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 3 kg, moves with a velocity of 45 m/s in the negative x-direction, and a second piece, also of mass 3 kg, moves with a velocity of 38 m/s in the negative y-direction. The third piece has a mass of 8 kg.

The question is broken down into 4 parts, please answer each part.

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?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 counter-clockwise from the positive x-axis.)

Explanation / Answer

let i and j are unit vectors along +ve x axis and +ve y axis.

let x component of velocity of third piece be Vx along +ve x axis.

let y component of velocity of third piece be Vy along +ve y axis.

as there is no external force,total momentum is conserved.

as the bomb was at rest in the beginning, initial momentum was 0.

so sum total of momentum of all the parts of the bomb will be 0.

equating x component of total momentum to 0:

3*(-45)+3*0+8*Vx=0

==>Vx=3*45/8=16.875 m/s

part b:

equating y component of total momentum to 0:

3*(-38)+8*Vy=0

==>Vy=3*38/8=14.25 m/s

part c:

magnitude of velocity =sqrt(Vx^2+Vy^2)

=sqrt(16.875^2+14.25^2)

=22.087 m/s

part d:

angle with +ve x axis=arctan(Vy/Vx)

=arctan(14.25/16.875)

=40.179 degrees

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