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Byju's Answer

Standard XII

Physics

Direction of Instantaneous Velocity

A mass m 1 ...

Question

A mass $m_{1}$ with initial speed $v_{0}$ in the positive $x$ -direction collides with a mass $m_{2} = 2 m_{1}$ which is initially at rest at the origin, as shown in figure. After the collision $m_{1}$ moves off with speed $v_{1} = v_{0} / 2$ in the negative $y$ - direction, and $m_{2}$ moves off with speed $v_{2}$ at angle $θ$ . Find the velocity (magnitude and direction) of the centre of mass after the collision :

v_{0} / 3

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v_{0} / 2

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v_{0} / 5

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v_{0} / 6

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Solution

The correct option is C $v_{0} / 3$
Since there is no external force acting on the 2 mass system, the momentum of the system and the velocity of the centre of mass remains constant.
Therefore, $V_{c m}$ is the same before and after the collision, i.e;
$V_{c m, x} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$ , as $m_{2} = 2 m_{1}$ & $V_{c m, y} = 0.$
$V_{c m, x} = \frac{m_{1} v_{o} + 2 m_{1} \times 0}{m_{1} + 2 m_{1}} = \frac{v_{o}}{3} .$

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