Help with concept problem~

1. Nov 28, 2004

sinas

Consider two situations:

1) You are driving 50 mph and crash head on into an identical car also going 50 mph.
2) You are driving 50 mph and crash head on into a stationary brick wall.

In neither case does your car bounce off the thing it hits, and the collision time is the same in both cases. Which of these two situations would result in the greatest impact force? (a-hitting the car, b-hitting the wall, c-forces equal, d-not enough info)

My intuition tells me that it's A, but I've learned not to trust my intuition :grumpy:

2. Nov 28, 2004

Gokul43201

Staff Emeritus
Yes, it's not a good idea to trust intuition, when there's something better : definition !!

How would you calculate the average impact force on your car ? Start from the definition.

3. Nov 28, 2004

ceptimus

Assuming the two cars are identical, and travelling at the same speed, but in opposite directions

Imagine the vertical plane that defines where the two cars meet. Would any part of either car penetrate this invisible plane? In practice, with real world cars, they probably would, as some of the hard bits of one car would punch their way into the softer bits of the other one, and vice versa.

But if we assume the cars are made of uniform stuff, say we replace the cars in our thought experiment with solid identical lumps of steel, then there is no reason to think that any part of either lump would pass through the invisible boundary. In which case we can replace the invisible boundary with an inpenetrable, unmovable wall, and not lessen the impact in any way.

So I would say with the sorts of cars that populate the world of physics questions, there is no difference between the two impacts. But if you asked me to choose whether I would rather drive my own car into an identical car coming the other way at the same speed, or into a solid wall, I would choose the other car every time.

4. Nov 28, 2004

sinas

average force = impulse/time

If the times are the same, then it just depends on the impulse, which is change in momentum...

For the two cars..
m(50)+m(-50)=2m(0)
delta p = m(50)

?

For the wall..

5. Nov 28, 2004

Gokul43201

Staff Emeritus
Good.

You are only interested in the change in momentum of your car in both cases. You're nearly there. How is change in momentum defined ?

(After doing this, look into ceptimus' visualization approach and see if that agrees with what you get this way)

6. Nov 28, 2004

sinas

So if your car has the same momentum going into each collision, and has 0 final momentum (mass*0 velocity?), then delta p for your car (and thus the force) is the same for each?

7. Nov 28, 2004

sinas

BTW thanks both of you for the help