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The momentum of a body is increased by 50%. What is the percentage change in its K.E.? |
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Answer» When the momentum is increased by 50%, velocity increase by 50%, i.e., velocity becomes \(\frac{3}{2}\) times, so, K.E. becomes \(\frac{9}{4}\) times, i.e. \(\frac{9}{4}\) × 100 = 225%. Hence increase in K.E. = 225 – 100 = 125%. |
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