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

  • Thread starter Thread starter Sarah
  • Start date Start date
  • Tags Tags
    Force Impact
Click For Summary
The discussion revolves around calculating the force of impact when a 1000 kg car traveling at 20 m/s collides with a steel wall over 0.5 seconds. Participants clarify that the initial momentum is 20,000 kg m/s and the final momentum is 0 kg m/s, leading to a change in momentum of -20,000 kg m/s. The average force during the collision is determined using the formula F = Δp/Δt, resulting in a force of 40,000 N, with the negative sign indicating direction. The conversation emphasizes understanding the implications of vector quantities and the relationship between forces exerted on the car and the wall, ultimately concluding that the magnitude of the force is what is typically required.
  • #31
Sarah said:
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?
 
  • Like
Likes Sarah
Physics news on Phys.org
  • #32
to the left?
 
  • #33
Sarah said:
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.
 
  • #34
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!
 
  • #35
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.
 
  • #36
It says to find the resulting force on the impact
 
  • #37
"Force on of the impact" is not specific enough to know which particular force is asked for. But it probably means that you are only expected to give the magnitude of the force. I harassed you with the sign business mainly to get you to think about the meaning of the signs in your calculation.
 
  • #38
ahhh ok that makes sense, and no again you helped a lot and now i can know the why behind it so it really helps
 
  • #39
Great. I'm glad I could help.
 

Similar threads

Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 32 ·
2
Replies
32
Views
1K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 9 ·
Replies
9
Views
976
Replies
20
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
794
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K