If momentum is increased by `20%` then K.E. increases by

Options

A. 0.44
B. 0.55
C. 0.66
D. 0.77

Answer

A

Explanation

`E=(P^(2))/(2m)`. If m is constant then `E prop P^(2)`
`rArr (E_(2))/(E_(1))=((P_(2))/(P_(1)))^(2)=((1.2P)/(P))^(2)=1.44`
`rArr E_(2)=1.44 E_(1)=E_(1)+0.44 E_(1)`
`E_(2)=E_(1)+44%` of `E_(1)`
i.e. the kinetic energy will increase by 44%