Momentum of a body is doubled. By what percent does its KE change?

Explanation

K.E. `=(p^(2))/(2m)` when p is doubled K.E. becomes 4 times.
`:.` % Increase in K.E. `=(DeltaK.E.)/(K.E)xx100=(4K.E-K.E.)/(K.E.)xx100`
`=3xx100=300%`