Kinetic energy of a particle is increased by 300 %.Find the percentage increase in momentum.

Explanation

Kinetic energy `E=(1)/(2)mv^(2)`, Momentum,p=mv
When E is increased by 300%
`E'=E+3E=4E=4((1)/(2)mv^(2))=2mv^(2)` If v' is velocity of body,then`(1)/(2)m(v')^(2) =2mv^(2)`
`impliesv'=2 v` So, p'=mv'=2mv
hence , percentage change in momentum
`=(2mv-mv)/(mv)xx100=100%`