A Three-Point Explosion Problem

Question

A 100 g projectile is traveling east at 20.0 m/s when it suddenly explodes into three pieces. A 33.0 g fragment is shot due north at 10 m/s while another 21.0 g fragment travels 30° South of East at 20 m/s. What is the speed and direction of the third fragment?

Answer 1

We have vector equation from law of coservation of momentum: mv = m₁v₁ + m₂v₂ + m₃v₃

Let x-axis looking to east and y-axis to north.

Progection on x-axis: mv = m₂v₂cos30° + m₃v_3x (1)

Projection on axis y: 0 = m₁v₁ - m₂v₂sin30° + m₃v_3y (2)

From (1) v_3x = (100·20 - 21·20cos30°)/46 = 35.57 m/s

From (2) v_3y = (21·20sin30° - 33·10)/46 = - 2.61 m/s

From here v₃ = (35.57² +(- 2.61)²)^1/2 = 35.67 m/sec

tanφ = v_3y/v_3x = - 2.61/35.57 = - 4.20°

Another words m₃ = 46 g piece will move south of east 4.20°

_.

Answer 2

to solve this problem you use conservation of momentum and use vector analysis to get the rhird vector