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If two particles of masses $m_{1}$ and $m_{2}$ move with velocities $v_{1}$ and $v_{2}$ towards each other on a smooth horizontal plane, what is the velocity of their centre of mass. ?

A

V = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}

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B

V = \frac{m_{1} v_{1} - m_{2} v_{2}}{m_{1} + m_{2}}

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C

V = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} - m_{2}}

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D

V = \frac{m_{1} v_{1} - m_{2} v_{2}}{m_{1} - m_{2}}

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Solution

The correct option is A $V = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$
As per center of mass of two objects say 1 and 2 we get then
$X = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$
$where x_{1} a n d x_{2} are position of obeject 1 and 2 from reference point respectively$
Position of object 1 and 2 at time =t
$x_{1} = v_{1} t + x_{u_{1}} x_{2} = v_{2} t + x_{u_{2}}$
where $x_{u_{1}} a n d x_{u_{2}}$ are the initial position of the objects respectively.
Now the center of mass of bjects 1 and 2 at time = t
$X (t) = \frac{{m_{1} (v_{1} t + x_{u_{1}}) + m_{2} (v_{2} t + x_{u_{2}})}}{m_{1} + m_{2}}$
Now differentiating with respect to time t, we get
$V (t) = \frac{d}{d t} {\frac{{m_{1} (v_{1} t + x_{u_{1}}) + m_{2} (v_{2} t + x_{u_{2}})}}{m_{1} + m_{2}}} V = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

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