Firework Explosion: Find the Speed of 3rd Piece

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The firecracker explodes into three equal mass pieces at its highest point, where its initial velocity is zero. Two pieces move at 120 m/s at right angles, creating a momentum that must be balanced by the third piece. Since momentum is conserved, the total momentum of the system remains zero after the explosion. The momentum of the first two pieces can be calculated using vector addition, leading to the conclusion that the third piece must move in a specific direction to maintain momentum conservation. By solving for the third piece's velocity, it can be determined that it moves at a speed of 120 m/s in a direction that balances the momentum of the other two pieces.
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A firecracker is tossed straight up into the air. It explodes into three pieces of equal mass just as it reaches the highest point. Two pieces move off at 120 m/s at right angles to each other. How fast is the third piece moving?

So, we know that since the firework is at its highest point when it explodes that Vfirework = 0. Two pieces move away from each other at a right angle at 120 m/s. What does this mean for the third piece of the firework as it explodes? I'm confused as to how I'm supposed to know its direction in relation to the other two pieces.
 
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What is conserved as the three pieces separate?
 
Momentum would be conserved, the initial momentum would be equal to final momentum... I thought the third piece might be moving at the same momentum as the two which fired off at 120m/s, but this assumption is wrong.

So, momentum is conserved, but I'm still unsure of how to solve this one..
 
The total momentum is zero. What's the total momentum of the first two pieces? Use this to find the momentum of the third piece.
 
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