Inelastic Collision of drunken driver

[rose3]

Homework Statement

A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision), with the locked wheels of the parked car leaving skid marks 8.0 m in length.
If the mass of the moving car is 2290kg and the mass of the parked car is 1150 kg how fast was the first car traveling when it hit the parked car?
Assume that the driver of the moving car made no attempt to brake and that the coefficient of friction between the tires and the road is 0.28.

Homework Equations

F=μN=μmg
ma=μmg
vf^2=vi^2+2ad
m1v1=m2v2

The Attempt at a Solution

I found the speed of the two cars after the collision.
a=μg= (0.28)(-9.8)=-2.744 m/s^2
vi^2=vf^2-2ad = 0-2(-2.744)(8.0)
vi=6.626m/s
[/B]
I found the speed of the moving car using the momentum equation.
v1=(m2v2)/m1=((3440kg)(6.626))/2290kg=9.95m/s

Apparently this answer is wrong.

 

[ehild]

Homework Statement

A drunken driver crashes his car into a parked car that has its brakes set. The two cars move off together (perfectly inelastic collision), with the locked wheels of the parked car leaving skid marks 8.0 m in length.
If the mass of the moving car is 2290kg and the mass of the parked car is 1150 kg how fast was the first car traveling when it hit the parked car?
Assume that the driver of the moving car made no attempt to brake and that the coefficient of friction between the tires and the road is 0.28.

Homework Equations

F=μN=μmg
ma=μmg
vf^2=vi^2+2ad
m1v1=m2v2

The Attempt at a Solution

I found the speed of the two cars after the collision.
a=μg= (0.28)(-9.8)=-2.744 m/s^2
vi^2=vf^2-2ad = 0-2(-2.744)(8.0)
vi=6.626m/s
[/B]
I found the speed of the moving car using the momentum equation.
v1=(m2v2)/m1=((3440kg)(6.626))/2290kg=9.95m/s

Apparently this answer is wrong.

Hi Rose, welcome to PF!
What does the sentence mean "Assume that the driver of the moving car made no attempt to brake" ?

 

[rose3]

Hi Rose, welcome to PF!
What does the sentence mean "Assume that the driver of the moving car made no attempt to brake" ?

I believe it just means that the speed of the moving car was constant before the collision, he didn't make an attempt to slow down. I'm not sure how this is supposed to affect the solution.

 

[ehild]

The coefficient of kinetic friction refers to sliding tyres. If the wheels of the car roll, there can be rolling resistance which is much smaller than the sliding friction, it can be ignored. The tyres slide if the brake is applied, but it happened to the parked car only.

 

[rose3]

The coefficient of kinetic friction refers to sliding tyres. If the wheels of the car roll, there can be rolling resistance which is much smaller than the sliding friction, it can be ignored. The tyres slide if the brake is applied, but it happened to the parked car only.

Ok so does this mean that friction works only on the parked object after the collision? How would this affect the final answer and how would I incorporate that into my calculations?

 

[ehild]

Ok so does this mean that friction works only on the parked object after the collision? How would this affect the final answer and how would I incorporate that into my calculations?

Yes, the friction works only on the parked car. How much is that force of friction?
As the cars are in contact after the collision, that force acts on both of them, and slows them down.

 

[rose3]

Ok so I calculated the normal force to be -3155.6 N and then divided this number by 3440 kg(the combined mass) to find the acceleration which was -0.9173 m/s^2. I then used this value to find the speed of the cars after the collision (which was 3.83 m/s). I used this value in the momentum equation to find the initial speed of the car (5.75 m/s). Apparently this is also incorrect. What am I doing wrong? I only have one more chance to get the right answer.

 

[ehild]

Your answer is correct, but they may want the speed in km/h . In real life, m/s is not the usual unit for the speed of cars.