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4 December, 07:02

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

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  1. 4 December, 07:27
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    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
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