Momentum of a particle is increased by `50%`. By how much percentage kinetic energy of particle will increase?

Explanation

`K=(1)/(2)m upsilon^(2)=(p^(2))/(2m)`
When momentum is increased by `50%`.
`p^(')=(3)/(2)p`
`K^(')=(p'^(2))/(2m)=((3//2p)^(2))/(2m)=(9)/(4)((p^(2))/(2m))`
`K^(')-K=(9)/(4)((p^(2))/(2m))-(p^(2))/(2m)=(5)/(4)((p^(2))/(2m))=1.25K`
`%` age increase in KE `=((K^(')-K)100)/(K)`
`(1.25K)/(K)xx100=125%`