By accident, a large plate is dropped and breaks into three pieces. The pieces f

Question

By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor, with v1 = 2.90 m/s and v2 = 1.60 m/s. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing, find the masses of pieces 1 and 2. m1 = kg m2 = kg

Explanation / Answer

You might as well start by decomposing the velocities into x and y-components:

vx(piece 1) = v1,x = -v1*sin(25) (its moving in the -x direction)
vy(piece 1) = v1,y = v1*cos(25)

vx(piece 2) = v2,x = v2*cos(45)
vy(piece 2) = v2,y = v2*sin(45)

vx(piece 3) = v3,x = 0
vy(piece 3) = v3,y = -v3 (its moving in the -y direction)

Now find the net momentum in the x-direction:
px,net = px1 + px2 + px3 = m1*v1,x + m2*v2,x + m3*v3,x
= -m1*v1*sin(25) + m2*v2*cos(45) + m3*0
= -m1*v1*sin(25) + m2*v2*cos(45)
You know that this is equal to zero (stated in the problem), so set it equal to zero:
m2*v2*cos(45) - m1*v1*sin(25) = 0
??? m1*v1*sin(25) = m2*v2*cos(45)
This equation has 2 unknowns (m1 & m2), so get one in terms of the other:
???m1 = (v2/v1)*[cos(45)/sin(25)]*m2
m1 = (1.70/2.85)*[cos(45)/sin(25)]*m2 = 0.998*m2
(evidently, m1 and m2 are practically equal).

Now find the net momentum in the y-direction:
py,net = py1 + py2 + py3 = m1*v1,y + m2*v2,y + m3*v3,y
= m1*v1*cos(45) + m2*v2*sin(45) - (1.30 kg)*(3.07 m/s)
= m1*v1*cos(45) + m2*v2*sin(45) - 3.991 kg*m/s
Set this equal to 0, also:
m1*v1*cos(25) + m2*v2*sin(45) - 3.991 = 0
Again, group your unknowns:
m1*(2.85)*cos(25) + m2*(1.7)*sin(45) = 3.991
(2.58)*m1 + (1.2)*m2 = 3.991
Use the result from before to find m2:
(2.58)*(.998)*m2 + (1.2)*m2 = 3.991
m2*(2.57 + 1.2) = 3.991

m2 = 1.057 kg
and therefore:
m1 = (0.998)*(1.057) = 1.055 kg

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