Q: If the K.E. of a body is increased by 300 %, its momentum will increase by

(a) 100 %

(b) 150 %

(c) $ \displaystyle \sqrt{300} $ %

(d) 175 %

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Ans: (a)

Sol: K’ = K + K (300/100)

K’ = 4 K

Initial Momentum $ \displaystyle P = \sqrt{2 m K}$

Final Momentum $\displaystyle P’ = \sqrt{2 m K’}$

$ \displaystyle \frac{P’}{P} = \sqrt{\frac{K’}{K}} = 2 $

% Increase in Momentum is

$ \displaystyle \frac{P’-P}{P}\times 100 = (\frac{P’}{P}-1)\times 100 $

$ \displaystyle = (2-1)\times 100 $ = 100 %