Question

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?

Answer 1

Approximately
`3.1\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

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

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

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

Momentum of the two rail car before release was
`0` since neither car was moving and velocity of both car was
`0`.

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

• 
  `20\, v` for the
  `20\; {\rm t}` rail car, and
• 
  `80\, (v - 3.9)` for the
  `80\; {\rm t}` rail car.

Under the assumption:

`(\text{total final momentum}) = (\text{total initial momentum})`.

`20\, v + 80\, (v - 3.9) = 0`.

`\begin{aligned}v = ((80)\, (3.9))/((100)) \approx 3.1\end{aligned}`.

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