Two balls with masses m₁ = 1kg and m₂ = 2kg are slipped on a thin smooth horizontal rod as shown. The balls are interconnected by a light spring of stiffness K = 24 N /m . The left hand ball (m₁) m is imparted the initial velocity 12 cm/s towards second ball. The amplitude of oscillations made by the arrangement is...................

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3 cm

2 cm

1 cm

4 cm

answer is B.

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Detailed Solution

Total mechanical energy of system = kinetic energy of mass $m_{1} = \frac{1}{2} m_{1} v_{1}^{2} . . . . (1)$

But total mechanical energy = kinetic energy due to motion of centre of mass + energy of oscillation. $\Rightarrow$ energy of oscillation.

$\begin{array}{l} = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} (m_{1} + m_{2}) {[\frac{m_{1} v_{1}}{m_{1} m_{2}}]}^{2} = \frac{1}{2} μ v_{1}^{2} = \frac{1}{2} μ (A w)^{2} \\ \Rightarrow v_{1} = A w \Rightarrow A = \frac{v_{1}}{w}; v_{1} = 12 cm / s w = \sqrt{\frac{k}{μ}} \end{array}$