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

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

he made 11 calls in one day

unbalanced force

explanation: