A stationary bomb explodes into three pieces. Two pieces of masses 2 kg and 1 kg fly off at right angles to each other with velocities of 15 m s ⁻¹ and 40 m s ⁻¹ respectively. If the third piece flies off with a velocity of 10 m s ⁻¹ , the mass of the third piece is

Step-by-step solution

Initial momentum of the bomb is zero. Momentum of the first piece, p₁ = m₁ v₁ = 2 15 = 30 kg m s ⁻¹ . Momentum of the second piece, p₂ = m₂ v₂ = 1 40 = 40 kg m s ⁻¹ . Since the two pieces fly off at right angles, the magnitude of their resultant momentum is p₁₂ = p₁^2 + p₂^2 = 30^2 + 40^2 = 50 kg m s ⁻¹ . To conserve momentum, the third piece must have an equal and opposite momentum. Thus, p₃ = 50 kg m s ⁻¹ . Given the velocity of the third piece v₃ = 10 m s ⁻¹ , its mass is m₃ = p₃ v₃ = 50 10 = 5 kg . Answer: 5 kg