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 5 kg, moves with a velocity of 43 m/s in the negative x-direction, and a second piece, also of mass 5 kg, moves with a velocity of 38 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.)

Explanation / Answer

here,

m1 = 5 kg

v1 = - 43 i m/s

m2 = 5 kg

v2 = 38 j m/s

m3 = 7 kg

(a)

as the bomb is initially at rest,

the initial momentum of the bomb is zero

(b)

let the x component of the velocity of the third particle be v3

using conervation of momentum

initial momentum = final momentum

0 = m1*v1 + m2*v2 + m3*v3

- 5*43 + 5*0 + 7 * v3x = 0

v3x = 30.71 i m/s

(c)

using conservation of momentum along y axis

0 = 0 = m1*v1y + m2*v2y + m3*v3y

7 * v3y = 5 * 38

v3y = 27.14 m/s

(d)

magnitude of the velocity , v3 = sqrt(v3x^2 + v3y^2)

v3 = sqrt( 27.14^2 + 30.71^2)

v3 = 41 m/s

the magnitude of the velocity of the third particle is 41 m/s

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