The momentum of a body is doubled. By what percentage does its kinetic energy increase?

Explanation

`K= (p^(2))/(2m) rArr (K_(1))/(K_(2))= (p_(1)^(2))/(p_(2)^(2))`
Let `k_(1)= K, p_(1)= p " then " p_(2)= 2p, K_(2)`=?
`(K)/(K_(2)) = (p^(2))/((2p)^(2)), K_(2)= 4K`
% increase in kinetic energy= `("Increase in kinetic energy")/("Initial kinetic energy")xx 100`
`(K_(2)-K_(1))/(K_(1)) xx 100 = (4K-K)/(K) xx 100= 300%`