# Linear Momentum: Rocket Explosion

## Homework Statement

A 975-kg two-stage rocket is traveling at a speed of 5.80 x 103 m/s with respect to Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a relative speed (relative to each other) of 2.20 x 103 m/s along the original line of motion. What is the speed and direction of each section (relative to earth) after the explosion?

## Homework Equations

$$m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b$$
where m = 975 kg, v = 5.80 x 103 m/s relative to earth, and ma = mb = 487.5 kg

## The Attempt at a Solution

At first I thought that special relativity and the addition of velocities could be used to find the speed of the rockets relative to earth. Then I realized that linear momentum is in chapter 7, while relativity was in chapter 27. Therefore, that idea didn't make sense. One student suggested the following:

$$m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b$$
(975)(5.80 x 103) = (487.5)(2.20 x 103)(va) + (487.5)(vb)

His logic was that one could then set va and vb equal to each other, since you had accounted for the difference in velocities relative to each other. I didn't think this made sense either, but I'm at a loss for any other way to approach the problem. Any and all help is welcomed and appreciated.