JEE MainPhysicsToughCenter of mass

Three particles each of mass m can slide on fixed frictionless
horizontal circular tracks in the same horizontal plane as shown in the
figure. The coefficient of restitution being e=0.5. Assuming that
m_2and m_3 are at rest initially and lie along a radial line before
impact and the spring is initially unstretched, then maximum extension
in spring in subsequent motion.

Show answer and solution

Velocity just after collision

V_{0} = V_{1} + V_{2}

.....(1)

\Rightarrow \frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}

- V_{1} + V_{2} = \frac{V_{0}}{2}

.....(2)
From (1) & (2)

\Rightarrow V_{1} = \frac{V_{0}}{4}\& V_{2} = \frac{3V_{0}}{4}

When maximum extension occurs then angular speed with rest to centre
for mass m_2& m_3 are same. Using angular momentum conservation
about centre of circle.

m\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\omega + m(2R)^{2}\omega

\Rightarrow \omega = \frac{3}{10}\frac{V_{0}}{R}

So velocity of

m_{2} = \frac{3}{5}V_{0}

& velocity of

m_{3} = \frac{3}{10}V_{0}

When extension is maximum using energy conservation.

\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx

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