# Homework Help: Newton's Third Law and Car Crashes

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1. Feb 29, 2016

### Balsam

1. The problem statement, all variables and given/known data
In terms of Newton's Third Law, why is it better for 2 cars of the same mass, both going 50km/h to crash into each other than for 1 car going 100km/h to crash into a wall?

2. Relevant equations
Newton's Third Law
F=ma

3. The attempt at a solution
I have no idea. I'm really confused

2. Feb 29, 2016

### giokrutoi

what do you mean in better

3. Feb 29, 2016

### RUber

If there are two cars, each one will be subjected to the other's opposing force. If there is one car, and you assume the wall is strong enough to fully stop the car, it will be subjected to the larger resisting force of the wall.

Let's say that $F_1$ is generated by car 1 and $F_2$ is generated by car 2, then $F_1= -F_2$ for a perfect head-on crash ending in full stop.
Let's say that $F_3$ is generated by the fast (100km/hr) car 3 and $F_4$ is generated by the wall, then $F_3= -F_4$ for a car-on-wall crash ending in full stop.

4. Feb 29, 2016

### Balsam

Less damaging

5. Feb 29, 2016

### Balsam

Of the 2 cars, each exerts a lesser force since their collision velocity is less. Whereas if the car crashed into the wall, the wall would exert a stronger reaction force on the car because of the increased velocity at the point of impact? Is this true?

6. Feb 29, 2016

### RUber

That's about right. I would assume that in all cases, it takes the same time for the cars to come to a stop. Thus, acceleration would be proportional to velocity.
More velocity brought down to zero velocity over the same time results in more force.
There are finer points when you deal with impulses and collisions, but as it refers to F=ma, this is the most fundamental explanation.
More velocity, same mass, same time => more force.

7. Feb 29, 2016

### Balsam

What if the masses were different for the 2 cars? I'm not sure how mass affects this situation

8. Feb 29, 2016

### RUber

Then you would not expect both cars to come to a full stop.