Question

If the momentum of a body is doubled, by what percentage does the kinetic energy change

Answer 1

initial kinetic energy =p^2/2m ,if momentum is double then kinetic energy=(2p)^2/2m =4p^2/2m

change in kinetic energy=3p^2/2m

%change in kinetic energy=

∆k.E/initial k.E =3p^2/2m/p^2/2m×100 =300%