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The momentum of a body is doubled. By wh...

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

Text Solution

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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) "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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