Your automobile of mass m1=900 kg collides at a traffic circle with another auto

Question

Your automobile of mass m1=900 kg collides at a traffic circle with another automobile of mass m2=1200, just before the collision, your auto was moving due east and the other auto was moving 40 deg south of east. After collision the autos remain entangled (im assuming inelastic collision here) while they skid with locked wheels, until coming to rest. Your speed before collision was 14 m/s. THe length of the skid marks is 18.2 m, and the coefficient of kinetic friction between tire and pavement is 0.85.

What is speed of the other auto before the collision?

Explanation / Answer

You can calculate the speed after the collision using work - energy

?*(M + m)*g*d = 1/2*(M + m)*v^2

so v after = sqrt(2*?*g*d) = sqrt(2*0.85*9.8*18.2) = 17.41m/s

Now use conservation of momentum in both east and south directions

East ---900*14 + 1200*v*cos(40) = (900+1200)*vx

or vx = 0.4377*v + 6

South 1200*v*sin(40) = 2100*vy or vy = 0.3673*v

Now square both eqn and add

vx^2 = 0.4377^2*v^2 + 5.2524*v + 36 = 0.1916v^2 + 5.2524V + 36

vy^2 = .3673^2*v^2 = 0.1349*v^2

Now adding we get vx^2 + vy^2 = vafter^2 =17.4^2 = 0.1916*v^2 + 0.1349*v^2 + 5.2524*v + 36

so 0.3265*v^2 + 5.2524*v - 267 = 0

Solving we get v =( -5.2524 + sqrt(5.2524^2 - 4*0.3265*(-267)))/(2*0.3265) = 21.6m/s

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