Question

A firework has been launched into the air and explodes into 3 pieces. Magically, the pieces explode in a two-dimensional plane parallel to the surface of the earth. A 75 gram piece flies off straight south at 23 m/s, a 25 gram piece flies off straight west at 13 m/s. If we knew the whole firework had a mass of 150 grams, what then is the speed and direction of the last piece?

Answer 1

Mass of the three masses are -

m1 = 75 g

m2 = 25 g

So, m3 = 150 - (m1 + m2) = 150 - (75 + 25) = 50 g

Now write the velocities of the masses in unit vector notations -

v1 = 23(-j) m/s

v2 = 13(-i) m/s

v3 = ?

Initial momentum, p1 = 0

Final momentum, p2 = m1v1 + m2v2 + m3v3 = 75*23(-j) + 25*13(-i) + 50*v3

Apply conservation of momentum -

p1 = p2

=> 0 = 75*23(-j) + 25*13(-i) + 50*v3

=> 50*v3 = 325i + 1725j

=> v3 = 6.5i + 34.5j

So, magnitude of the velocity, |v3| = sqrt[6.5^2 + 34.5^2] = 35.1 m/s

And its direction, = tan^-1(34.5 / 6.5) = 79.3^o from the positive direction of x- axis means west of east.