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 T1= 2.59s, given the following details for the motion of particles 1,2,and 3: Particle 1: m1= 1.93 kg,                    v1(t) = (6.97 m/s)+(0.211 m/s^2)*t Particle 2: m2= 3.29 kg, v2(t) = (8.05 m/s)+(0.441m/s^2)*t Particle 3: m3= 4.61 kg, v3(t) = (6.33 m/s)+(0.321 m/s^2)*t                    Particle 4: m4= 5.39 kg, v4= m/s

Explanation / Answer

The net momentum of the system will be zero because the center of the mass is at rest and there is no external force.

m1v1+m2v2+m3v3+m4v4 = 0.

V4 = - (1.93*(6.97+0.211*2.59) + 3.29*(8.05+0.441*2.59) + 4.61*(6.33+0.321*2.59) )/ 5.39 = -14.42m/s

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