firework physics question?

Question

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?

Answer 1

Use the conservation of momentum.

Just before the firecracker explodes, it is it at its highest point, which means it is momentarily not moving, which means its momentum is zero.

The firecracker breaks apart because the 3 fragments are pushing against each other. This is a case of "internal forces," and when all the forces are internal, the momentum of the system is conserved.

That means the total momentum after the explosion must be the same as the total momentum just before the explosion -- namely, zero.

So, this means if you add up the three momentum vectors of the three fragments, their vector sum must be zero. (If you use the head-to-tail method of adding vectors, that means the head of the 3rd vector will reach back to touch the tail of the first vector).

So, draw a diagram. You should end up with a triangle consisting of three vectors. If you note what the problem says about the first 2 vectors being at right angles, it should be obvious how to figure out the length of the third vector.

(Of course, those are momentum vectors, not velocity vectors: but just divide them all by "m" (the mass of each of the three pieces), and you've then got velocity vectors. They're all to the same scale.)