Let there are three equal masses situated at the vertices of an equilateral triangle, as shown in Fig. Now particle `A` starts with a velocity `v_(1)` towards line `AB`, particle `B` starts with the velocity `v_(2)`, towards line `BC` and particle `C` starts with velocity `v_(3)` towards line `CA`. Find the displacement of the centre of mass of the three particles `A, B` and `C` after time `t`. What would it be if `v_(1)=v_(2)=v_(3)`?

Explanation

First we write the three velocities in vectorial form, taking right direction as positive `x`-axis and upwards as positive `y`-axis.
`vec_(1)=-1/2v_(1)hati-sqrt3/2v_(1)hatj`
`vecv_(2)=v_(2)hati, vecv_(3)=1/2v_(3)hati+sqrt3/2v_(3)hatj`
Thus the velocity of centre of mass of the system is
`vecv_(CM)=(vecv_(1)+vecv_(2)+vecv_(3))/3`
`=(v_(2)-1/2v_(1)-1/2v_(3))hati+sqrt3/2(v_(3)-v_(1))hatj` which can be written as `vecv_(CM)=v_(x)hati+v_(y)hatj`
Thus displacement of the centre of mass in time `t` is `/_\vecr=v_(x)thati+v_(y)thatj`
If `v_(1)=v_(2)=v_(3)=v` we have
`vecv_(CM)=0`
Therefore there is no displacement of centre of mass of the system.