Calculate the velocity of the centre of mss of the system of particles shown in figure.

Explanation

`m_1=1.0kg`
`V_1=(-15 cos 37^0veci-1.5sin37^0vecj)`
`m_2=1.2kg`
`V_2=0.4vecj`
`m_3=1.5kg`
`V_3=-1.0cos37^0veci+1.0sin37^0vecj`)`
`m_4=0.50kg`
`V_4=3.0veci`
`m_5=1.0kg`
`V_5=2.0cos37^0veci-2.0sin37^0vecj`
`V_(cm)=(m_1 v_1+m_2v_2+m_2v_3+m_4v_4+m_5v_5)/(m_1+m_2+m_3+mK_4+m_5)`
`1/(1.0+1.2+1.5+1.0+0.50)^([1.0(-1.5xx4/5veci-1.5xx3/5vecj)+...-2.0xx3/5vecj])`
`=0.20m/s at 45^0` below the direction towards right,