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JEE Mains · Physics · STD 11 - Work,energy,power and collision
  • JEE Mains 2021
  • Hard
  • MCQ

A block moving horizontally on a smooth surface with a speed of \(40\, {m} / {s}\) splits into two parts with masses in the ratio of \(1: 2\). If the smaller part moves at \(60\, {m} / {s}\) in the same direction, then the fractional change in kinetic energy is :-

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Answer & Solution

Correct Answer

(C) \(\frac{1}{8}\)

Step-by-step Solution

Detailed explanation

\(3 {MV}_{0}=2 {MV}_{2}+{MV}_{1}\)

\(3 {V}_{0}=2 {V}_{2}+{V}_{1}\)

\(120=2 {V}_{2}+60 \Rightarrow {V}_{2}=30\, {m} / {s}\)

\(\frac{\Delta {K} {E}}{{K} . {E} .}=\frac{\frac{1}{2} {MV}_{1}^{2}+\frac{1}{2} 2 {M} V_{2}^{2}-\frac{1}{2} 3 {M} V_{0}^{2}}{\frac{1}{2} 3 {M} V_{0}^{2}}\)

\(=\frac{{V}_{1}^{2}+2 {V}_{2}^{2}-3 {V}_{0}^{2}}{3 {V}_{0}^{2}}\)

\(=\frac{3600+1800-4800}{4800}\)

\(=\frac{1}{8}\)

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