If momentum of a body increases by 100% then what will be percentage increase in its kinetic energy ?

Options

A. `200%`
B. `100%`
C. `300%`
D. `400%`

Answer

C

Explanation

`KE_(0)=(p^(2))/(2m)`
Where is momentum and m is mass. Momentum is increased by 100%, then
`KE'=((2p)^(2))/(2m)=(4p^(2))/(2m)`
Percentage increase in K.E. is
`=(((4p^(2))/(2m)-(p^(2))/(2m)))/((p^(2))/(2m))xx100((p^(2))/(2m)(4-1))/((p^(2))/(2m))=300%`