A bomb at rest at the origin of an xy-coordinate system explodes into three piec

Question

A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass 4 kg, moves with a velocity of 48 m/s in the negative direction, and a second piece, also of mass 4 kg, moves with a velocity of 40 m/s in the negative y-direction. The third piece has a mass of 10 kg. What is the initial momentum of the bomb before the explosion? What is the x-component of the velocity of the third piece just after the explosion? What is the y-component of the velocity of the third piece just after the explosion? What is the magnitude of the velocity of the third piece? Just after the explosion, in what direction is the third piece moving? (Give your answer as an angle measured clockwise from the positive x-axis.)

Explanation / Answer

here ,

initial momentum of system = m * 0

initial momentum of system = 0 kg.m/s

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as there is no external force on the bomb

using conservation of momentum in x - direction

4 * ( -48 ) + 10 * Vx = 0

Vx = 19.2 m/s

the x-component of the velocity of the third piece just after the explosion is 19.2 m/s

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for the conservation of momentum in y - direction

4 * (-40) + 10 * Vy = 0

Vy = 16.0 m/s

the y-component of the velocity of the third piece just after the explosion is 16.0 m/s

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magnitude of velocity = sqrt(vx^2 + vy^2)

magnitude of velocity = sqrt(19.2^2 + 16.0^2)

magnitude of velocity = 24.99 m/s

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theta = arctan(vy/vx)

theta = arctan(16/19.2)

theta = 39.8 degree

the angle of velocity of third piece is 39.8 degree

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