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Question

${P}_{{i}}={P}_{{f}}$

${m} 	imes 40=\frac{{m}}{2} 	imes {v}+\frac{{m}}{2} 	imes 60$

$40=\frac{{v}}{2}+3$

$\Rightarrow {v}=20$

$({K} . {

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$1$

Vedclass · Answer

${P}_{{i}}={P}_{{f}}$

${m} 	imes 40=\frac{{m}}{2} 	imes {v}+\frac{{m}}{2} 	imes 60$

$40=\frac{{v}}{2}+3$

$\Rightarrow {v}=20$

$({K} . {E} .)_{{I}}=\frac{1}{2} {m} 	imes(40)^{2}=800\, {m}$

$({K} . {E} .)_{{f}}=\frac{1}{2} \frac{{m}}{2} \cdot(20)^{2}+\frac{1}{2} \cdot \frac{{m}}{2}(60)^{2}=1000\, {m}$

$|\Delta {K} . {E} .|=|1000\, {m}-800 \,{m}|=200\, {m}$

$\frac{\Delta {K} . {E}}{({K} . {E} .)_{{i}}}=\frac{200 \,{m}}{800\, {m}}=\frac{1}{4}=\frac{{x}}{4}$

${x}=1$

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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