Answer:
a. 228° counterclockwise
b. 1.8 kg
Explanation:
a. From the law of conservation of momentum,
momentum before explosion = momentum after explosion.
We resolve the momenta into its components.
Let m₁, m₂ and m₃ be the masses of the fragments after explosion and v₁, v₂ and v₃ be their velocities. Let be the angle between the third fragment and the horizontal direction.
So, m₁v₁ - m₃v₃sinθ = 0 (1) horizontal component of momentum. Also
m₂v₂ - m₃v₃cosθ = 0 (2) Vertical component of momentum
From (1) m₁v₁ = m₃v₃sinθ (3)
From (2) m₂v₂ = m₃v₃cosθ (4)
Dividing (3) by (4)
m₃v₃sinθ/m₃v₃cosθ = m₁v₁/m₂v₂
sinθ/cosθ = m₁v₁/m₂v₂
tanθ = m₁v₁/m₂v₂
θ = tan⁻¹(m₁v₁/m₂v₂).
Let m₁ = 2.0 kg and v₁ = 20 m/s and m₂ = 3.0 kg and v₂ = 12 m/s
θ = tan⁻¹(2 kg × 20 m/s/3 kg × 12 m/s) = tan⁻¹(40/36) = 48.01° ≅ 48°
Since the third piece flies off in the third quadrant, its direction is 180 + 48 = 228° counterclockwise
b. From (3), m₃ = m₁v₁/v₃sinθ v₃ = 30 m/s. Substituting the other values, we have
m₃ = 2 kg × 20 m/s ÷ (30 m/s ×sin48°) = 1.79 kg ≅ 1.8 kg for the mass of the third piece
