Consider a system of two particles in the xy plane: m1 = 2.60 kg is at r1 = (1.0

Question

Consider a system of two particles in the xy plane: m1 = 2.60 kg is at r1 = (1.00 i+ 2.00 j) m and has velocity (3.00 i + 0.500 j) m/s; m2 = 3.80 kg is at r2 = (- 4.00 i - 3.00 j) m and has velocity (3.00 i - 2.00 j) m/s.
(d) What is the total linear momentum of the system?

I need help with Part D.
(b) Find the position of the center of mass of the system and mark it on the grid.

(c) Determine the velocity of the center of mass and also show it on the diagram.

Consider a system of two particles in the xy plane: m1 = 2.60 kg is at r1 = (1.00 i+ 2.00 j) m and has velocity (3.00 i + 0.500 j) m/s; m2 = 3.80 kg is at r2 = (- 4.00 i - 3.00 j) m and has velocity (3.00 i - 2.00 j) m/s. (d) What is the total linear momentum of the system? I need help with Part D. (b) Find the position of the center of mass of the system and mark it on the grid. V CM = (3.0 m/s i -0.98 m/s j) Correct Answer R CM =(-1.97 m i -0.97 m j) Correct Answer (c) Determine the velocity of the center of mass and also show it on the diagram.

Explanation / Answer

Mass  m 1 = 2.60 kg
Velocity of mass m 1 is u 1 = (3.00 i + 0.500 j) m/s Mass m 2 = 3.80 kg Velocity of mass m 2 is u 2 = (3.00 i - 2.00 j) m/s.
(d) The total linear momentum of the system P =  m 1 u 1+ m 2 u 2                                                                       = 7.8 i + 1.3 j + 11.4 i - 7.6 j                                                                       = 19.2 i -6.3 j Magnitude of P = [ 19.2 2 + (-6.3) 2 ]                         = 20.2 kg m/ s                                                                       = 7.8 i + 1.3 j + 11.4 i - 7.6 j                                                                       = 19.2 i -6.3 j Magnitude of P = [ 19.2 2 + (-6.3) 2 ]                         = 20.2 kg m/ s

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