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A boxcar traveling at 8 m/s approaches a string of 4 identical boxcars sitting stationary on the track. The moving boxcar collides and links with the stationary cars and they all move off together along the track. What is the final speed of the cars immediately after the colision? (You may take the mass of each boxcar to be 18,537 kg.)

Respuesta :

Answer:

[tex]v_f = 1.6 m/s[/tex]

Explanation:

As per momentum conservation we know that initial momentum of one box car must be equal to the momentum of all box cars together

here we know that there is no external force on the system of all box cars so there momentum conservation is applicable

[tex]m_1v_{i} = (m_1 + m_2 + m_3 + m_4 + m_5)v_{f}[/tex]

so above equation is momentum of one box car is equal to the momentum of all box cars together

So here we will have

[tex]v_f = \frac{m_1 v_i}{m_1 + m_2 + m_3 + m_4 + m_5}[/tex]

[tex]v_f = \frac{m(8)}{5m}[/tex]

[tex]v_f = \frac{8}{5} = 1.6 m/s[/tex]