Question

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

Answer 1

The answer is `44%`

Method
If Momentum of the body increases by `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%`