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Kinetic energy of a particle is increase...

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

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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%`
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  • If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

    A
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    B
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    C
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