Calculate the velocity of the centre of mass of the system of particles shown in figure. 
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$\text{m}_1=1\text{kg},\ \vec{\text{v}}_1=-1.5\cos37\hat{\text{i}}-1.55\sin37\hat{\text{j}}=-1.2\hat{\text{i}}-0.9\hat{\text{j}}$ $\text{m}_2=1.2\text{kg},\ \vec{\text{v}}_2=0.4\hat{\text{j}}$ $\text{m}_3=1.5\text{kg},\ \vec{\text{v}}_3=-0.81\hat{\text{i}}+0.6\hat{\text{j}}$ $\text{m}_4=0.5\text{kg},\ \vec{\text{v}}_4=3\hat{\text{i}}$ $\text{m}_5=1\text{kg},\ \vec{\text{v}}_5=1.6\hat{\text{i}}-1.2\hat{\text{j}}$ 
So, $\vec{\text{v}}_{\text{c}}=\frac{\text{m}_1\vec{\text{v}}_1+\text{m}_2\vec{\text{v}}_2+\text{m}_3\vec{\text{v}}_3+\text{m}_4\vec{\text{v}}_4+\text{m}_5\vec{\text{v}}_5}{\text{m}_1+\text{m}_2+\text{m}_3+\text{m}_4+\text{m}_5}$ $=\frac{1(-1.2\hat{\text{i}}-0.9\hat{\text{j}})+1.2(0.4\hat{\text{j}})+1.5(-0.8\hat{\text{i}}+0.6\hat{\text{j}})+0.5(3\hat{\text{i}})+1(1.6\hat{\text{i}}-1.2\hat{\text{j}})}{5.2}$ $=\frac{-1.2\hat{\text{i}}-0.9\hat{\text{j}}+4.8\hat{\text{j}}-1.2\hat{\text{i}}+0.90\hat{\text{j}}+1.5\hat{\text{i}}+1.6\hat{\text{i}}-1.2\hat{\text{i}}}{5.2}$ $=\frac{0.7\hat{\text{i}}}{5.2}-\frac{0.72\hat{\text{j}}}{5.2}$
So, $\vec{\text{v}}_{\text{c}}=\frac{\text{m}_1\vec{\text{v}}_1+\text{m}_2\vec{\text{v}}_2+\text{m}_3\vec{\text{v}}_3+\text{m}_4\vec{\text{v}}_4+\text{m}_5\vec{\text{v}}_5}{\text{m}_1+\text{m}_2+\text{m}_3+\text{m}_4+\text{m}_5}$ $=\frac{1(-1.2\hat{\text{i}}-0.9\hat{\text{j}})+1.2(0.4\hat{\text{j}})+1.5(-0.8\hat{\text{i}}+0.6\hat{\text{j}})+0.5(3\hat{\text{i}})+1(1.6\hat{\text{i}}-1.2\hat{\text{j}})}{5.2}$ $=\frac{-1.2\hat{\text{i}}-0.9\hat{\text{j}}+4.8\hat{\text{j}}-1.2\hat{\text{i}}+0.90\hat{\text{j}}+1.5\hat{\text{i}}+1.6\hat{\text{i}}-1.2\hat{\text{i}}}{5.2}$ $=\frac{0.7\hat{\text{i}}}{5.2}-\frac{0.72\hat{\text{j}}}{5.2}$
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