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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")XX 100` `(K_(2)-K_(1))/(K_(1)) xx 100 = (4K-K)/(K) xx 100= 300%` |
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