# Homework Help: Accident Analysis.

1. Sep 28, 2008

### AnNe11

1. The problem statement, all variables and given/known data

A 1400 sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.39 west and 6.55 south of the impact point.
Part A -
How fast was sedan traveling just before the collision?
Part B
How fast was SUV traveling just before the collision?
2. Relevant equations

3. The attempt at a solution

2. Sep 28, 2008

### tiny-tim

Welcome to PF!

Hi Anne! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

Last edited: Sep 28, 2008
3. Oct 23, 2009

### ifyjo

Hi, it look like it's been a while since this thread was posted in, but I happen to have the same question.

I will post my attempts at solving it below.
First, if you don't mind, I would like to supplement my own figures for the question instead of the original posters since it seems he (or she) has lost interest in the question.

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.57 m west and 6.28 m south of the impact point.​

I believe the momentum before the collision will be the same after the collision.
To find the momentum after the collision, I need the mass of the enmeshed cars (3600 kg I believe) and its speed. Since I was given the coefficient of kinetic friction and the distance traveled until rest, I think I calculate the acceleration, then pair it with distance in the equation V2 = V02 + 2ad, where d is the length of the hypotenuse between 5.57 m and 6.28 m (8.3942 m). I

I'm not sure about the acceleration, but my attempt has put the net force equal to the coefficient of friction multiplied by the mass and the acceleration due to gravity
(ma = kmg). This would yield 7.35 for the acceleration, giving me 11.1083 as the initial velocity when plugged into the equation in the preceding paragraph.

However, I am not sure how go from there.

Any help would be much appreciated, thanks!

4. Oct 23, 2009

### tiny-tim

Welcome to PF!

Hi ifyjo! Welcome to PF!

Yes, the deceleration is µg, = .735, and the distance is 8.3942, so the "initial" velocity is 11.108 in the direction given.

Now, you know the direction of the two cars just before the collision, so call their speeds u and w, resolve 11.108 into west and south components and use conservation of momentum.

(incidentally, since you'll be resolving the "initial" velocity into west and south components anyway, you could have used the same deceleration on each component separately, without bothering about the hypotenuse!)

5. Oct 25, 2009

### ifyjo

Thanks so much for your help tiny-tim. I got it all to work out fine.