In: Physics

# A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just...

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 45 m/s in the negative x-direction, and a second piece, also of mass 4 kg, moves with a velocity of 36 m/s in the negative y-direction. The third piece has a mass of 7 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.)

## Solutions

##### Expert Solution

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 * ( -45 ) + 7 * Vx = 0

Vx = 25.7 m/s

the x-component of the velocity of the third piece just after the explosion is 25.7 m/s

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for the conservation of momentum in y - direction

4 * (-36) + 7 * Vy = 0

Vy = 20.6 m/s

the y-component of the velocity of the third piece just after the explosion is 20.6 m/s

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

magnitude of velocity = sqrt(20.6^2 + 25.7^2)

magnitude of velocity = 32.94 m/s

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theta = arctan(vy/vx)

theta = arctan(20.6/25.7)

theta = 38.7 degree

the angle of velocity of third piece is 38.7 degree

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