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The momentum of a body doubled. By what ...

The momentum of a body doubled. By what percentage does its kinetic energy increases ?

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`K=(p^(2))/(2m)rArr (K_(1))/(K_(2))=(p_(1)^(2))/(p_(2)^(2))`
Let `K_(1)=K, p_(1)=p` then `p_(2)=2p, K_(2)=?`
`(K)/(K_(2))=(p^(2))/((2p)^(2)), K_(2)=4K`.
% increase in kinetic energy = `("Increase in kinetic energy")/("Initial kinetic energy")xx100`
`(K_(2)-K_(1))/(K_(1))xx100=(4K-K)/(K)xx100=300%`
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