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physics question

hi, please could I have some help on this question? I'm really confused about momentum. I thought that in an elastic collision, relative speed of approach= relative speed of separation. In this case, both objects are moving in the same direction so relative velocity would be 20-8=12ms-1 and then they still carry on moving in the same direction so 10-v=8 so v should be 2?
question: https://ibb.co/pZ5VmPt
thanks!
Reply 1
Original post by anonymous56754
hi, please could I have some help on this question? I'm really confused about momentum. I thought that in an elastic collision, relative speed of approach= relative speed of separation. In this case, both objects are moving in the same direction so relative velocity would be 20-8=12ms-1 and then they still carry on moving in the same direction so 10-v=8 so v should be 2?
question: https://ibb.co/pZ5VmPt
thanks!

It would have to be 18. Its the (absolute) relative speeds that are equal, and obv (basic physical argument) y must be faster than x.
(edited 1 month ago)
Original post by mqb2766
It would have to be 18. Its the (absolute) relative speeds that are equal, and obv (basic physical argument) y must be faster than x.

sorry, I still don't really understand. What do you mean by absolute and why does my method not work? thanks!
Original post by anonymous56754
hi, please could I have some help on this question? I'm really confused about momentum. I thought that in an elastic collision, relative speed of approach= relative speed of separation. In this case, both objects are moving in the same direction so relative velocity would be 20-8=12ms-1 and then they still carry on moving in the same direction so 10-v=8 so v should be 2?
question: https://ibb.co/pZ5VmPt
thanks!

See equation 3 in this link.
https://www.miniphysics.com/types-of-collision.html

Absolute values mean modulus or just the magnitude of the number without the minus sign as shown in this pdf on page 2 after C) Elastic Collision Examples.
https://courses.physics.illinois.edu/phys211/su2012/Text/ch13.pdf
Reply 4
Original post by anonymous56754
sorry, I still don't really understand. What do you mean by absolute and why does my method not work? thanks!

Before collision the absolute value of the relative velocty is
|20-12| = 8m/s
Ater collision this could be either
8 = |10-18|
8 = |10-2|
but y must be travellling faster than x, so 18. If y is travelling at 2m/s, x would have to pass/teleport through x. Also as x loses momentum in the collision, y must gain momentum for it to be conserved.
(edited 1 month ago)
Original post by mqb2766
Before collision the absolute value of the relative velocty is
|20-12| = 8m/s
Ater collision this could be either
8 = |10-18|
8 = |10-2|
but y must be travellling faster than x, so 18. If y is travelling at 2m/s, x would have to pass/teleport through x. Also as x loses momentum in the collision, y must gain momentum for it to be conserved.

ohhh, I see, that makes a lot more sense, thank you 🙂
I see you included a modulus sign, can this be applied for momentums?
Reply 6
Original post by anonymous56754
ohhh, I see, that makes a lot more sense, thank you 🙂
I see you included a modulus sign, can this be applied for momentums?

No. Using gcse physics terminology, momentum is a vector quantity (as its essetially velocity, so sign indicates direction) whereas relative speed is a scalar quantity so is positive. The absolute value or modulus function is implied when you talk about relative speeds being equal and is how resitution is usually stated
https://en.wikipedia.org/wiki/Coefficient_of_restitution
so
|u-v| = |U-V|
where lower case is pre impact velociities and upper case is post impact velocities. Note that either u-v or U-V will be negative, but not both, as the objects must be approaching pre impact, but seperating post impact. so there is a sign change in the relative velocity in the collision. So your 20-12=8 and 10-#=8 must be wrong, as the first one is positive so the latter must be negative or 10-#=-8, so #=18.

So we express the relative speed relationship using the modulus or absolute value function around the relative velocities. Its an easy thing to overlook and using speeds (modulus / absolute value function) means that we dont care about whether the relative velocities are positive or negative or ... but we have to be a bit careful about applying it.

The relative speeds are equal (perfectly elastic) relationship is derived in
https://en.wikipedia.org/wiki/Elastic_collision
and its just a few lines from conserving KE and rearranging, dots, and dividing by momentum to get there. Its worth being clear that its really just the conservation of KE in another, simpler form. When you do it, you get something like
u-v = V-U
so the difference / relative velocity is flipped pre/post collision. Putting a modulus / absolute value function round each side means they represent relative speed as both are positive.
(edited 1 month ago)
Original post by mqb2766
No. Using gcse physics terminology, momentum is a vector quantity (as its essetially velocity, so sign indicates direction) whereas relative speed is a scalar quantity so is positive. The absolute value or modulus function is implied when you talk about relative speeds being equal and is how resitution is usually stated
https://en.wikipedia.org/wiki/Coefficient_of_restitution
so
|u-v| = |U-V|
where lower case is pre impact velociities and upper case is post impact velocities. Note that either u-v or U-V will be negative, but not both, as the objects must be approaching pre impact, but seperating post impact. so there is a sign change in the relative velocity in the collision. So your 20-12=8 and 10-#=8 must be wrong, as the first one is positive so the latter must be negative or 10-#=-8, so #=18.
So we express the relative speed relationship using the modulus or absolute value function around the relative velocities. Its an easy thing to overlook and using speeds (modulus / absolute value function) means that we dont care about whether the relative velocities are positive or negative or ... but we have to be a bit careful about applying it.
The relative speeds are equal (perfectly elastic) relationship is derived in
https://en.wikipedia.org/wiki/Elastic_collision
and its just a few lines from conserving KE and rearranging, dots, and dividing by momentum to get there. Its worth being clear that its really just the conservation of KE in another, simpler form. When you do it, you get something like
u-v = V-U
so the difference / relative velocity is flipped pre/post collision. Putting a modulus / absolute value function round each side means they represent relative speed as both are positive.

Sorry, I'm confused by this line "but not both, as the objects must be approaching pre impact, but seperating post impact. so there is a sign change in the relative velocity in the collision."
In the question above, particle with 20ms-1 carried on moving in the same direction at 10ms-1 instead of seperating and its velocity didn't become negative?
Reply 8
Original post by anonymous56754
Sorry, I'm confused by this line "but not both, as the objects must be approaching pre impact, but seperating post impact. so there is a sign change in the relative velocity in the collision."
In the question above, particle with 20ms-1 carried on moving in the same direction at 10ms-1 instead of seperating and its velocity didn't become negative?

The relative velocity changed sign. So the relative velocity of Y (to X) pre collision is
12-20 = -8
as relative to X, Y is moving backwards (relative displacement is decreasing). The relative velocity of Y (to X) post collision is
18-10 = 8
as relative to X, Y is moving forwards (relative displacement is increasing). Its the relative velocities thats important, not the individual velocities. Obv the relative displacement of Y (to X) is positive.

Obviously youd make the same argument flipping X and Y as 20-12=8, 10-18=-8.
(edited 1 month ago)
Original post by mqb2766
The relative velocity changed sign. So the relative velocity of Y (to X) pre collision is
12-20 = -8
as relative to X, Y is moving backwards (relative displacement is decreasing). The relative velocity of Y (to X) post collision is
18-10 = 8
as relative to X, Y is moving forwards (relative displacement is increasing). Its the relative velocities thats important, not the individual velocities. Obv the relative displacement of Y (to X) is positive.
Obviously youd make the same argument flipping X and Y as 20-12=8, 10-18=-8.

i see, thank you!

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