Question

When momentum of a body increases by 200% its Kinetic energy increases by?

Answer 1

answer is 800% ( using K = p^2 / 2m )

let's see how-

let initial kinetic energy be K1 = (p^2) / 2m

new kinetic energy K2 = { p + p ( 200 / 100 ) }^2 / 2m

solving further.... we get-

K2 = 9 p^2 / 2m

Now,

let x be the % increase in kinetic energy.

K1 + { K1 * ( x / 100 ) } = K2

you will get-

( p^2 / 2m ) * ( x / 100 ) = ( 9 p^2 / 2m ) - ( p^2 / 2m )

solve this. . . .

to get

(x / 100) = (4 p^2 / m) * (2m / p^2 )

x/100 = 8

x=800%


OR



Momentum,p = mv
P increases by 200%
P1 = 200/100P + P = 3P
m1 = m
v1 = 3v

Increase in K.E = {1/2m(3v)^2 - 1/2mv^2} / 1/2mv^2 x 100%
= 9 - 1 x 100%
= 800%