Distance traveled during a head on collision

  • #31
phinds said:
1500*(20)+2400*(-12.5) = (1500+2400)*0

Makes it look like he is equating 20 - 12.5 with 0
Sorry, I still can't see how you can read it that way. Are you suggesting some interpretation of the asterisks other than multiplication?
 
Physics news on Phys.org
  • #32
haruspex said:
Sorry, I still can't see how you can read it that way. Are you suggesting some interpretation of the asterisks other than multiplication?
See post #29 for someone who had an immediate response identical to mine.
 
  • #33
phinds said:
See post #29 for someone who had an immediate response identical to mine.
Well, I can't see how that fits with what you wrote in post #30, and jbriggs' "physics" interpretation is how I read it. Mr. Krock took the equation m1v1i+m2v2i=(m1+m2)vf, plugged in the known values and found that vf must be 0. He then posted that fact in the form of a solved equation. I'm sure it was not intended to represent the logic by which he arrived at the result. As a way of illustrating where he was up to, I have no problem with it.
 
  • #34
haruspex said:
Well, I can't see how that fits with what you wrote in post #30,
You're right. Post #30 was an incorrect way to express my dismay at his intermediate result.

I'm sure it was not intended to represent the logic by which he arrived at the result.
Well, I'm not a mind reader. I was commenting on what he posted, not what he meant (although actually, I addressed that as well).

As a way of illustrating where he was up to, I have no problem with it.
Well, I do. Let's agree to disagree.
 
  • #35
phinds said:
You're right. Post #30 was an incorrect way to express my dismay at his intermediate result.

Well, I'm not a mind reader. I was commenting on what he posted, not what he meant (although actually, I addressed that as well).

Well, I do. Let's agree to disagree.
Ok, all good.
 
  • #36
It would help if we could get him to write the two equations.
To list the number of unknowns
To determine if there are enough equations to solve for the unknowns
To rearrange the equations to solve for the unknowns

Then to give the results for the three conditions 1) fully elastic. 2) inelastic 3) partially elastic.

Then to answer the question two ways. 1) for what was asked - distance travelled. Which is not knowable since no stopping forces were provided.

2) for what they probably wanted to ask - vehicle velocities.
 
  • #37
NickAtNight said:
It would help if we could get him to write the two equations.
To list the number of unknowns
To determine if there are enough equations to solve for the unknowns
To rearrange the equations to solve for the unknowns

Then to give the results for the three conditions 1) fully elastic. 2) inelastic 3) partially elastic.

Then to answer the question two ways. 1) for what was asked - distance travelled. Which is not knowable since no stopping forces were provided.

2) for what they probably wanted to ask - vehicle velocities.
No progress can be made at all until we are clear what is meant by the distance a vehicle travels during a collision. See post #25.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
335
Views
19K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
1K