The momentum of a body is increased by 100% .what is the percentage increase in kinetic energy ??

Dear students Kinetic energy = p^2/2m     If the momentum increases by 100% , then the p increases to 2p. Substitute p =1 K before increase in momentum = p^2/2m = 1/2m                  - (1) K' after increase in momentum by 100% = 2p^2/2m= 4/2m     - (2) Let p = 1  Divide (2) by (1) and substitute p=1 K'/K = (4/2m) / (1/2m)             = 4 ( 2m gets divided by 2m to give 1)       K'= 4K ( cross-multiplication) Percentage increase = (K'-K / K) * 100                                   = (4K-K/K) *100                                   = (3K /K) *100                                   = 3*100                                   = 300% When momentum of a body increases by 100% then its kinetic energy will increase by 300%. Regards