Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car, [tex]v_{a}[/tex] = 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
[tex]mv_{a} + 2mv_{b} = (m + 2m)v'[/tex] (1)
where,
[tex]v_{a}[/tex] = velocity of the first car
[tex]v_{b}[/tex] = velocity of the 2 coupled cars after collision
Now, from eqn (1)
[tex]v' = \frac{v_{a} + 2v_{b}}{3}[/tex]
[tex]v' = \frac{3.00 + 2\times 1.20}}{3}[/tex]
v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s
