A 20 ton rail car and a 80 ton rail car, initially at rest, are connected together with a giant but massless compressed spring between them. When released, the 20 ton car is pushed away at a speed of 3.9 m/s relative to the 80 ton car. What is the speed of the 20 ton car relative to the ground?

When an object of mass [tex]m[/tex] travels with a velocity of [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].

Assume that momentum of the two rail cars are conserved immediately after release. Let [tex]v[/tex] denote the velocity of the [tex]20\; {\rm t}[/tex] rail car. Since the two rail cars are moving away from one another, the velocity of the [tex]80\; {\rm t}[/tex] rail car will be [tex](v - 3.9)[/tex].

(Note that the velocity of the [tex]80\; {\rm t}[/tex] rail car will be negative since that rail car is moving away from the [tex]20\; {\rm t}[/tex] rail car.)

Momentum of the two rail car before release was [tex]0[/tex] since neither car was moving and velocity of both car was [tex]0[/tex].

Momentum of the two rail cars right after release will be:

[tex]20\, v[/tex] for the [tex]20\; {\rm t}[/tex] rail car, and
[tex]80\, (v - 3.9)[/tex] for the [tex]80\; {\rm t}[/tex] rail car.

Under the assumption:

[tex](\text{total final momentum}) = (\text{total initial momentum})[/tex].

[tex]20\, v + 80\, (v - 3.9) = 0[/tex].

[tex]\begin{aligned}v = \frac{(80)\, (3.9)}{(100)} \approx 3.1\end{aligned}[/tex].

Therefore, the speed of the [tex]20\; {\rm t}[/tex] rail car would be approximately [tex]3.1\; {\rm m\cdot s^{-1}}[/tex] right before release.