using law of conservation of momentum
take east-west as the X-axis and north-south as-axis
then along x-axis
m1*u1 = (m1+m2)*Vx
Vx = (m1*u1)/(m1+m2) = (1500*50)/(1500+2200)
Vx=20.3 m/sec
and along Y-axis
m2*u2 = (m1+m2)*Vy
(2200*66) =(1500+2200)*Vy
Vy = 39.3 m/sec
a) then speed immediatelyu after collision is V = sqrt(Vx^2+Vy^2) = sqrt(20.3^2+39.3^2) = 44.3 m/sec
b) direction is theta = tan^(-1)(Vy/Vx) = tan^(-1)(39.3/20.3) =
62.7 degrees above the east direction
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