Question 96 of 183

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 $x=$ ________.

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Answer: 1

$2m\times 40=60\times m+mv$
$V=20m/s$
Now $\Delta KE=K{E}_{f}-K{E}_{i}=\dfrac{1}{2}m\times (20{)}^{2}+\dfrac{1}{2}m\times (60{)}^{2}-\dfrac{1}{2}2m\times (40{)}^{2}$
$\Delta KE=\dfrac{1}{2}m(800)\Rightarrow K{E}_{i}=\dfrac{1}{2}2m\times (40{)}^{2}=\dfrac{1}{2}m\times 3200$
Fractional change $=\dfrac{\Delta KE}{K{E}_{i}}=\dfrac{1}{4}$