A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any objects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

The velocity of the center mass of the particles is calculated as;

The velocity of particle 1 at time, t = 2.83 s;

The velocity of particle 2 at time, t = 2.83 s;

The velocity of particle 3 at time, t = 2.83 s;

The velocity of the particle 4 at time, t = 2.83 s;

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

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v = - 14.08 m / s

Explanation:

The definition of center of mass is

        = 1 /M  ∑sun

where M is the total mass of the system and and are the position and mass of each component.

The velocity of the center of mass can be found by deriving this expression with respect to time

          = 1 / M ∑ m_{i} vi

let's find the total mass

          M = m₁ + m₂ + m₃ + m₄

          M = 1.45 + 2.81 +3.89 + 5.03

          m = 13.18 kg

let us substitute in the velocity of the center of mass = 0

          0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)

          v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄

let's substitute the given values

v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03

They ask us for the calculations for a time t = 2.83 s

          v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03

          v = - 14.08 m / s

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