Question

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

Answer 1

Answer:

v = - 14.08 m / s

Explanation:

The definition of center of mass is

        $x_{cm}$ = 1 /M  ∑sun $x_{i} m_{i}$

where M is the total mass of the system and $x_{i}$ and $m_{i}$ 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

         $v_{cm}$ = 1 / M ∑ m_{i} $v_{i}$ 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 $v_{cm}$ = 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