JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
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 \(BC\). Time it takes to reach to the point \(C\) from point \(A\) is \(t (\sqrt{2}+1) s\). The value of \(t\) is........(use \(g =10 m / s ^{2}\) )

- A \(8\)
- B \(4\)
- C \(2\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
From E.C. \(=\frac{1}{2} mv _{0}^{2}= mgh\) \(v _{0}=10 \sqrt{2}\) For \(A \rightarrow B\) at \(B , v =0\) \(a =- g \sin 45^{\circ}=\frac{-10}{\sqrt{2}}\) \(v = u + at _{1} \Rightarrow 0=10 \sqrt{2}-\frac{10}{\sqrt{2}} t _{1} \Rightarrow t _{1}=2\,sec\) For \(B \rightarrow C\)…
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