A bomb initially at rest on a smooth, horizontal surface is exploded into three pieces. Two pieces fly off at 90o to each other, a 2.0 kg piece at 20 m/s and a 3.0 kg piece at 12 m/s. The third piece flies off at 30 m/s.
(a) Determine the direction of motion for the third piece.
(b) What is its mass?

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

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