Answer:
Magnitude of the third piece of the vessel after collision is (v√2) m/s
Explanation:
According To Newton's 2nd law, in such a system, momentum is conserved
Momentum before explosion = Momentum after explosion
Total mass of the vessel before explosion = m + m + 3m = 5m
Velocity = (0î + 0j) m/s since it was at rest.
Momentum = 5m (0î + 0j)
Momentum of the first piece after the explosion = m(-vî)
Momentum of the 2nd piece after explosion = m(-vj)
For the momentum of the 3rd piece,
Mass = 3m, let the velocity be (Vₓî + Vᵧj) m/s
Momentum of 3rd piece after collision = 3m (Vₓî + Vᵧj)
Momentum before explosion = Momentum after explosion
5m (0î + 0j) = m(-vî) + m(-vj) + 3m (Vₓî + Vᵧj)
The term m can be factories out,
Grouping the x and y components together,
0î + 0j = (-vî + Vₓî) + (-vj + Vᵧj)
-v + Vₓ = 0, Vₓ = v
-v + Vᵧ = 0, Vᵧ = v
So, the velocity of the 3rd fragment,
(Vₓî + Vᵧj) = (vî + vj) m/s
Magnitude = √(v² + v²) = √(2v²) = v√2 m/s
