Question #1182

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.

Expert's answer

Let's write the Momentum conservation law for the horizontal and vertical projections:

V: 3MV = MV_{1} + MV_{3v}

H: 0 = MV_{2} + MV_{3h}

where V = 277 m/s, V_{2} = 475 m/s

Thus |V_{2}| = |V_{3h}| and |V_{3v}| = |3V-V_{1}|

V_{3} = √ (V_{3h}^{2} + V_{3v}^{2}) = √ ( 475^{2} + (3*277 - 212)^{2}) = 780.25

V: 3MV = MV

H: 0 = MV

where V = 277 m/s, V

Thus |V

V

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