If momentum of a body is increased by 20 %, t | vedclass

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Question

$P ^{\prime}= P +\frac{20}{100} P =1.2 P$

$\%$ change in $KE =\frac{ K ^{\prime}- K }{ K } 	imes 100$

$=\left(\frac{\frac{ P ^{2}}{2 m }-\frac{

Vedclass · Accepted Answer

$44$

Vedclass · Answer

$P ^{\prime}= P +\frac{20}{100} P =1.2 P$

$\%$ change in $KE =\frac{ K ^{\prime}- K }{ K } 	imes 100$

$=\left(\frac{\frac{ P ^{2}}{2 m }-\frac{ P ^{2}}{2 m }}{\frac{ P ^{2}}{2 m }}ight) 	imes 100$

$= {\left[(1.2)^{2}-1ight] 	imes 100 }$

$= 44\%$

If momentum of a body is increased by $20 \%$, then its kinetic energy increases by$....\%$

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