Consider a system of two particles in the xy-plane. For the first particle, Its mass is m₁ = 1.30 kg Its position is 7₁ = (1.202 + 2.203) m Its velocity is ₁ = (2.200€ + 0.1003) m/s For the second particle, Its mass is m₂ = 2.90 kg Its position is 72 = (-3.60% - 2.403) m Its velocity is ₂ = (2.2002 - 2.000)) m/s a. Find the position of the center of mass of the system. 7CM = ₁m+m b. Determine the velocity of the center of mass. OM = m/s + 3 m/s 2 c. What is the total linear momentum of the system? PT = kg-m/s + kg-m/s
Consider a system of two particles in the xy-plane. For the first particle, Its mass is m₁ = 1.30 kg Its position is 7₁ = (1.202 + 2.203) m Its velocity is ₁ = (2.200€ + 0.1003) m/s For the second particle, Its mass is m₂ = 2.90 kg Its position is 72 = (-3.60% - 2.403) m Its velocity is ₂ = (2.2002 - 2.000)) m/s a. Find the position of the center of mass of the system. 7CM = ₁m+m b. Determine the velocity of the center of mass. OM = m/s + 3 m/s 2 c. What is the total linear momentum of the system? PT = kg-m/s + kg-m/s
Related questions
Question
Transcribed Image Text:Consider a system of two particles in the xy-plane.
For the first particle,
Its mass is m₁ = 1.30 kg
Its position is 7¹₁ = (1.202 + 2.203) m
Its velocity is ₁ = (2.2002 + 0.100)) m/s
For the second particle,
Its mass is m₂ = 2.90 kg
Its position is 7¹2 = (-3.60% - 2.403) m
Its velocity is v₂ = (2.2001 - 2.000)) m/s
a. Find the position of the center of mass of the system.
7CM =
im+m
b. Determine the velocity of the center of mass.
UCM = 2 m/s + m/s
c. What is the total linear momentum of the system?
Pr =
kg-m/s + kg-m/s
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
