Question

A bomb explodes into three identical pieces. One flies in the + y-direction at 14.4m/s, a second in the - x-direction at 17.6m/s. What's the magnitude of the velocity of the third piece? And What's the direction of the velocity of the third piece? Count counterclockwise from the + x axis?

Answer 1

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Answer 2

By conservation of momentum, the momentum before the explosion = momentum after the explosion.

Now momentum before the explosion = 0, because the bomb had no velocity.

Hence the momentum after the explosion is also in total zero.

Answer 3

Now we are told that each piece is identical, so let us write that each piece has mass m. The momentum of the first piece is
p1 = m.v1
where v1 is its velocity:
p1 = m <0,14,4> = <0,14.4m>
because we're told it moves in the +y direction

The momentum of the second piece is
p2 = m.v2 = m <-17.6, 0> = <-17.6m, 0>
because we're told it moves in the negative x direction.

So far so good?

Answer 4

The last step is we know the net momentum after the explosion is zero. That is

p1 + p2 + p3 = 0.

We now know p1 and p2 so we can solve for p3.

Make sense?

Answer 5

I got it, Thanks!!