A spring is held compressed so that its stored energy is 2.4 J. Its ends are in contact with masses 1 g and 48 g placed on frictionless table. When the spring is released, the heavier mass will acquired a speed of

\frac{2.4}{49} m s^{- 1}

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\frac{2.4 \times 48}{49} m s^{- 1}

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\frac{10}{7} c m s^{- 1}

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\frac{10}{7} m s^{- 1}

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Solution

The correct option is D $\frac{10}{7} m s^{- 1}$
$m_{2} = 48 g$ , $m_{1} = 1 g$
$\frac{m_{1} v_{1}^{2}}{2} + \frac{m_{2} v_{2}^{2}}{2} = 2.4$ ......... (1)
Conserving momentum
$m_{1} v_{1} = m_{2} v_{2}$
$v_{2} = \frac{10}{7} m s^{- 1}$

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A spring is held compressed so that its stored energy is 2.4 J. Its ends are in contact with masses 1 g and 48 g placed on frictionless table. When the spring is released, the heavier mass will acquired a speed of

The correct option is D 107ms−1m2=48g, m1=1gm1v212+m2v222=2.4 ......... (1)Conserving momentumm1v1=m2v2v2=107ms−1

The correct option is D $\frac{10}{7} m s^{- 1}$
$m_{2} = 48 g$ , $m_{1} = 1 g$
$\frac{m_{1} v_{1}^{2}}{2} + \frac{m_{2} v_{2}^{2}}{2} = 2.4$ ......... (1)
Conserving momentum
$m_{1} v_{1} = m_{2} v_{2}$
$v_{2} = \frac{10}{7} m s^{- 1}$