A 1500-kg car traveling east at 50 m/s collides with a 2200-kg truck traveling north at...

Question

Question

Answer 1

Answer #1

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

Answer 2

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