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 2 kg, moves with a velocity of 45 m/s in the negative x-direction, and a second piece, also of mass 2 kg, moves with a velocity of 32 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 clockwise from the positive x-axis.)

Explanation / Answer

M = m1+m2+m3, m1 = 2kg , m2 = 2 kg , m3 =9 kg

u =0 , v1 = -45i, v2 = -32j

p =mv

Initila momentum of bomb before explosion is zero.

From conservation of momentum along x direction

Mux = m1v1x+m2v2x+m3v3x

0 = -(2*45)+0 +(9*v3x)

v3x = 10 m/s

From conservation of momentum along y direction

Muy = m1v1y+m2v2y+m3v3y

0 = 0-(2*32)+0 +(9*v3y)

v3y = 7.11 m/s

v3 = (10^2+7.11^2)^0.5

magnitude of velocity of third particle v3 = 12.27 m/s

tan(theta) = 7.11/10

theta =35.4 degrees

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