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…

Question

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 $g = 10 m s^{- 2}$ )

MARKS Content Team · Accepted Answer

Correct answer: 2. From energy conservation at point A and B, 1/2mv02=mgh ⇒v_0=√gh=√(10×10) ⇒v_0=10√2 m s^(-1) For A→B At B, v=0 Acceleration of a particle moving on a smooth incline is gsinθ. Therefore, along AB, a=-gsin45°=-10√2 m s^(-2). Using equation of motion, v=u+at_1 0=10√2-(10√2t_1⇒t_1=2 sFor B→CUsing second equation of motion, s=ut_2+1)/2at22 ⇒10/(sin30°)=1/210sin30°t22 t_2=2√2 s So total time T=t_1+t_2 =2√2+2 =2√2+1 s

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