The velocity of B immediately after collision is along unit vector:

ˆ i

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\frac{ˆ i + ˆ j}{\sqrt{2}}

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\frac{\sqrt{3}}{2} ˆ i + \frac{1}{2} ˆ j

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none of these

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Solution

The correct option is D none of these
After collision B and C will move together because the collision is inelastic.
Applying the momentum conservation
$m_{c} v_{o} = (m_{b} + m_{c}) v$ where $v$ is the velocity of combined system "B+C".
$m v_{o} = 2 m v$
$v = \frac{v_{o}}{2}$
When the chord becomes tight, the velocity of B and A will be same along the the length of chord.
Velocity of B after collision will be along $-^j$ and magnitude is $v = \frac{v_{o}}{2}$ .

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