#8658; 10 sin 30 &#176; = 1 2 10 sin 30 &#176; t 2 2 t 2 = 2 2 &#160; s So total time T = t 1 + t 2 = 2 2 + 2 = 2 2 + 1 &#160; s

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane A…

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