Question

A 200 kg rocket in deep space moves with a velocity of (121 m/s)i + (38.0 m/s)ĵ. Suddenly, it explodes into three pieces, with

the first (82 kg) moving at -(214 m/s)i + (304 m/s)ĵ and the second (54 kg) moving at (20.0 m/s)î - (72.0 m/s)ĵ. Find the
velocity (in m/s) of the third piece. (Express your answer in vector form.)

Answer 1

We can use the conservation of momentum to solve this problem. The total momentum of the rocket before the explosion is equal to the total momentum of the three pieces after the explosion.

Let's start by finding the momentum of the rocket before the explosion:

p1 = m1 * v1 = 200 kg * (121 m/s)i + (38.0 m/s)ĵ

p1 = (24200 kg m/s)i + (7600 kg m/s)ĵ

Now let's find the momentum of the first piece after the explosion:

p2 = m2 * v2 = 82 kg * (-(214 m/s)i + (304 m/s)ĵ)

p2 = (-17548 kg m/s)i + (24848 kg m/s)ĵ

And the momentum of the second piece:

p3 = m3 * v3 = 54 kg * ((20.0 m/s)î - (72.0 m/s)ĵ)

p3 = (1080 kg m/s)î - (3888 kg m/s)ĵ

The momentum of the third piece is the remaining momentum:

p4 = p1 - p2 - p3

p4 = (30868 kg m/s)i - (21696 kg m/s)ĵ

Finally, we can find the velocity of the third piece by dividing its momentum by its mass:

v4 = p4 / m4 = p4 / (m1 - m2 - m3)

v4 = (30868 kg m/s)i - (21696 kg m/s)ĵ / (200 kg - 82 kg - 54 kg)

v4 = (30868 kg m/s)i - (21696 kg m/s)ĵ / 64 kg

Simplifying:

v4 = (483.6 m/s)i - (338.5 m/s)ĵ

So the velocity of the third piece is (483.6 m/s)i - (338.5 m/s)ĵ.