Question

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity v in the negative x direction, and a second piece, also of mass m, moves with the same velocity in the negative y direction. The third piece has mass 3m. Just after the explosion, (a) what is the magnitude of the velocity of the third piece in terms of the given variables

Answer 1

Magnitude of the third piece of the vessel after collision is (v√2) m/s

Step-by-step 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