- JEE Mains 2020
- Hard
- MCQ
Two particles of equal mass \(\mathrm{m}\) have respective initial velocities \(u\hat{i}\) and \(u\left(\frac{\hat{\mathrm{i}}+ \hat{\mathrm{j}}}{2}\right) .\) They collide completely inelastically. The energy lost in the process is
Answer & Solution
(B) \(\frac{1}{8} \mathrm{mu}^{2}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{mu\hat{i}}+\operatorname{mu}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right)=(\mathrm{m}+\mathrm{m}) \overline{\mathrm{v}}\)
\(\Rightarrow \overline{\mathrm{v}}=\frac{3}{4} \mathrm{u} \hat{\mathrm{i}}+\frac{\mathrm{u}}{4} \hat{\mathrm{j}}\)
\(\Rightarrow|\mathrm{v}|=\frac{\mathrm{u}}{4} \sqrt{10}\)
Final kinetic energy \(=\frac{1}{2} 2 \mathrm{m}\left(\frac{\mathrm{u}}{4} \sqrt{10}\right)^{2}=\frac{5}{8} \mathrm{mu}^{2}\)
Initial kinetic energy
\(=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{m}\left(\frac{\mathrm{u}}{\sqrt{2}}\right)^{2}=\frac{6}{8} \mathrm{mu}^{2}\)
Loss in \(\mathrm{K.E.}=\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{1}{8} \mathrm{mu}^{2}\)
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