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Question: Answered & Verified by Expert
If a lighter body of mass ' $\mathrm{M}_1$ ' and velocity ' $\mathrm{V}_1$ ' and a heavy body (mass $\mathrm{M}_2$ and velocity $\mathrm{V}_2$ ) have the same kinetic energy then
PhysicsCenter of Mass Momentum and CollisionMHT CETMHT CET 2023 (10 May Shift 1)
Options:
  • A $\mathrm{M}_2 \mathrm{~V}_2 < \mathrm{M}_1 \mathrm{~V}_1$
  • B $\mathrm{M}_2 \mathrm{~V}_2=\mathrm{M}_1 \mathrm{~V}_1$
  • C $\mathrm{M}_2 \mathrm{~V}_1 < \mathrm{M}_1 \mathrm{~V}_2$
  • D $\mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1$
Solution:
1073 Upvotes Verified Answer
The correct answer is: $\mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1$
$\begin{array}{ll} & \mathrm{KE}_1=\mathrm{KE}_2 \\ & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \quad \ldots .(\because \mathrm{p}=\mathrm{mv}) \\ \therefore \quad & \mathrm{p}=\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})} \\ \therefore \quad & \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{\sqrt{\mathrm{M}_1}}{\sqrt{\mathrm{M}_2}} \\ & \text { But } \mathrm{M}_2>\mathrm{M}_1 \\ \therefore \quad & \mathrm{p}_2>\mathrm{p}_1 \\ & \text { i.e., } \mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1\end{array}$

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