# Inelastic collision

1. Oct 23, 2005

### 7tongc5

A 1000kg toyota collides into a 2200kg Cadillac at rest. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8m before stopping. The coefficient of friction btwn tires and road = .4 ... what was the speed of the toyota at impact?
I dont know how to start it off (i dont know how i should set up equations while taking friction into account).
Any hints would be helpful. Thanks. :)

2. Oct 24, 2005

### ehild

You should consider the problem as two separate process. The first is collision. It happens in such a very short time that both cars stay at the same place where they were just before the collision process. After the collision went on, both cars move with the same velocity. This velocity is determined by the law of conservation of linear momentum.
The second process is motion under the effect of friction. The wreck of the cars moves as a single body of mass 2200 kg + 1000 kg. You can calculate the force of friction, the deceleration due to friction, the displacement, the final velocity whis is zero, so you can figure out the initial velocity of the wreck. From that you get the initial momentum of the wreck which is equal to the initial momentum of the toyota as the cadillac had been in rest. From this momentum, dividing it by the mass of the toyota you get the velocity of the toyota before collision.

ehild

3. Oct 24, 2005

### 7tongc5

it's totally awesome that you laid out the steps, but i still managed to screw it up somehow: the answer's supposed to be 20 m/s

1) force of friction = .4(3200)(9.8) = 12,544
2) F = ma
12544 = 3200a --> a = -3.92 m/s2
3) v = v<initial>^2 + 2(-3.92)(2.8)
21.952 = v<initial> ^2
4.69 m/s = v<initial>
4) total momentum of wreck = 4.69 (3200) = 14992.948
4) 14992.948 = m<toyota> x v
v = 14.99 m/s

really appreciate your help, 7tongc5

Last edited: Oct 24, 2005
4. Oct 24, 2005

### ehild

I got the same result as you.

ehild