JEE Advanced · Physics · 7. COM & Collisions
Paragraph:
A small block of mass \(M\) moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from \(60^{\circ}\) to \(30^{\circ}\) at point \(B\). The block is initially at rest at \(A\). Assume that collisions between the block and the incline are totally inelastic \(\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)\)

Question:
The speed of the block at point \(C\), immediately before it leaves the second incline is
- A \(\sqrt{120} \mathrm{~m} / \mathrm{s}\)
- B \(\sqrt{105} \mathrm{~m} / \mathrm{s}\)
- C \(\sqrt{90} \mathrm{~m} / \mathrm{s}\)
- D \(\sqrt{75} \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{105} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Height fallen by the block from \(B\) to \(C\). \(h_2=3 \sqrt{3} \tan 30^{\circ}=3 \mathrm{~m}\)
Let \(v_3\) be the speed of block, at point \(C\), just before it leaves the second incline, then
\(
\begin{aligned}
v_3 & =\sqrt{v_2^2+2 g h_2} \\
& =\sqrt{45+2 \times 10 \times 3} \\
& =\sqrt{105} \mathrm{~ms}^{-1}
\end{aligned}
\)
\(\therefore\) correct option is (b).
Let \(v_3\) be the speed of block, at point \(C\), just before it leaves the second incline, then
\(
\begin{aligned}
v_3 & =\sqrt{v_2^2+2 g h_2} \\
& =\sqrt{45+2 \times 10 \times 3} \\
& =\sqrt{105} \mathrm{~ms}^{-1}
\end{aligned}
\)
\(\therefore\) correct option is (b).
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