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

Options

A. `50%`
B. `100%`
C. `125%`
D. `25%`

Answer

C

Explanation

Let initial momentum `=p_(i)` ,
`:.` final momentum `=p_(f)`
`p_(f)=p_(i)+(50)/(100)p_(i)=(3)/(2)p_(i)` or `(p_(f))/(p_(i))=(3)/(2)`
as `K=(p^(2))/(2m):. K alphap^(2)`
or `(K_(f))/(K_(i))=((p_(f))/(p_(i)))^(2)=(9)/(4)`or `4K_(f)=9K_(i)`
or `K_(f)=(9)/(4)K_(i)`
So, `K_(f)-K_(i)=(5)/(4)K_(i)`
`%` increase in KE
`=(K_(f)-K_(i))/(K_(i))xx100=((5)/(4)K_(i))/(K_(i))xx100=(5)/(4)xx100`
`=125%`