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Three particles of equal masses are plac...

Three particles of equal masses are placed at the corners of an equilateral triangle as shown in the figure. Now particle `A` starts with a velocity `v_(1)` towards line `AB`, particle `B` starts with a velocity `v_(2)` towards line `BC` and particle `C` starts with velocity `v_(3)` towards line `CA`. The displacement of `CM` of three particle `A, B` and `C` after time `t` will be (given if `v_(1)=v_(2)=v_(3))`

Text Solution

Verified by Experts

First we write the three velocities in vectorial form, taking right direction as positive x-axis and upwards as positive y-axis.
`v_(1)=-(1)/(2)v_(1)hat"i"-(sqrt3)/(2)v_(1)hat"j"`
`v_(2)=v_(2)hat"i"`
`v_(3)=-(1)/(2)v_(3)hat"i"+(sqrt3)/(2)v_(3)hat"j"`
Thus the velocity of certre of mass of the system is, `vecv=(m_(1)vecv_(1)+m_(2)vecv_(2)+m_(3)vecv_(3))/(m_(1)+m_(2)+m_(3))`
`vec"v"_("cm")=(vecv_(1)+vecv_(2)+vecv_(3))/(3)`
`vec"v"_("cm")=((v_(2)-(1)/(2)v_(1)-(1)/(2)v_(3))hati+(sqrt3)/(2)(v_(3)-v_(1))j)/(3)`
Which can be written as `vec"v"_(cm)=v_(x)hat"i"+v_(y)hat"j"`
Thus displacement of the centre of mass in time `t` is `Deltar=v_(x)that"i"+v_(y)that"j"`
If `v_(1)=v_(2)=v_(3)=v` we have `vec"v"_(cm)=0`
Therefore no displacement of centre of mass of the system.
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