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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

JEE Advanced 2008 · Paper 1 · Physics · Center of Mass, Momentum and Collision

Paragraph: A small block of mass M moves on a frictionless surface...

Paragraph: A small block of mass MM moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from 60∘to30∘60^{\circ} to 30^{\circ} at point BB. The block is initially at rest at AA. Assume that collisions between the block and the incline are totally inelastic (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)Paragraph: A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle
Question: The speed of the block at point CC, immediately before it leaves the second incline is

Options

  1. A120 m/s\sqrt{120} \mathrm{~m} / \mathrm{s}
  2. B105 m/s\sqrt{105} \mathrm{~m} / \mathrm{s}
  3. C90 m/s\sqrt{90} \mathrm{~m} / \mathrm{s}
  4. D75 m/s\sqrt{75} \mathrm{~m} / \mathrm{s}

Correct answer

B. 105 m/s\sqrt{105} \mathrm{~m} / \mathrm{s}

Step-by-step solution

Height fallen by the block from BB to CC. h2=33tan⁡30∘=3 mh_2=3 \sqrt{3} \tan 30^{{}^{\circ}}=3 \mathrm{~m} Let v3v_3 be the speed of block, at point CC, just before it leaves the second incline, then v3=v22+2gh2=45+2×10×3=105 ms−1\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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