Two particles A and B initially at rest move | vedclass

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Question

Force $F_{A}$ on particle $A$ is given by

$F_{A}=m_{A} a_{A}=\frac{m_{A} v}{t} \quad \ldots$ $(1)$

similarly $F_{B}=m_{B} a_{B}=\frac{m_{B} 	im

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Vedclass · Answer

Force $F_{A}$ on particle $A$ is given by

$F_{A}=m_{A} a_{A}=\frac{m_{A} v}{t} \quad \ldots$ $(1)$

similarly $F_{B}=m_{B} a_{B}=\frac{m_{B} 	imes 2 v}{t}$           $...(2)$

Now $\frac{m_{A} v}{t}=\frac{m_{B} 	imes 2 v}{t}\left(\because F_{A}=F_{B}ight)$

So $m_{A}=2 m_{B}$

For the centre of mass of the system

$v=\frac{m_{A} v_{A}+m_{B} v_{B}}{m_{A}+m_{B}}$

${	ext { Or } v=\frac{2 m_{B} v-m_{B} 	imes 2 v}{2 m_{B}+m_{B}}=0}$

Negative sign is used bacause the particles are travelling in opposite directions.

Alternatively, if

 we consider the two masses in a system then no external force is acting on the system. Mutual forces are internal forces. since the centre of mass is initially at rest, it well remain at rest.

Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed of $B$ is $2v$, the speed of centre of mass of the system is

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