A block moving horizontally on a smooth surface with a speed of \(40\, {ms}^{-1}\) splits into two equal parts. If one of the parts moves at \(60\, {ms}^{-1}\) in the same direction, then the fractional change in the kinetic energy will be \(x: 4\) where \(x=..... .\)

\({m} \times 40=\frac{{m}}{2} \times {v}+\frac{{m}}{2} \times 60\)

\(40=\frac{{v}}{2}+3\)

\(\Rightarrow {v}=20\)

\(({K} . {E} .)_{{I}}=\frac{1}{2} {m} \times(40)^{2}=800\, {m}\)

\(({K} . {E} .)_{{f}}=\frac{1}{2} \frac{{m}}{2} \cdot(20)^{2}+\frac{1}{2} \cdot \frac{{m}}{2}(60)^{2}=1000\, {m}\)

\(|\Delta {K} . {E} .|=|1000\, {m}-800 \,{m}|=200\, {m}\)

\(\frac{\Delta {K} . {E}}{({K} . {E} .)_{{i}}}=\frac{200 \,{m}}{800\, {m}}=\frac{1}{4}=\frac{{x}}{4}\)

\({x}=1\)