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 1 2.99 s, given the following details for the motion of particles 1, 2 and 3: m/s)+(0.365 Particle 1: m,-2.17 kg, v, Particle 2: m, = 2.81 kg, v2(r)=(8.93 m/s)+(0.357 m/s*)×t Particle 3: m,-497 kg, v,li-(787 m/s)+0.489 m/sx, Particle 4: m,-4.91 kg, (r)=(6.97 m/s*)×t (8.93 m/s)+(0.357 m/sx Number m/s

Explanation / Answer

m1v1 + m2 v2 + m3 v3 + m4v4 = 0

v4(t) = -[m1v1 + m2 v2 + m3 v3] / m4

v4(t) = -[2.17 * (6.97 + 0.365 t) + 2.81 *(8.93 + 0.357 t) + 4.97 *(7.87 + 0.489 t)] / 4.91

v4(t) = -[15.12 + 0.792 t + 25.09 + 1.00 t + 39.11 + 2.43 t] / 4.91

v4(t) = -[15.12 + 0.792 t + 25.09 + 1.00 t + 39.11 + 2.43 t] / 4.91

v4(t) = -[79.32 + 4.22 t] / 4.91 = -[16.15 + 0.859 t]

t = 2.99 s

v4 = -[16.15 + 0.859 * 2.99]

v4 = - 18.7 m/s

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