# HELP Cars collision problem

1. Oct 29, 2008

### Kudo Shinichi

HELP!!!Cars collision problem

1. The problem statement, all variables and given/known data
You are a police officer analyzing the scene of a collision. On a rainy afternoon, a Porsche Carrera, of mass 1380kg, was travelling due north along Hwy38 near Verona, Ontario when a 2304kg SUV pulled out of a driveway (travelling due west) in front of the Porsche. As soon as the vehicles collided, they stuck together. Both drivers immediately applied their brakes, causing the wheel to lock. You measure the skid marks and see that the two vehicles skidded together at an angle of 40 degree west of the Porsche’s original due north direction for 11.3m. the coefficient of kinetic friction between wet concrete and rubber is 0.5.

The SUV driver claims that the Porsche was speeding. How fast was the Porsche actually travelling? Knowing that the speed limit along Hwy38 is 80km/h, should you charge the Porsche driver for speeding?

2. Relevant equations
ma=mg*sin$$\theta$$-$$\mu$$(k)*mg*cos$$\theta$$
m1v1+m2v2=m1v1'+m2v2'

3. The attempt at a solution
I found out the acceleration for both after both cars collide by using
a=(mg*sin40-mu(k)*mg*cos 40)/m
where, m=3684 g=9.8
I know that according to the Law of conservation of Momentum P=P'
Then, I found out the equation for the problem:
mass1*velocity1+mass2*velocity2=(mass1+mass2)*velocity
(mass1+mass2)*velocity because both cars stick together
Now my question is how can I find all of the velocity for the equation, since I don't have time I can't use v=v+at
Thank you for helping

2. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

You can use conservation of energy.
For the simple model here the friction gives a constant force (coeff * weight) and energy = force*distance.
Assuming all the original energy was ke of the porsche and it all went into friction of the skid you can work out the ke and so the speed.

3. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

so,
Frictional force=mu(k)*ma=0.5*3684*2.55=4697.1N
Energy=force*distance=4697.1N*11.3m=53077.23J
KE=1/2*m*v^2
53077.23=0.5*3684*v^2
v=5.368

m2v2=(m1+m2)v
1380*v2=3684*5.368
therefore, v2=14.33m/s

Sorry, but the answer I got is a bit wierd...Is it the correct approach for this question?

4. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

The frictional force = normal force * coeff = weight * coeff = mg*coeff
Think about it - the magnitude of the friction doesn't care about the direction.

5. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

Frictional force=0.5*3684*9.8=18051.6N
Energy=18051.6*11.3=203983.08J
KE=203983.08
therefore,
203983.03=0.5*3684*v^2
v=10.510.524
m2v2=(m1+m2)v
1380*v2=3684*10.524
v2=28.1m/s=101.16km/h
therefore, we should charge Porsche driver for speeding

I think this is a pretty reasonable solution, but I am still not 100% sure whether it is correct or not...

6. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

Almost. If you asssume the SUV was stationary when hit, all the KE was the Porsche so the M in the KE part is only the M of the porsche.

7. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

Energy=18051.6*11.3=203983.08J
KE=203983.08
therefore,
203983.03=0.5*1380*v^2
v=45.901
m2v2=(m1+m2)v
1380*v2=3684*45.901
v2=45.901m/s=165.242km/h (a bit over twice as fast as the Hwy speed limit...)
therefore, we should charge Porsche driver for speeding

I think this should be the final answer for this problem

8. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

Looks ok

Not sure what you are trying to show with this bit:
m2v2=(m1+m2)v
1380*v2=3684*45.901

9. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

Sorry, there were some typo
Energy=18051.6*11.3=203983.08J
KE=203983.08
therefore,
203983.03=0.5*1380*v^2
v=17.194
m2v2=(m1+m2)v
1380*v2=3684*17.194
v2=45.901m/s=165.242km/h (a bit over twice as fast as the Hwy speed limit...)
therefore, we should charge Porsche driver for speeding

It should be good...sorry about the mistake again...

10. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

I just re-read the question more carefully.
It doesn't say the SUV was stationary.

You also have to use conservation of momentum to work out how much of the original momentum was porsche going north and SUV going west.

When they collide and move off at 40deg you can resolve this momentum into North and West components. ie, V*3684*cos(40) and V*3684*sin(40).
Momentum is conserved in each component so you know all the North component of momentum came from the Porsche going North.
You can work out V from knowing the total KE after the collision, it's just the energy that went into the skid we already found.
Then you know the total North momentum must be the same as the initial Porsche moment, and you knwo the Porsche mass so you have it's speed

Last edited: Oct 29, 2008
11. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

Sorry, I still don't quite get how you get the initial speeds for both cars. Do you use cos40=adj/11.3 to find out the distance that Porche will travel and sin40=oppo/11.3 to find out the distance that SUV will travel. and use the acceleration i got previously for 11.3m to find out the velocity for both cars.

12. Oct 29, 2008

### mgb_phys

Re: HELP!!!Cars collision problem

I think I edited my answer while you were writing yours.
The important thing is that momentum North-South and East-West are conserved separately.
You can resolve the resulting momentum vector of the collision into NS and EW which corresponds to the original momentum of the vehicles movign NS and EW

13. Oct 29, 2008

### Kudo Shinichi

Re: HELP!!!Cars collision problem

KE=203983.08
therefore,
203983.03=0.5*3684*v^2
v=10.524
North=v*cos40*3684
Momentum=10.524*cos40*3684=29699.862Kgm/s
29699.862=1380*V
V=21.52m/s=77.47km/h
the speed is below Hwy speed limit therefore no charge required

Is it is the correct way of solving this problem? I am not really sure what do you mean by total KE, I assumed that the KE you said is the one with two masses