Force on the car in a collision

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Homework Help Overview

The discussion revolves around a collision scenario involving two cars, each with a mass of 1000 kg, both accelerating towards each other at 1 m/s². The original poster seeks to determine the force on car B at the moment of collision, referencing Newton's laws of motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of Newton's third law, questioning the forces involved in the collision and the definitions of force in the context of the problem. There are attempts to clarify the nature of the forces acting on each car before and during the collision.

Discussion Status

The conversation is ongoing, with participants raising questions about the assumptions made in the original problem statement. Some have suggested that more detail is needed regarding the forces and conditions of the collision, while others are attempting to clarify the implications of Newton's laws in this context.

Contextual Notes

Participants note the absence of information regarding the velocities at the moment of collision and the duration of the impact, which complicates the assessment of forces involved. There is also a discussion about the proper interpretation of force in relation to momentum and acceleration.

  • #31
Nile Anderson said:
how do we define a force again, I think my error lies there , is it not the change of momentum rather than the momentum
It's neither.
A force is something one body exerts on another. A consequence of a force acting over a period of time can be a change in momentum.
Nile Anderson said:
I am thinking really on a resultant
Sure, but you have to identify the body acted on. If you fix on one of the cars, the other car exerts a force on it during the impact, but it does not exert a force on itself, so the resultant is not zero.
To get a zero resultant here you need to introduce a third object for both cars to act on.
 
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  • #32
haruspex said:
It's neither.
A force is something one body exerts on another. A consequence of a force acting over a period of time can be a change in momentum.

Sure, but you have to identify the body acted on. If you fix on one of the cars, the other car exerts a force on it during the impact, but it does not exert a force on itself, so the resultant is not zero.
To get a zero resultant here you need to introduce a third object for both cars to act on.
Newton's Second Law ?
 
  • #33
Nile Anderson said:
Newton's Second Law ?
Applied here how, exactly?
 
  • #34
F=mv-mu/t=m(v-u)/t=ma
 
  • #35
Nile Anderson said:
F=mv-mu/t=m(v-u)/t=ma
I asked exactly how you are applying it in the present context. Which body has mass m? What force acting on the body is represented by F? What time period does t stand for?
Anyway, F=m(v-u)/t is not quite right. That will give you the average force over the time interval t. Likewise, your a is the average acceleration.,
 
  • #36
I see , interpretation then
 
  • #37
Nile Anderson said:
I see , interpretation then
No, not interpretation, application. It doesn't mean anything to quote a law in relation to a problem if you cannot state how the entities in the law relate to those in the problem.
 
  • #38
haruspex said:
No, not interpretation, application. It doesn't mean anything to quote a law in relation to a problem if you cannot state how the entities in the law relate to those in the problem.
I say interpretation to say that you are completely right my friend , I have thought about it and that is what I meant you are right , I started with in inaccurate point, trying to defend that point , but deduction can truly get you know where with an incorrect base
 
  • #39
Nile Anderson said:
I say interpretation to say that you are completely right my friend , I have thought about it and that is what I meant you are right , I started with in inaccurate point, trying to defend that point , but deduction can truly get you know where with an incorrect base
OK.
 
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