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Q.

A spring is held compressed. Its stored energy is 2. 4 joule. Its ends are in contact with masses of 1 gm and 48 gm placed on a smooth horizontal surface. When the spring is released, the mass will acquire a velocity of

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a

$\frac{\bar{2} \cdot 40}{149} m / s$

b

$\frac{24 \times 49}{48} m / s$

c

$\frac{10}{7} m / s$

d

$\frac{10^{4}}{7} m / s$

answer is C.

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Detailed Solution

$\frac{1}{2} \times (\frac{1}{1000}) \times v_{1}^{2} + \frac{1}{2} \times (\frac{48}{1000}) \times v_{2}^{2} = 2 \cdot 4 or v_{1}^{2} + 48 v_{2}^{2} = 2 \cdot 4 \times 2000 = 4800 Applying the law of conservation of momentum \frac{1}{1000} v_{1} = \frac{48}{1000} v_{2} or v_{1} = 48 v_{2} Frorn eq . [1), we have {(48 v_{2})}^{2} + 48 v_{2}^{2} = 4800 Solving, we get v_{2} = (10 / 7) m / s$

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