Question

Three identical train cars, coupled together are rolling east at 2.0 m/s. A fourth car traveling east at 4.0 m/s catches up with the three and couples to make a fourcar train. A moment later the train cars hit a fifth car that was at rest on the tracks, and it couples to make a five car train. What is the speed of the five car train?

Answer 1

Answer:

The value is  $v = 2 \ m/s$

Explanation:

From the question we are told that

   The velocity of the each of the three cars is $u_1 = u_2 = u_3 = 2 \ m/s$

    The velocity of the fourth car is  $u_4 = 4 \ m/s$

    The initial velocity of the fifth car $u_5 = 0 \ m/s$

Generally from the law of momentum conservation we have that

    $m_1 u_1 + m_2 u_2 + m_3 u_3 +m_4u_4 + m_5u_5 = [m_1 + m_2 + m_3 +m_4+ m_5]v$

Given that the cars are identical then their mass will be the same

i.e

    $m_1 =m_2 = m_3 = m_4 = m_5 = m$

=>   $[u_1 + u_2 + u_3 +u_4 + u_5]m = 5mv$

=>   $2+ 2 + 2 +4 + 0 = 5v$

= >   $v = 2 \ m/s$