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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • 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

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Answer & Solution

Correct Answer

(B) \(\frac{1}{8} \mathrm{mu}^{2}\)

Step-by-step Solution

Detailed explanation

From momentum conservation

\(\operatorname{mu\hat{i}}+\operatorname{mu}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right)=(\mathrm{m}+\mathrm{m}) \overline{\mathrm{v}}\)

\(\Rightarrow \overline{\mathrm{v}}=\frac{3}{4} \mathrm{u} \hat{\mathrm{i}}+\frac{\mathrm{u}}{4} \hat{\mathrm{j}}\)

\(\Rightarrow|\mathrm{v}|=\frac{\mathrm{u}}{4} \sqrt{10}\)

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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