# A Firework Exploding

1. Oct 24, 2008

### JJones_86

1. The problem statement, all variables and given/known data
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 130 m/s at right angles to each other. How fast is the third piece moving?

2. Relevant equations
P = m*v
Pi = Pf

3. The attempt at a solution

So the initial momentum is 0 because it is not moving when it explodes, so doesn't that mean that final momentum is 0 because of the conservation of momentum? I tried this, but I'm not getting the right answerl...

2. Oct 24, 2008

### LowlyPion

What you have are two vectors at right angles. You are right about momentum being conserved. Hence you have the third momentum that must balance the first 2.

3. Oct 24, 2008

### JJones_86

Ok here is my attempt at a solution

Particle 1
P = m * v
P = m * 130 m/s

Particle 2
P = m * v
P = m * 130 m/s

Particle 3(Unknown)
P = m * v

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So in order for it to equal 0 I used this equation...

P1 + P2 + P3 = 0
m(130 m/s) + m(130 m/s) + m*v = 0

So I'm obviously using the wrong equation because of the two unknowns..

4. Oct 24, 2008

### LowlyPion

What you failed to do is treat the vectors by their x,y components.

Your equation is not a scalar equation, it is a vector equation.

$$\vec P_1 + \vec P_2 + \vec P_3 = 0$$

5. Oct 24, 2008

### JJones_86

Ok, I got it now, Thank you very much!