Question

KE of a body is increased by 44%.What is the percent increase in the momentum?

Answer 1

the relation between kinetic energy (E) and momentum (p) is given as

E = p2/2m

so,

as the increase in percentage is 300%, then



(E2/E1) X 100 = 300

or

E2 = 3E1



now, here mass remains constant

thus,

p2/p1 = (2mE2)1/2 / (2mE1)1/2

so,

p2/p1 = (E2/E1)1/2

or

p2/p1 = (3)1/2 = 1.732



thus, the percentage increase in momentum would be

p2/p1 X 100 = 173.2 %