# Speeds regarding 2D inelastic collision

• Negus
In summary, you are called to the site of a collision between a small car and a half ton truck on a lonely country road. The truck has left a skid mark that is 25.7 m long and the car has left a skid mark that is 20.3 m long, with the truck's skid mark being due North and the car's skid mark being due East. The debris traveled 10.6 m 18.4 degrees E of N and the collision appears to have been completely inelastic. The estimated mass of the truck and contents was 1250 kg, while the estimated mass of the car and contents was 750 kg. The speed limit was 90 km/hr and miraculously no
Negus

## Homework Statement

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 traveled 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. Homework 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)

## 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.

Negus said:
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.
no, you need the momentum equations here.

haruspex said:
no, you need the momentum equations here.
So would the speeds I found through the momentum equations correct? If not, why?

Negus said:
So would the speeds I found through the momentum equations correct?
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.

Negus said:
So would the speeds I found through the momentum equations correct?
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.

haruspex said:
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.

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)

Negus said:
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)
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.

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?

Pedro Delano said:
And then we discover the velocity just before impact by using the work-energy theorem.
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.

## 1. What is a 2D inelastic collision?

A 2D inelastic collision is a type of collision in which two objects collide and stick together, resulting in a loss of kinetic energy. In this type of collision, the objects do not bounce off each other like in an elastic collision.

## 2. How is momentum conserved in a 2D inelastic collision?

In a 2D inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. This means that the combined mass and velocity of the objects will remain the same before and after the collision.

## 3. What is the formula for calculating the final velocity of an object after a 2D inelastic collision?

The formula for calculating the final velocity of an object after a 2D inelastic collision is v = (m1v1 + m2v2) / (m1 + m2), where m1 and v1 are the mass and velocity of the first object, and m2 and v2 are the mass and velocity of the second object.

## 4. Can the coefficient of restitution be used to calculate the final velocity in a 2D inelastic collision?

No, the coefficient of restitution is only applicable in elastic collisions, where the objects bounce off each other. In a 2D inelastic collision, the objects stick together and there is no rebound, so the coefficient of restitution cannot be used.

## 5. How do you determine the direction of the final velocity after a 2D inelastic collision?

The direction of the final velocity after a 2D inelastic collision can be determined by using the law of conservation of momentum. The direction of the final velocity will be in the same direction as the total momentum of the objects before the collision.

• Introductory Physics Homework Help
Replies
10
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
6K
• Introductory Physics Homework Help
Replies
4
Views
5K
• Introductory Physics Homework Help
Replies
20
Views
2K
• Introductory Physics Homework Help
Replies
14
Views
1K