ExamBro
ExamBro
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:
If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point \(B\), immediately after it strikes the second incline is

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

Answer & Solution

Correct Answer

(C) zero

Step-by-step Solution

Detailed explanation

In elastic collision, component of \(v_1\) parallel to \(B C\) will remain unchanged, while component perpendicular to \(B C\) will remain unchanged in magnitude but its direction will be reversed.


\(
\begin{aligned}
& v_{\|}=v_1 \cos 30^{\circ}=(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1} \\
& v_{\perp}=v_1 \sin 30^{\circ}=(\sqrt{60})\left(\frac{1}{2}\right)=\sqrt{15} \mathrm{~ms}^{-1}
\end{aligned}
\)
Now vertical component of velocity of block :
\(
\begin{aligned}
v & =v_{\perp} \cos 30^{\circ}-v_{\|} \cos 60^{\circ} \\
& =(\sqrt{15})\left(\frac{\sqrt{3}}{2}\right)-(\sqrt{45})\left(\frac{1}{2}\right)=0
\end{aligned}
\)
\(\therefore\) correct option is (c).
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app