A body of mass $m_{1} = 4 k g$ moves at $5 i m / s$ and another body of mass $m_{2} = 2 k g$ moves at $10 i m / s$ . The kinetic energy of centre of mass is

\frac{200}{3} J

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\frac{500}{3} J

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\frac{400}{3} J

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\frac{800}{3} J

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Solution

The correct option is C $\frac{400}{3} J$
$v_{c} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$v_{c} = \frac{40}{6} m s^{- 1}$

Kinetic energy $= \frac{1}{2} m v^{2} = \frac{1}{2} \times 6 \times (\frac{40}{6})^{2}$

$K_{c} = \frac{400}{3} J$

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A body of mass m1=4 kg moves at 5i m/s and another body of mass m2=2 kg moves at 10i m/s. The kinetic energy of centre of mass is

The correct option is C 4003Jvc=m1v1+m2v2m1+m2vc=406ms−1Kinetic energy =12mv2=12×6×(406)2Kc=4003J

A body of mass $m_{1} = 4 k g$ moves at $5 i m / s$ and another body of mass $m_{2} = 2 k g$ moves at $10 i m / s$ . The kinetic energy of centre of mass is

The correct option is C $\frac{400}{3} J$
$v_{c} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$v_{c} = \frac{40}{6} m s^{- 1}$

Kinetic energy $= \frac{1}{2} m v^{2} = \frac{1}{2} \times 6 \times (\frac{40}{6})^{2}$

$K_{c} = \frac{400}{3} J$