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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 \(B\) immediately after it strikes the second incline is

  1. A \(\sqrt{60} \mathrm{~m} / \mathrm{s}\)
  2. B \(\sqrt{45} \mathrm{~m} / \mathrm{s}\)
  3. C \(\sqrt{30} \mathrm{~m} / \mathrm{s}\)
  4. D \(\sqrt{15} \mathrm{~m} / \mathrm{s}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\sqrt{45} \mathrm{~m} / \mathrm{s}\)

Step-by-step Solution

Detailed explanation

Between \(A\) and \(B\), height fallen by block \(h_1=\sqrt{3} \tan 60^{\circ}=3 \mathrm{~m}\).
\(\therefore\) speed of block just before striking the second incline,
\(
v_1=\sqrt{2 g h_1}=\sqrt{2 \times 10 \times 3}=\sqrt{60} \mathrm{~ms}^{-1}
\)



In perfectly inelastic collision, component of \(v_1\) perpendicular to \(B C\) will become zero, while component of \(v_1\) parallel to \(B C\) will remain unchanged.
\(\therefore\) speed of block \(B\) immediately after it strikes the incline is,
\(
\begin{aligned}
v_2 & =\text { component of } v_1 \text { along } B C \\
& =v_1 \cos 30^{\circ} \\
& =(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}
\end{aligned}
\)
\(\therefore\) correct option is (b)
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