# Help with Physics Elastic Collisions

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Plane350

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#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.

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.

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aviatoruk

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#2

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#2

(Original post by

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.

**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.

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Plane350

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#3

(Original post by

Can you explain what the question wants, what are you looking to find out?

**aviatoruk**)Can you explain what the question wants, what are you looking to find out?

‘Find the velocity v𝟣 of A immediately after the collision with B, in terms of 𝑢

and 𝑚.’

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aviatoruk

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#4

(Original post by

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.’

**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.’

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

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Plane350

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#5

(Original post by

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

**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

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.

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#6

**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

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