# Speeds regarding 2D inelastic collision

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1. Dec 29, 2015

### Negus

1. The problem statement, all variables and given/known data
You are called to the site of a collision between a small car and a half ton truck on a lonely country road. You are given the following information:
a) On clean dry pavement at a temperature of 20 C, the coefficient of kinetic friction is 0.600 on locked wheels.
b) The truck has left a skid mark that is 25.7 m long
c) The car has left a skid mark that is 20.3 m long
d) The estimated mass of the truck and contents was 1250 kg
e) The estimated mass of the car and contents was 750 kg
f) The truck’s skid mark is due North
g) The car’s skid mark is due East
h) The debris travelled 10.6 m 18.4 degrees E of N
i) The collisions appears to have been completely inelastic
j) The speed limit was 90 km/hr
k) Miraculously no one was seriously hurt

From this data can you work out the speed of the two vehicles?

2. Relevant equations

∑F = mTa

Ffric = μkmg

KE = (1/2)mv2 (I do not believe this is required since Ein ≠ Eout)

Momentum has not been covered in the course yet (next chapter), but I know how to use these type of equations:
mtruckvitruck = mdebrisvfsin(θf)
mcarvicar = mdebrisvfcos(θf)

3. The attempt at a solution
In calculating the acceleration on the debris of the collision (which I assumed included the combined vehicle mass), I used the Ffric and ∑F formulas to get

a = Ffric/mtotal
a = μ(mtruck + mcar)g/(mtruck + mcar)
a = μg

With the kinematics equation,

and knowing that vf = 0,

-v02 = 2μgd
v0 = √-2μgd
v0 = √-2(0.6)(-9.8)(10.6)
v0 = 11.16m/s = 40.2km/hr

After getting this value, I used the logic that the initial speed of the debris should be the final speed of the cars at the moment of the collision. Since the distances of the vehicles's skid marks were given, I plugged them into the same kinematics equation.

For the truck, I got 74.4km/hr, and for the car, I got 68.6km/hr. However, I tried solving the problem with the momentum equations just to make sure and I got different results (61.4km/hr for the truck and 32.1km/hr for the car). Not only that, but I was confused whether the skid mark distance implied that final velocities for both vehicles were 0. Any help would be appreciated; clearing any confusion works much better than giving answers.

2. Dec 29, 2015

### haruspex

no, you need the momentum equations here.

3. Dec 29, 2015

### Negus

So would the speeds I found through the momentum equations correct? If not, why?

4. Dec 29, 2015

### haruspex

yes.
I did not understand your comment about skid marks and final velocity. The skid marks would be on the approach to the collision point, so no, the final velocity there is not zero. There probably are skid marks from collision to final resting place too, but those are not mentioned.

5. Dec 29, 2015

### haruspex

yes.
I did not understand your comment about skid marks and final velocity. The skid marks would be on the approach to the collision point, so no, the final velocity there is not zero. There probably are skid marks from collision to final resting place too, but those are not mentioned.

6. Dec 29, 2015

### Negus

I also noticed how there was no mention of skid marks by the debris in the information given by the problem. I thought that meant friction had taken its course and the vehicles may have stopped to a point, which didn't make much sense. Momentum seems to be the best way to solve the problem; nevertheless, I feel like there is some other way to approach the problem knowing that we haven't encountered momentum problems yet (or even go over two-dimensional collisions)

7. Dec 29, 2015

### haruspex

Not aware of any method other than momentum, or equivalent.
I get a rather higher number for the truck's initial speed. Please post the details of your working.

8. Dec 29, 2015

### Pedro Delano

We can't calculate the work of the friction force? And then we discover the velocity just before impact by using the work-energy theorem. For this to work we must consider the initial velocity 90Km/h. Is it wrong?

9. Dec 29, 2015

### haruspex

Never apply that unless you have reason to believe work is conserved (or can assess the work lost by some means). In this case you have neither. Much work will be lost in the collision, and we do not know how much, yet. Momentum conservation is the only way to figure out the velocities just before impact.