(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 975-kg two-stage rocket is traveling at a speed of 5.80 x 10^{3}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 10^{3}m/s along the original line of motion. What is the speed and direction of each section (relative to earth) after the explosion?

2. Relevant equations

[tex]m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b[/tex]

where m = 975 kg, v = 5.80 x 10^{3}m/s relative to earth, and m_{a}= m_{b}= 487.5 kg

3. 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:

[tex]m\vec{v} = m_a\vec{v}_a + m_b\vec{v}_b[/tex]

(975)(5.80 x 10^{3}) = (487.5)(2.20 x 10^{3})(v_{a}) + (487.5)(v_{b})

His logic was that one could then set v_{a}and v_{b}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.

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# Homework Help: Linear Momentum: Rocket Explosion

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