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Two blocks of masses 3kg and 6kg respectively are placed on a smooth horizontal surface. They are connected by a light spring of force constant k = 200 N/m. Initially the spring is unstretched. The indicated velocities are imparted to the blocks. The maximum extension of the spring will be

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20 cm

25 cm

15 cm

30 cm

answer is A.

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Detailed Solution

From conservation of momentum,

Initial momentum = momentum of center of mass

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) V_{cm} \Rightarrow 6 \times 2 - 3 \times 1 = (6 + 3) V_{cm} \Rightarrow V_{cm} = 1 m / s$

$From conservation of energy,$

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} {kx}^{2} + \frac{1}{2} (m_{1} + m_{2}) V_{cm}^{2}$

$\begin{array}{l} \Rightarrow \frac{1}{2} (3) (1)^{2} + \frac{1}{2} (6) (2)^{2} = \frac{1}{2} {Kx}^{2} + \frac{1}{2} (9) V^{2} \end{array}$

$\Rightarrow \frac{3}{2} + \frac{24}{2} = \frac{1}{2} (200) x^{2} + \frac{1}{2} 9 (1)^{2}$

$∴ 100 x^{2} = 9 \Rightarrow x = \sqrt{\frac{9}{100}} = 0 .3 m$

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