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Standard XII

Physics

Velocity of COM

Two bodies of...

Question

Two bodies of different masses $2 k g$ and $4 k g$ are moving with velocities $2 m / s$ and $10 m / s$ towards each other due to mutual gravitational attraction. Then the velocity of the centre of mass is

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Solution

Taking direction of $4 k g Q s + V e$ ,

$V_{C M} = \frac{(m_{1} \times v_{1}) + (m_{2} \times v_{2})}{m_{1} + m_{2}}$

$= \frac{2 \times (- 2) + 4 \times 10}{2 + 4}$

$\Rightarrow \frac{- 4 + 40}{6} = \frac{36}{6} = 6 m / s$

$6 m / s$ in direction of $4 k g$ block.

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Two bodies of different masses 2 kg and 4 kg are moving with velocities 2 m/s and 10 m/s towards each other due to mutual gravitational attraction. Then the velocity of the centre of mass is

Taking direction of 4kgQs+Ve,VCM=(m1×v1)+(m2×v2)m1+m2=2×(−2)+4×102+4⇒−4+406=366=6m/s6m/s in direction of 4kg block.

Two bodies of different masses $2 k g$ and $4 k g$ are moving with velocities $2 m / s$ and $10 m / s$ towards each other due to mutual gravitational attraction. Then the velocity of the centre of mass is

Taking direction of $4 k g Q s + V e$ ,

$V_{C M} = \frac{(m_{1} \times v_{1}) + (m_{2} \times v_{2})}{m_{1} + m_{2}}$

$= \frac{2 \times (- 2) + 4 \times 10}{2 + 4}$

$\Rightarrow \frac{- 4 + 40}{6} = \frac{36}{6} = 6 m / s$

$6 m / s$ in direction of $4 k g$ block.