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In a smooth circular tube of radius R, a mass of 3m at rest splits into two fragments of m and 2m respectively. The smaller mass attains a speed of $2 V_{0}$ tangentially at that point and continues to move in circular path. The other fragment also moves in the circular tube. The arrangement is present in a horizontal plane. Which of the following is/are correct.

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instantaneously the heavier fragment acquires a speed of $2 V_{0}$

instantaneously the heavier fragment acquires a speed of $V_{0}$

the two meet each other after a time interval of $2 π R / 3 V_{0}$

the two meet each other after a time interval of $2 π R / 4 V_{0}$

answer is B, C.

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Detailed Solution

From conservation of linear momentum

$3 m (0) = m (2 V_{0}) + 2 m \bar{v} \Rightarrow \bar{v} = - V_{0}$

Instantaneously the heavier fragment acquires a speed of $V_{0}$ .

Both travel a distance of $2 π R$ in a time “t”.

$t = \frac{2 π R}{3 v_{0}}$

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