Here's an interesting exercise:

 

 

The entire scenario takes place in one dimension of space.

 

Ball B is at rest. Ball A has momentum p_a.

 

Ball A makes a head-on perfectly elastic collision with Ball B. Ball B then moves off with a momentum of p_b.

 

 

Prove that:

 

[math]|\frac{p_b}{p_a}|<2[/math]

Here's an interesting exercise:

 

 

The entire scenario takes place in one dimension of space.

 

Ball B is at rest. Ball A has momentum p_a.

 

Ball A makes a head-on perfectly elastic collision with Ball B. Ball B then moves off with a momentum of p_b.

 

 

Prove that:

 

[math]|\frac{p_b}{p_a}|<2[/math]

This looks like a homework question.

 

What have you done to try and solve it?

Right now I'm too tired, I'll try to think about it tomorrow and check what other people said. Good luck with that one!

Edited December 8, 201213 yr by MindShadowfax

Ordinarily, you would need to know if ball A is at rest after the collision, but since the required answer is requested in absolute form, and assuming 'positive' values to be in the direction of the original momentum P_a... it's safe to say we can assume ball A to be at rest post-collision. Given that assumption, I'd probably answer the question essentially in english, as a perfectly elastic collision maintains both momentum and kinetic energy. The question therefore gives the answer in itself, in that P_a = P_b, therefore P_a/P_a = 1, and 1 < 2.

 

That said, if it is (or 'was'... old thread I know) a homework question and some working were expected to be shown, you can prove the conservation of momentum via the equation for the conservation of kinetic energy, as both KE and P are proportional to both mass and velocity. This is largely academic though, because all you end up proving is that m_av_a = m_bv_b, even though m_a and m_b need not necessarily be equal.

Ordinarily, you would need to know if ball A is at rest after the collision, but since the required answer is requested in absolute form, and assuming 'positive' values to be in the direction of the original momentum P_a... it's safe to say we can assume ball A to be at rest post-collision. Given that assumption, I'd probably answer the question essentially in english, as a perfectly elastic collision maintains both momentum and kinetic energy. The question therefore gives the answer in itself, in that P_a = P_b, therefore P_a/P_a = 1, and 1 < 2.

 

That said, if it is (or 'was'... old thread I know) a homework question and some working were expected to be shown, you can prove the conservation of momentum via the equation for the conservation of kinetic energy, as both KE and P are proportional to both mass and velocity. This is largely academic though, because all you end up proving is that m_av_a = m_bv_b, even though m_a and m_b need not necessarily be equal.

This would be true if, and only if, the balls had the same mass.

This would be true if, and only if, the balls had the same mass.

Are you saying that to assume an elastic collision resulting in Ball A being at rest (newton's cradle style) would be to assume equivalent masses?

 

Would the easiest solution to the problem simply be to rearrange conservation of momentum into P_a/P_b=?

 

A quick correction to my post as well...

I should have written:

m_av_a² = m_bv_b²

Are you saying that to assume an elastic collision resulting in Ball A being at rest (newton's cradle style) would be to assume equivalent masses?

 

Would the easiest solution to the problem simply be to rearrange conservation of momentum into P_a/P_b=?

 

A quick correction to my post as well...

I should have written:

m_av_a² = m_bv_b²

I think they are after a proof that

 

 

regardless of any differences in the masses

note: you should have gotten Pa=P+Pb from the conservation of momentum, and the knowledge that |P|<|Pa| since it's an head on elastic collision. I guess that's done then with additional mathematical manipulation.

1)Working from tbe momentum eqtn will certainly have you assume that velocity of A is 0 after collision.Thus Pa will become equal to Pb and a ratio of 1 which therefore is less than 2.

2)Working from the conservation of Kinetic Energy,will Pa/Pb=Vb/Va.Vb cannot be twice Va[ That we all know] since Vb is influenced by Va.

3)One can also create: Pa = [-(Pa-Pb)] + Pb

This will lead to Pa=Pb again.If one does not want to apply that A is stationary after collision then Pa after collision can be regarded as [-(Pa-Pb)].But thesevis somehow bizzarre.

Actually,if you say Pa=Pb then Ma=Mb which is a necessary condition,which also indicate the ratio is 1.if they don't have same mass then Pa is greater than Pb because some momentum is still there in A because it can't simply stop after the collision,which clearly show the ratio is less than 1 -> less than 2.

Pa=Pb does not imply Ma=Mb. It implies MaVa=MbVb, and for the same reason that 1x4=2x2, 4 does not equal 2.

Edited April 3, 201312 yr by Mellinia