If the momentum of a body is doubled, its KE. increases by 

(A) 50% 

(B) 300% 

(C) 100% 

(L) 400%

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2 Answers

Correct option is: (B) 300%

Kinetic energy , K = \(\frac{P^2}{2m}\)

When momentum p is doubled , the kinetic energy becomes

\(K^1 = \frac{(2p)}{2m}\)

\(K^1 = 4 (\frac{p^2}{2m})\)

\(K^1 = 4K\)

increases in kinetic energy = \(\frac{K^1 - K}{K}\)

\(= \frac{4K - K}{K} \times 100\)

\(\left(\frac{3K}{K}\right) \times 100\)

300%

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Correct option is: (B) 300%

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