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 42 m/s in the negative x-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 9 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 counter-clockwise from the positive x-axis.)

Explanation / Answer

a) Since the bomb is at rest initially,the momentum is zero

b)Considering the momentum conservation in x direction,we get

9*vx - (4*42)=0

vx = 18.67m/s

c)Considering the momentum conservation in y direction

9*vy-(4*40)=0

vy = 17.78 m/s

d)manitude of velocity = sqrt(18.672+17.782)=25.78 m/s

e)The angle the piece makes with x axis counterclockwise = tan-1(17.78/18.67)=43.6o

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