Question

A system consists of two particles. Particle 1 with mass 2.0 kg is located at (2.0 m, 6.0 m) and has a velocity of (2.1 m/s, 4.1 m/s). Particle 2 with mass 4.5 kg is located at (4.0 m, 1.0 m) and has a velocity of (4.6 m/s, 4.6 m/s). Determine the position and the velocity of the center of mass of the system.

I know the answers are:

Position: 3.38x + 2.54y
Velocity: 3.83x + 4.45y

How does one arrive at these answers?

Answer 1

m1=2 kg, x1=2m, y1 6 m, v1x =2.4 m/s, v1y =4.1 m/s,
m2 = 4.5 kg, x2=4 m, y2 = 1 m, v2x=4.5 m/s, v2y =4.6 m/s.

x(c) =(m1•x1+m2•x2)/(m1+m2) =(2•2+4.5•4)/6.5 = 3.38 m,
y(c) =(m1•y1+m2•y2)/(m1+m2) =(2•6+4.5•1)/6.5 = 2.54 m.

v(cx) =(m1•v1x+m2•v2x)/(m1+m2) =(2•2.1+4.5•4.6)/6.5 = 3.83 m/s,
v(cy) =(m1•v1y+m2•v2y)/(m1+m2) =(2•4.1+4.5•4.6)/6.5 = 4.45 m/s.

Position of C.M. 3.38x+2.54y,
Velocity of C.M. 3.83x+4.45y.