A system of four particles moves along one dimension. The center of mass of the

Question

A system of four particles moves along one dimension. The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system. Find the velocity, V4, of particle 4 at t 2.83 s, given the following details for the motion of particles 1, 2 and 3: m/s,×, m / s 2) x 1 m/s*)×1 Particle 1: m1=2.05 kg, Particle 2: m,-2.93 kg, Particle 3: m3 = 4.25 kg, Particle 4 m4-4.67 kg, v,A=(6.09 v2 (r) = (7.83 v3(d= (7.87 m/s) + (0.321 m / s) + (0.399 m/s) + (0.321 Number m/ s

Explanation / Answer

Given m1 = 2.05 kg, m2 = 2.93 kg, m3 = 4.25 kg, m4 = 4.67 kg

At time t1 = 2.83 s

v1 = 6.09 + (0.321 * 2.83) = 6.99 m/s

v2 = 7.83 + (0.39 * 2.83) = 8.93 m/s

v3 = 7.87 + (0.321 * 2.83) = 8.77 m/s

v4 = ?

the center of mass of velocity of the system is given by

vcm = (m1*v1 + m2*v2 + m3*v3 + m4*v4) / (m1 + m2 + m3 + m4)

0 = (2.05*6.99 + 2.93*8.93 + 4.25*8.77 + 4.67*v4) / (2.05 + 2.93 + 4.25 + 4.67)

v4 = - 16.65 m/s

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