As isolated body of mass M=60kg and velocity V=<40,0> meters per second explodes into two pieces one with mass m1=20kg going 30 degrees above the horizontal with velocity v1=? and the other with mass m2=40kg going 60 degrees below the horizontal with velocity v2=?...using the law of conservation of momentum, find the velocities of the two pieces.
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Initial, Vx = 40, Vy = 0
momentum Px = 40•60, Py = 0
final, first piece
V1x = V1cos30 = (0.866)V1
P1x = 17.32 V1
V1y = V1sin30 = (0.5)V1
P1y = 10V1
second piece
V2x = V2cos60 = 0.5V2
P2x = 20V2
V2y = V2sin60 = –(0.866)V2
P2y = –34.64V2
Set the momentum in each direction equal
x:
2400 = 17.32 V1 + 0.5V2
y:
0 = (0.5)V1 – 34.64V2
two equations in two unknowns, you can do the rest
momentum Px = 40•60, Py = 0
final, first piece
V1x = V1cos30 = (0.866)V1
P1x = 17.32 V1
V1y = V1sin30 = (0.5)V1
P1y = 10V1
second piece
V2x = V2cos60 = 0.5V2
P2x = 20V2
V2y = V2sin60 = –(0.866)V2
P2y = –34.64V2
Set the momentum in each direction equal
x:
2400 = 17.32 V1 + 0.5V2
y:
0 = (0.5)V1 – 34.64V2
two equations in two unknowns, you can do the rest