Calculating Force of Impact: Solving for F in a Collision with a Steel Wall

[Sarah]

Hey everyone so I'm having trouble figuring out this problem, it's for HW and I don't want to get an answer I just need help having it explained. I wasn't sure if I needed to use the momentum equation to figure it out.
Suppose a 1000kg car is traveling at 20 meters/second before experiencing a collision. (a) If the car runs into a steel wall and the collision occurs over the course of 0.5 seconds, then what is the resulting force of impact?


So I'm trying to figure out F right? So m=1000 kg, a= 20 m/s or do I need momentum? I'm just confused and need help breaking it down. I don't know where to plug in the 0.5 sec. The car is going from 20 m/s and is stopping but where do I put in the 0.5 sec?
Thanks!

 

[TSny]

Welcome to PF.

car is traveling at 20 meters/second ...

So m=1000 kg, a= 20 m/s

Does the 20 m/s represent a velocity or an acceleration?

 

[Sarah]

Oh! I think velocity!

 

[TSny]

Oh! I think velocity!

Yes.

 

[Sarah]

would the correct formula be F x change in time = change in momentum
then J=change in momentum ?

 

[Sarah]

and would 0.5 sec be considered the change in time?

 

[TSny]

would the correct formula be F x change in time = change in momentum

Yes. F is the average force of the collision.

 

[Sarah]

so then F=ma therefore it would be (1000 kg) (change in velocity/time) = change in momentum?
Thank you for your help:)

 

[TSny]

so then F=ma therefore it would be (1000 kg) (change in velocity/time) = change in momentum?
Thank you for your help:)

No. You're mixing the two formulas F = ma and F Δt = change in momentum.

You can just work with F Δt = change in momentum. Can you calculate the change in momentum of the car?

 

[Sarah]

Would the change in momentum be p=MV so (1000 kg) (20 m/s) ? so 20000 kg m/s?

 

[TSny]

I can't tell if you're thinking about this correctly.

So just to be clear, note that "change in momentum" means the difference between the final momentum and the initial momentum:
change in momentum = Δp = p_f - p_i.

What is the initial momentum?
What is the final momentum?
What is the change in momentum?

 

[Sarah]

ok. this is getting much clearer I think...haha
ok so then the pi would = 20000 kg m/s and pf= 0 m/s therefore it would be -20000 kg m/s for the final momentum?

 

[TSny]

ok. this is getting much clearer I think...haha
ok so then the pi would = 20000 kg m/s and pf= 0 m/s therefore it would be -20000 kg m/s for the final momentum?

Yes. Good. Except your final number is the change in momentum, right?

 

[Sarah]

No because I'm trying to find the final force right? So where the heck does the 0.5 sec fit? change of time??

 

[Sarah]

thank you for helping me btw

 

[TSny]

You are trying to determine the (average) force, F, that acts on the car during the time interval, Δt, of the collision.
(Once the collision is over, there is no force. So, the final force is zero.)

The formula states that the impulse F Δt is equal to the change in momentum. Here, F is the average force during the collision (which is what you want to find). You correctly found the initial and final momenta. You are given the time interval of the collision. So, you're almost there.

 

[Sarah]

oh! so would the formula be f 0.5 sec=-20000 kg/ms and then do the calculations to it?

 

[TSny]

Yes.

 

[Sarah]

yay! so subtract 0.5 sec to -20000?

 

[TSny]

Subtract?

 

[Sarah]

divide! sorry its been a while with math and science for me lol

 

[Sarah]

F=40000 kg m/s

 

[TSny]

OK. I think you have it. If you get a negative answer for F, you should try to interpret the meaning of the negative answer.

 

[Sarah]

yes meant to put a negative so would it be F= -40000

 

[TSny]

OK. In post # 22 you did not quite get the correct units and you didn't specify the units for F in your last post.

Should the negative result for F be of concern?

 

[Sarah]

No it shouldn't shoot haha, I wrote down in my notebook that units should be N/ S does that seem right?

 

[TSny]

No it shouldn't shoot haha, I wrote down in my notebook that units should be N/ S does that seem right?

You're calculating a force. What is the SI unit for force?

 

[Sarah]

N ??

 

[TSny]

Yes. You should get Newtons. To see how this comes about, note that solving the formula F Δt = Δp for F gives F = Δp/Δt. Δp has units of mass times velocity, or kg m/s. When you divide by Δt (which has units of seconds), you get that Δp/Δt has units of kg m/s². You should be able to see that these units are equivalent to Newtons (Think F = ma).

 

[Sarah]

I'm not quite sure how to make it not negative

 

[TSny]

I'm not quite sure how to make it not negative

Velocity, momentum, and force are all vector quantities. They have directions. For problems where an object moves along a straight line, we often use signs to indicate direction. So, if we take positive to be toward the right, then the initial velocity of +20 m/s means that the car was initially moving to the right. It was stopped by a force. What must be the direction of the force in order to stop the car?

 

[Sarah]

to the left?

 

[TSny]

to the left?

Sure. If the force were to the right it would speed up the car rather than stop it.

So, the negative sign is telling you that the force of the collision on the car is in the opposite direction of the initial velocity of the car.

It is not clear in the statement of the problem whether or not you should include the negative sign in your answer. It could be that you are only meant to specify the magnitude of the force. (Magnitudes of vectors are always positive or zero). If so, you would drop the negative sign and state that the magnitude of the force is 40,000 N.

 

[Sarah]

thank you so so much for all your help I feel like I get such a better understanding and can go on and do my other hw! Thank you!

 

[TSny]

OK. Just one more point. The problem doesn't specify whether or not the force that is asked for is the force that the wall exerts on the car or the force that the car exerts on the wall.

You found the force on the car. By Newton's third law, the force on the wall would be equal in magnitude but opposite in direction.

Force on car = -40,000 N
Force on wall = +40,000 N

Both forces have the same magnitude of 40,000 N.