Consider a system of two particles in the xy-plane.For the first particle,Its mass is m₁ = 1.30 kgIts position is 7¹₁ = (1.202 + 2.203) mIts velocity is ₁ = (2.2002 + 0.100)) m/sFor the second particle,Its mass is m₂ = 2.90 kgIts position is 7¹2 = (-3.60% - 2.403) mIts velocity is v₂ = (2.2001 - 2.000)) m/sa. Find the position of the center of mass of the system.7CM =im+mb. Determine the velocity of the center of mass.UCM = 2 m/s + m/sc. What is the total linear momentum of the system?Pr =kg-m/s + kg-m/s

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