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  • Two particles A and B start moving due to their mutual interaction only. If at any time 't;, <img alt="\vec{a}_{A} \; and\; \vec{a}_{B}" src="https://learn.careers360.com/latex-image/?%5Cvec%

Two particles A and B start moving due to their mutual interaction only. If at any time 't;, \vec{a}_{A} \; and\; \vec{a}_{B} are their respective accelerations, \vec{v}_{A} \; and\; \vec{v}_{B} are their respective velocities and upto that time {W}_{A} \; and\; {W}_{B} are the work done on A & B respectively by the mutual force , {m}_{A} \; and\; {m}_{B} are their masses respectively , then which of the following is always correct.
Option: 1

\vec{V}_{A}+\vec{V}_{B}=0


Option: 2

m_{A}\; \vec{V}+m_{B}\; \vec{V}_{B}=0


Option: 3 W_{A}+ W_{B}=0

Option: 4

\vec{a}_{A}+\vec{a}_{B}=0


Answers (1)

As we have learnt in

 

Law of Consevation of Momentum -

 \vec{F}=\frac{\vec{dp}}{dt}

\vec{F}=0            then \vec{p}=constant

\vec{p}=\vec{p}_{1}+\vec{p}_{2}+\cdots =const

- wherein

\ast Independent of frame of reference

 

 

Since,\sum \vec{F}_{ext}=\vec{0}

\            Momentum of system will remain conserved, in this case Momentum of system will remain  zero.

Posted by

Sumit Saini

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