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

Answer 2

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

The given parameters;

$m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg$

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

$M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}$

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

$v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s$

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

$v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s$

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

$v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s$

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

$v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s$

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

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