If the momentum of an object increases by $20 %$ $20%$ , what will be the percent increase in its kinetic energy?

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20 %

$20%$ , what will be the percent increase in its kinetic energy?"

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Answer 1 · 2015-05-11T21:04:26

The answer is $44 %$ $44%$

Method
If Momentum of the body increases by $20 %$ $20%$ it means that
$DeltaP=20% " of " P_("before")=0.2P_("before")=0.2m u$

$=>DeltaP=0.2m u$

where $DeltaP$ is the increase in momentum
$m$ is mass
$u$ is initial velocity
$v$ is final velocity

$=>mv-m u=0.2m u$

Simplify that and you'll get,

$v=1.2u$

We were asked to find the percentage increase in $KE$ ,

$DeltaKE= xKE_("before")$

And we're looking for that $x$

$=> x=(DeltaKE)/(KE_("before")$

$=>x=(1/2mv^2-1/2m u^2)/(1/2m u^2)=(cancel(1/2m)(v^2-u^2))/(cancel(1/2m) u^2)=(v^2-u^2)/u^2$

Remember $v=1.2u$

$=>x=((1.2u)^2-u^2)/u^2=(0.44u^2)/u^2=0.44$

$x=44%$

If the momentum of an object increases by 20%20%20%, what will be the percent increase in its kinetic energy?