An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be

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2 m/s

8 m/s

1 m/s

4 m/s

answer is A.

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Detailed Solution

$\frac{1}{2} {Kx}^{2} = {KE}_{A} + {KE}_{B}$ and $m_{A} v_{A} = m_{B} v_{B}$

$\Rightarrow v_{A} = 2 v_{B}$ ...(i)

$\frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} {(V_{B})}^{2} = 12$

$\frac{1}{2} m_{A} \times 4 V_{B}^{2} + \frac{1}{2} m_{B} {(V_{B})}^{2} = 12$

on substituting the values of masses

we get $v_{B} = 2 m / s$

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