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)`?

Options

A. zero
B. `(v_(1)+v_(2)+v_(3))/3t`
C. `(v_(1)+(sqrt(3))/2v_(2)+v_(3)/2)/3 t`
D. `(v_(1)+v_(2)+v_(3))/4 t`

Answer

A

Explanation

Using `vecv_(CM)=hativ_(x)+hatjv_(y)`
where `v_(x)=(mv_((x)_(1))+mv_((x)_(2))+mv_((x)_(2)))/(3m)=0`
`v_(y)=(mv_(y_1)+mv_(y_2)=mv_(x_3))/(3m)=0`
net velocity of zero (using vector theory) i.e. `CM` is at rest. So, displcement of `CM` of zero. So choices `b, c` and `d` are wrong.