Two bodies of mass $4 k g$ and $5 k g$ are moving along east and north directions with velocities $5 m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is:

\frac{25}{9} m / s

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\frac{9}{25} m / s

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\frac{41}{9} m / s

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\frac{16}{9} m / s

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Solution

The correct option is C $\frac{25}{9} m / s$
Let east direction be x axis and north be y axis. Then, velocity of first body is $v_{1} = 5 i m / s$ and that of body 2 will be $v_{2} = 3 j m / s$ . Since velocity of center of mass is given by

$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}} = \frac{4 (5 i) + 5 (3 j)}{4 + 5} = \frac{20}{9} i + \frac{15}{9} j m / s$

Hence, magnitude of velocity is $| v_{c m} | = \frac{\sqrt{400 + 225}}{9} m / s = \frac{25}{9} m / s$ .

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Two bodies of mass 4 kg and 5 kg are moving along east and north directions with velocities 5 m/s and 3 m/s respectively. Magnitude of the velocity of centre of mass of the system is:

Two bodies of mass $4 k g$ and $5 k g$ are moving along east and north directions with velocities $5 m / s$ and $3 m / s$ respectively. Magnitude of the velocity of centre of mass of the system is: