# Help with Physics Elastic Collisions

Watch
Announcements
#1
Hi. I have been trying to figure out how to answer a question for ages now and still can’t seem to work it out. I have managed to complete part a, b, and c but can’t work out d. If anyone is able to help that would be much appreciated!

The question is as follows:
‘Three particles A, B and C lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with a velocity u. Particle A collides with B which then collides with C. Each of the particles has mass m and the collisions are elastic.’

Part D: ‘Now consider the same scenario but this time the masses of A, B and C are m, 2m and 3m respectively.’

I really don’t know how to go about working this one out so I would be thankful for anyone who could help.
Thanks.
0
1 year ago
#2
(Original post by Plane350)
Hi. I have been trying to figure out how to answer a question for ages now and still can’t seem to work it out. I have managed to complete part a, b, and c but can’t work out d. If anyone is able to help that would be much appreciated!

The question is as follows:
‘Three particles A, B and C lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with a velocity u. Particle A collides with B which then collides with C. Each of the particles has mass m and the collisions are elastic.’

Part D: ‘Now consider the same scenario but this time the masses of A, B and C are m, 2m and 3m respectively.’

I really don’t know how to go about working this one out so I would be thankful for anyone who could help.
Thanks.
Can you explain what the question wants, what are you looking to find out?
1
#3
(Original post by aviatoruk)
Can you explain what the question wants, what are you looking to find out?
Oh yeah, sorry I forgot to add that.

‘Find the velocity v𝟣 of A immediately after the collision with B, in terms of 𝑢
and 𝑚.’
0
1 year ago
#4
(Original post by Plane350)
Oh yeah, sorry I forgot to add that.

‘Find the velocity v𝟣 of A immediately after the collision with B, in terms of 𝑢
and 𝑚.’
Take momentum, so p=mu= m(A+B) (Call velocities of A and B A and B respectively)
u = A + B
Then you can consider energy which is constant across the system
0.5 m u^2 = 0.5m A^2 + 0.5 m B^2
Cancelling gives:
u^2=A^2 +B^2
Then you from momentum you know u^2 = (A+B)^2 = A^2 +B^2 +2AB
At this point you know either A or B has to be zero otherwise this falls apart, and because they're going from left to right, and it was A with momentum, A must be zero.
It's worth having a look here if you want to visualise it: https://phet.colorado.edu/sims/colli...on-lab_en.html
For the second case, try and do the same but you'll not get the case where A or B is 0
1
#5
(Original post by aviatoruk)
Take momentum, so p=mu= m(A+B) (Call velocities of A and B A and B respectively)
u = A + B
Then you can consider energy which is constant across the system
0.5 m u^2 = 0.5m A^2 + 0.5 m B^2
Cancelling gives:
u^2=A^2 +B^2
Then you from momentum you know u^2 = (A+B)^2 = A^2 +B^2 +2AB
At this point you know either A or B has to be zero otherwise this falls apart, and because they're going from left to right, and it was A with momentum, A must be zero.
It's worth having a look here if you want to visualise it: https://phet.colorado.edu/sims/colli...on-lab_en.html
For the second case, try and do the same but you'll not get the case where A or B is 0
Thanks so much for the help! However, I tried this with the part D and it says it was incorrect. Not sure where I went wrong, but this is what I did:
Using conservation of momentum since initial speed of B is 0:
mu = mA + 2mB.
u = A + 2B.
u^2 = A^2 + 4AB + 4B^2.
Using conservation of energy since the collision is elastic:
0.5mu^2 = 0.5mA^2 + mB^2 (since B has a mass of 2m this time).
u^2 = A^2 + 2B^2.

Then I set them equal to each other and solved:
A^2 + 2B^2 = A^2 + 4AB + 4B^2.
0 = 4AB + 2B^2.
4AB = -2B^2.
4A = -2B.
A = -1/2B.

I entered this into the answer box and it is apparently incorrect - although it does say it wants it in terms of u and m but I’m not sure what to do.
0
#6
(Original post by aviatoruk)
Take momentum, so p=mu= m(A+B) (Call velocities of A and B A and B respectively)
u = A + B
Then you can consider energy which is constant across the system
0.5 m u^2 = 0.5m A^2 + 0.5 m B^2
Cancelling gives:
u^2=A^2 +B^2
Then you from momentum you know u^2 = (A+B)^2 = A^2 +B^2 +2AB
At this point you know either A or B has to be zero otherwise this falls apart, and because they're going from left to right, and it was A with momentum, A must be zero.
It's worth having a look here if you want to visualise it: https://phet.colorado.edu/sims/colli...on-lab_en.html
For the second case, try and do the same but you'll not get the case where A or B is 0
Disregard my last message, I’ve got it now! Thanks so much for your help!
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How would you describe the quality of the digital skills you're taught at school?

Excellent (8)
10%
Okay (26)
32.5%
A bit lacking (22)
27.5%
Not good at all (24)
30%