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Question

$3 {MV}_{0}=2 {MV}_{2}+{MV}_{1}$

$3 {V}_{0}=2 {V}_{2}+{V}_{1}$

$120=2 {V}_{2}+60 \Rightarrow {V}_{2}=30\, {m} / {s}$

$\frac{\Delta {K} {E}}{{

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

$3 {MV}_{0}=2 {MV}_{2}+{MV}_{1}$

$3 {V}_{0}=2 {V}_{2}+{V}_{1}$

$120=2 {V}_{2}+60 \Rightarrow {V}_{2}=30\, {m} / {s}$

$\frac{\Delta {K} {E}}{{K} . {E} .}=\frac{\frac{1}{2} {MV}_{1}^{2}+\frac{1}{2} 2 {M} V_{2}^{2}-\frac{1}{2} 3 {M} V_{0}^{2}}{\frac{1}{2} 3 {M} V_{0}^{2}}$

$=\frac{{V}_{1}^{2}+2 {V}_{2}^{2}-3 {V}_{0}^{2}}{3 {V}_{0}^{2}}$

$=\frac{3600+1800-4800}{4800}$

$=\frac{1}{8}$

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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