# Elastic collisions

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There's three balls on a smooth surface, A, B, and C, with masses m, 2m, and 3m respectively. Ball A is given a velocity u and collides elastically with B, which collides similarly with C.

I need to find the fraction of u that ball C has immediately after its impact with B.

I did the following:

0.5 * u^2 * m = 0.5 * ku^2 * 2m

1 = 2k

k = 1/2

Then:

0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m

2(u^2)/4 = 3ku^2

1/2 = 3k

k = 1/6

Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?

I need to find the fraction of u that ball C has immediately after its impact with B.

I did the following:

0.5 * u^2 * m = 0.5 * ku^2 * 2m

1 = 2k

k = 1/2

Then:

0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m

2(u^2)/4 = 3ku^2

1/2 = 3k

k = 1/6

Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?

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(Original post by

There's three balls on a smooth surface, A, B, and C, with masses m, 2m, and 3m respectively. Ball A is given a velocity u and collides elastically with B, which collides similarly with C.

I need to find the fraction of u that ball C has immediately after its impact with B.

I did the following:

0.5 * u^2 * m = 0.5 * ku^2 * 2m

1 = 2k

k = 1/2

Then:

0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m

2(u^2)/4 = 3ku^2

1/2 = 3k

k = 1/6

Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?

**pineneedles**)There's three balls on a smooth surface, A, B, and C, with masses m, 2m, and 3m respectively. Ball A is given a velocity u and collides elastically with B, which collides similarly with C.

I need to find the fraction of u that ball C has immediately after its impact with B.

I did the following:

0.5 * u^2 * m = 0.5 * ku^2 * 2m

1 = 2k

k = 1/2

Then:

0.5 * u^2/4 * 2m = 0.5 * ku^2 *3m

2(u^2)/4 = 3ku^2

1/2 = 3k

k = 1/6

Which I'm told I need to input as a decimal to 3 s.f. I typed in: 0.167 (no units) but it's wrong? Is my method wrong?

Also conscious of the fact that we're having to assume that A stops moving instantly after the collision as it doesn't actually say this

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(Original post by

I'm not really sure what method you're using but I would use the conservation of momentum formula and say for the first one that (m x u) + (2m x 0) = (m x 0) + (2m x ku). This gives k=1/2 for the first one however it doesn't give 1/6 as the second answer.

Also conscious of the fact that we're having to assume that A stops moving instantly after the collision as it doesn't actually say this

Posted from TSR Mobile

**smartalan73**)I'm not really sure what method you're using but I would use the conservation of momentum formula and say for the first one that (m x u) + (2m x 0) = (m x 0) + (2m x ku). This gives k=1/2 for the first one however it doesn't give 1/6 as the second answer.

Also conscious of the fact that we're having to assume that A stops moving instantly after the collision as it doesn't actually say this

Posted from TSR Mobile

What did you get as your second answer?

Also, that's a good point...

[edit

] It's a smooth surface, so assuming air resistance is negligible and considering the balls are close together as in the diagram I have, it's not unreasonable to assume velocity on impact is equal to initial velocity.

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(Original post by

I'm using the idea of kinetic energy being conserved to answer it, but your method seems fine.

What did you get as your second answer?

Also, that's a good point...

[edit

] It's a smooth surface, so assuming air resistance is negligible and considering the balls are close together as in the diagram I have, it's not unreasonable to assume velocity on impact is equal to initial velocity.

**pineneedles**)I'm using the idea of kinetic energy being conserved to answer it, but your method seems fine.

What did you get as your second answer?

Also, that's a good point...

[edit

] It's a smooth surface, so assuming air resistance is negligible and considering the balls are close together as in the diagram I have, it's not unreasonable to assume velocity on impact is equal to initial velocity.

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