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The momentum of a body is doubled. By what percentage does its kinetic energy increase ? |
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Answer» <P> Solution :`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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