MHT CET · Physics · Center of Mass Momentum and Collision
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
- 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 \(\mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1\)
Answer & Solution
Correct Answer
(D) \(\mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \mathrm{KE}_1=\mathrm{KE}_2 \\ & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \quad \ldots .(\because \mathrm{p}=\mathrm{mv}) \\ \therefore \quad & \mathrm{p}=\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})} \\ \therefore \quad & \frac{\mathrm{p}_1}{\mathrm{p}_2}=\frac{\sqrt{\mathrm{M}_1}}{\sqrt{\mathrm{M}_2}} \\ & \text { But } \mathrm{M}_2>\mathrm{M}_1 \\ \therefore \quad & \mathrm{p}_2>\mathrm{p}_1 \\ & \text { i.e., } \mathrm{M}_2 \mathrm{~V}_2>\mathrm{M}_1 \mathrm{~V}_1\end{array}\)
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