# Problem: Consider a system consisting of three particles: m1 = 2 kg, v1  &lt; 10, -5, 12 &gt; m/s m2 = 9 kg, v2 &lt;- 14, 3, -3&gt;m/s m3 = 2 kg, v3 &lt;- 23, 37, 20 &gt; m/s What Is the velocity of the center of mass of thls system? vcm = m/s

###### FREE Expert Solution

Velocity of the center of mass:

$\overline{){\stackrel{\mathbf{⇀}}{\mathbf{v}}}_{\mathbf{C}\mathbf{M}}{\mathbf{=}}\frac{\mathbf{\Sigma }{\mathbf{m}}_{\mathbf{i}}{\mathbf{v}}_{\mathbf{i}}}{\mathbf{\Sigma }{\mathbf{m}}_{\mathbf{i}}}}$

vCM,x = Σmivix/Σmi = (m1v1x + m2v2x + m3v3x)/(m1 + m2 + m3) = [(2)(10) + (9)(-14) + (2)(-23)]/(2 + 9 + 2) = -11.7 m/s

###### Problem Details

Consider a system consisting of three particles:

m1 = 2 kg, v1  < 10, -5, 12 > m/s

m= 9 kg, v<- 14, 3, -3>m/s

m= 2 kg, v<- 23, 37, 20 > m/s

What Is the velocity of the center of mass of thls system?

vcm = m/s