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14/02/2022 5:09 pm
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A block moving horizontally on a smooth surface with a speed of 40 m/s splits into two parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{8}\)
(4) \(\frac{1}{4}\)
1 Answer
1
14/02/2022 5:15 pm
Correct answer: (3) \(\frac{1}{8}\)
Explanation:
3MV0 = 2MV2 + MV1
3V0 = 2V2 + V1
120 = 2V2 + 60
⇒ V2 = 30 m/s
\(\frac{ΔK.E}{K.E}\) = \(\frac{\frac{1}{2}MV_1^2 + \frac{1}{2}2MV_2^2 - \frac{1}{2}3MV_0^2}{\frac{1}{2}3MV_0^2}\)
= \(\frac{V_1^2 + 2V_2^2 - 3V_0^2}{3V_0^2}\)
= \(\frac{3600 + 1800 - 4800}{4800}\)
= \(\frac{1}{8}\)