- #1

tul725

## Homework Statement

A two-stage rocket is traveling at 4500 m/s before the stages separate. The 3000-kg first stage is pushed away from the second stage with an explosive charge, after which the first stage continues to travel in the same direction at a speed of 3000 m/s . How fast is the 1900-kg second stage traveling after separation?[/B]

Rocket's mass = 4900 kg

Velocity of rocket = 4500 m/s

m1= 3000kg

v1 = 3000m/s

m2=1900kg

v2 = ?

## Homework Equations

Conservation of Momentum

## The Attempt at a Solution

The way I tried imagining it was by changing the inertial reference frame to the zero momentum frame, the frame would be going at 4500 m/s. In this situation m1v1 + m2v2 = 0. I then plugged in numbers to get v2 = -4736.84 m/s. Then to get the velocity relative to the Earth I added the velocity of the frame relative to the earth, which was 4500m/s. I then get 236.84 m/s. Which is incorrect.

After researching the answer, I came across a similar problem where the conservation of momentum equation was set up:

(MassofRocket)(VelocityofRocket) = m1v1 + m2v2

The answer I got for this, v2 = 6868 m/s was correct.

What I am struggling to understand is why is momentum conserved even if there is the explosive part of the collision. Wouldn't that cause an impulse to the situation and momentum would be changed? And why is my original method not proper?