What is the speed?

Question

A 1.1x10^3 kg Toyota collides into the rear end of a 3.3x10^3 kg Cadillac stopped at a red light. the bumpers lock, the brakes are locked, and the 2 cars skid forward 4.6 m before stopping. The coefficient of friction between the tires and the road is .50 What was the speed of the Toyota at impact?

Answer 1

oracle thinks it's necessary to assume an elastic collision. It'll really be inelastic. Let's see if I can do it for a completely inelastic collision.

But first, we need to find the velocity, v, of the pair immediately after the collision. The work done by the friction for 4.6 m equals the kinetic energy immediately after the collision. So
Work = KE
mu*(mt+mc)*g*d= (1/2)*(mt+mc)*v^2
where mu = coeffidient of friction, mt=mass toyota, mc=mass cadillac, and d = distance to stop
Cancell the masses
0.5*9.8 m/s^2*4.6 m = (1/2)*v^2
v^2 = 45 m^2/s^2
v = 6.7 m/s (This is probably what you got doing what oracle said to do.) It's not the toyota's original speed.

Momentum is conserved, so
mt*ut + mc*uc = (mt+mc)*v
where mt=mass toyota, mc=mass cadillac, ut and uc are initial velocity of toyota and cadillac respectively (uc = 0), v is the final velocity of the pair stuck together.
1.1*10^3 kg*ut + 3.3*10^3*0 = 4.4*10^3 kg*6.7 m/s
ut = 4.4*10^3 kg*6.7 m/s / 1.1*10^3 kg
ut = 26.9 m/s