Calculating Initial Speed in an Inelastic Collision with Friction

[Oliviam12]

Homework Statement

A drunk driver strikes a parked car. During the collision the cars become entangled and slide to a stop together. The drunk driver's car has a total mass of 742 kg, and the parked car has a total mass of 776 kg. If the cars slide 18 m before coming to rest, how fast was the drunk driver going? The coefficient of sliding friction between the tires and the road is 0.59.

Homework Equations

Err?

The Attempt at a Solution

Not any idea really? Just that I think its inelastic... and that I can't use the equation v1`=m1-m2/m1+m2 V1 to get the answer... How do I incorporate the friction?

 

[Doc Al]

Yes, it's a completely inelastic collision. (What's conserved during the collision?) Treat the problem in two parts: (1) the collision (What's the speed of the two cars immediately after the collision?) (2) the slowing down due to friction.

Work backwards from the given information.

 

[Oliviam12]

Energy is conserved, but I don't know how to find the speed of the two cars after the collision :/

 

[Doc Al]

Energy is not conserved (that's what inelastic means). Work backwards. Hint: What's the force of friction that slows the cars? The acceleration? The speed just after the collision?

 

[Oliviam12]

I am sorry but, I don't understand this problem at all... Can you redirect me to an example problem or a tutorial?

 

[saket]

Go through your text. You should find there that during inelastic collision, only thing that remains conserved is linear momentum. (Here, we are not concerned with rotaion and all.)
So assume initial speed to be 'v' and conserve momentum, to get speed of the two cars, just after the collision. (Note, after collision the two cars stick -- entangle -- to each other.) After obtaining this speed, you can apply work-energy theorem to know the work done by the friction. And, assuming uniform friction, you can get the distance required to stop the cars. Equate it with the given data (18m) to get initial speed 'v'.