Transfer of momentum problem

[Chain Mail]

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]

[swansont]

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?

[MindShadowfax]

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, 2012 by MindShadowfax

[theholykrael]

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.

[J.C.MacSwell]

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.

[theholykrael]

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

[J.C.MacSwell]

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

[Mellinia]

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.

[King Thando Mathe]

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.

[daniton]

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.

[Mellinia]

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, 2013 by Mellinia