Calculate speed from elastic and inelastic collisions? (momentum)

AI Thread Summary
The discussion revolves around calculating the speed of trolley B after a collision with trolley A, focusing on both elastic and inelastic scenarios. The user, Oscar, is confused about the elastic collision where trolley A stops and trolley B moves off, which is deemed impossible due to momentum conservation principles. The calculations provided suggest speeds of 4.8 m/s for the elastic case and 2.4 m/s for the inelastic case, with the latter being confirmed as correct. Participants emphasize the importance of checking momentum conservation and suggest that the question may contain a mistake regarding the elastic collision scenario. The conversation concludes with Oscar considering whether to address the potential error with the teacher.
OscarF
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Homework Statement
Question:

Trolley A, of mass 180 g, is moving at a velocity of 4 m/s towards trolley B which is stationary and has a mass of 120 g. Trolley A then collides with trolley B.

Calculate the speed of trolley B if:

i the collision is elastic (trolley A stops and trolley B moves off)

ii the collision is inelastic (the two trolleys join and move off together).
Relevant Equations
momentum = mass x velocity
So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated,

Oscar.
 
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OscarF said:
Homework Statement:: Question:

Trolley A, of mass 180 g, is moving at a velocity of 4 m/s towards trolley B which is stationary and has a mass of 120 g. Trolley A then collides with trolley B.

Calculate the speed of trolley B if:

i the collision is elastic (trolley A stops and trolley B moves off)

ii the collision is inelastic (the two trolleys join and move off together).
Relevant Equations:: momentum = mass x velocity

So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated,

Oscar.
How much can you do? You need to show us your work.

Note that:

OscarF said:
i the collision is elastic (trolley A stops and trolley B moves off)

Is not a possible scenario. Which you may be able to prove.
 
PeroK said:
How much can you do? You need to show us your work.

Note that:
Is not a possible scenario. Which you may be able to prove.

Well, from a bit more digging around I came up with these two calculations:

(For the elastic scenario)
V2f=(2m1/(m1+m2))*v1i

(For the inelastic scenario)
V2f=(m1v1)/(m1+m2)

Which led me to the answers:

i) 4.8 m/s

ii) 2.4 m/s
 
The answer to ii) looks right.

Did you check momentum conservation for i)? What happens to the 180g cart after the collision?
 
PeroK said:
The answer to ii) looks right.

Did you check momentum conservation for i)? What happens to the 180g cart after the collision?

Good point! Will check now, one second.
 
OscarF said:
Good point! Will check now, one second.
I just realized... obviously the momentum will not be the same... because for trolley B pre-collision the momentum would be 0 and afterwards it will be whatever 120*4.8 is...
 
OscarF said:
I just realized... obviously the momentum will not be the same... because for trolley B pre-collision the momentum would be 0 and afterwards it will be whatever 120*4.8 is...
Hmm... I personally wouldn't expect such a twist in a homework like this... Do you think the solution would be to write something like this event isn't possible because...
 
OscarF said:
Hmm... I personally wouldn't expect such a twist in a homework like this... Do you think the solution would be to write something like this event isn't possible because...
As I understand it, you are using a formula for the velocity of the second cart in an elastic collision. You must have a similar formula for the velocity of the first cart.

Perhaps whoever set the question made a mistake. In an elastic collision the moving cart only stops if the two carts have equal mass; otherwise, both carts are moving after the collision.
 
PeroK said:
As I understand it, you are using a formula for the velocity of the second cart in an elastic collision. You must have a similar formula for the velocity of the first cart.

Perhaps whoever set the question made a mistake. In an elastic collision the moving cart only stops if the two carts have equal mass; otherwise, both carts are moving after the collision.
I do have an equation for the first cart - let me just check what that gives as a result in this instance
 
  • #10
OscarF said:
I do have an equation for the first cart - let me just check what that gives as a result in this instance
PeroK said:
As I understand it, you are using a formula for the velocity of the second cart in an elastic collision. You must have a similar formula for the velocity of the first cart.

Perhaps whoever set the question made a mistake. In an elastic collision the moving cart only stops if the two carts have equal mass; otherwise, both carts are moving after the collision.
Hmm... The output results as 0.8 m/s if I do that... Due to the nature of the question, it just feels unlikely that it's a mistake - but you never know - so I have a few options. Either, I just keep it as it is - the answer fine, done and dusted or I add the answer but also add a little note to the side or I can try and talk to my teacher before the homework is due. If I were to right a note about why it's impossible, what would you suggest?
 
  • #11
OscarF said:
Hmm... The output results as 0.8 m/s if I do that... Due to the nature of the question, it just feels unlikely that it's a mistake - but you never know - so I have a few options. Either, I just keep it as it is - the answer fine, done and dusted or I add the answer but also add a little note to the side or I can try and talk to my teacher before the homework is due. If I were to right a note about why it's impossible, what would you suggest?
It depends whether your teacher is the sort of person who can accept that he/she made a mistake!

##0.8m/s## and ##4.8m/s## are definitely correct.
 
  • #12
PeroK said:
It depends whether your teacher is the sort of person who can accept that he/she made a mistake!

##0.8m/s## and ##4.8m/s## are definitely correct.
They didn't make the worksheet - It's an AQA physics gcse practice question. My teacher's a nice person though - would be fine if I wanted to contest the question!
 
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  • #13
OscarF said:
Homework Statement:: Question is below
Relevant Equations:: momentum = mass x velocity

Background:

Unfortunately, I missed my class' lesson on momentum. I've watched this video: https://www.khanacademy.org/science...lisions-ap/v/elastic-and-inelastic-collisions. To try and help me with the following homework - it helped but I still can't really get the answer.

Question:

Trolley A, of mass 180 g, is moving at a velocity of 4 m/s towards trolley B which is stationary and has a mass of 120 g. Trolley A then collides with trolley B.

Calculate the speed of trolley B if:

i the collision is elastic (trolley A stops and trolley B moves off)

ii the collision is inelastic (the two trolleys join and move off together).

Any help would be much appreciated,
Oscar.
You just need two general facts:
There being no external horizontal forces on the two trolley system, their total momentum is conserved. Just remember that velocity and momentum are vectors, so direction matters.
If the collision is elastic, total kinetic energy is conserved.

In case ii, KE is not conserved, but you know instead that the two have the same final velocity.

This is enough to find the final velocities. For any further help you must show some attempt.
 
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