Answer to Question #1182 in Mechanics | Relativity for Jon Allen

Question

A rocket is fired vertically upward.&
At theinstant it reaches an altitude of 1630 m and a
speed of 277 m/s, it explodes into three equal
fragments.&
One fragment continues to moveupward with a speed of 212 m/s following the
explosion.&
The second fragment has a speedof 475 m/s and is moving east right after the
explosion.&
What is the magnitude of the velocity ofthe third fragment?
Answer in units of m/s.

Answer 1

Let's write the Momentum conservation law for the horizontal and vertical projections:
V: 3MV = MV₁ + MV_3v
H: 0 = MV₂ + MV_3h
where V = 277 m/s, V₂ = 475 m/s
Thus |V₂| = |V_3h| and |V_3v| = |3V-V₁|
V₃ = √ (V_3h² + V_3v²) = √ ( 475² + (3*277 - 212)²) = 780.25