# Car Crash Work and Energy Problem

Tags:
1. Dec 7, 2016

### a1234

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

A red van came through and crashed into a green van. The driver of the red van claims that he was driving at a speed of 11 meters/second at the moment of the crash. You want to find out if his claim is true.

a. the mass of the red van is 1000 kg
b. the mass of the green van is 800 kg
c. the length of the skid mark from the red van is 20 meters
d. the length of the skid mark from the green van is 2.5 meters
e. the red van was moving at 8 meters/second when the crash happened
f. the coefficient of kinetic friction for the skidding is 0.4

1. Tell how much work the brakes did on the red van.
2. What was the change in energy of the red van before the crash?
3. What was the red van's initial kinetic energy?
4. Is the driver's claim correct?

2. The attempt at a solution

I don't see a use for most of the data given for the green van, but here are my calculations:

Frictional force = coefficient of friction * normal force
Ff = 0.4 * 1000 * 9.8 = 3920

Work = -Ff * d
W = -3920 * 20 = -78,400 J ---> work done on red van

KEf = 1/2mv^2
KEf = 1/2 * 1000 * 8^2 = 32,000 J

W = kinetic energy change
W = KEf - KEi
-78,400 = 32,000 - KEi
KEi = 46,400 J ---> initial kinetic energy of red van

And since we don't know the initial velocity...

KEi = 46,400
46,400 = 1/2mv^2
46,400 = 1/2 * 1000 * v^2
v^2 = 92.8
initial velocity is app. 9.63 meters/second ---> the driver was wrong about going at 11 m/s

The change in kinetic energy is 32,000 - 46,400 = -14,400 J.

2. Dec 7, 2016

### PhanthomJay

Method looks good, but check your work-energy equation you did the math wrong.

3. Dec 7, 2016

### a1234

You mean this?
W = KEf - KEi

Not sure how to work out the positive and negative signs.
Should 32,000 be negative too?

-78,400 = -32,000 - KEi
-46,400 = -KEi
KEi = 46,400 J

4. Dec 8, 2016

### TomHart

The red van originally had some initial kinetic energy. It lost 78,400 J of that from skidding. And after skidding, KEf = 32,000.
So KEi = 78,400 + 32,000 = 110,400

Last edited: Dec 8, 2016
5. Dec 8, 2016

### haruspex

I'm not making much sense of the problem statement.
At the instant before the crash, or before slamming on the brakes prior to the crash?
Is that skid leading up to the crash point or from it?
I guess that is the speed of the red van at the instant before the crash, and the 11m/s is something else - or should that say green van?
If all the deceleration occurred while skidding, the brakes did no work. It was all done by friction between road and tyres.

6. Dec 8, 2016

### TomHart

I took it to mean that 11 ms-1 was his claimed speed before he hit the brakes. But I agree that the wording of the problem is not at all clear.

7. Dec 8, 2016

### a1234

"...he was driving at a speed of 11 meters/second at the moment of the crash."
This is supposed to say that the driver claims he was going at 11 m/s when he felt he had to jam the brakes to avoid hitting the other car. But when the collision actually happened, he was going at 8 m/s. So we still need to find out if the driver's claim is true.
Sorry for the confusion.

8. Dec 8, 2016

### TomHart

What you wrote here was correct until the very last line. You just did the math wrong.

9. Dec 8, 2016

### a1234

-78,400 = 32,000 - KEi
KEi = 110,400 J

10. Dec 8, 2016

### TomHart

So then, was the driver telling the truth?

11. Dec 8, 2016

### a1234

KEi = 1/2mv^2
110,400 = 1/2 * 1000 * v^2
110,400/500 = v^2
initial velocity is app. 14.85 m/s. So the driver is still wrong.

32,000 - 110,400 = -78,400 J, the change in kinetic energy.

12. Dec 8, 2016

### TomHart

Looks good.

13. Dec 8, 2016

### haruspex

Looks right to me too.
Note that the mass of the vehicle is irrelevant for this part of the question. You can just use the SUVAT equation vf2=vi2+2as, where a=-μg.