What Is the Maximum Deflection Angle in a Partially Inelastic Collision?

[thenewbosco]

I had made an error on my previous post so i will take it back to the beginning. I can't get this one for some reason.

First look at the diagram: http://bosco.iwarp.com/diagram.jpg

OK here's the problem:

Two identical balls (mass m) undergo a collision. Initially one ball is stationary, the other has kinetic energy of 8J. The collision is partially inelastic with 2 J energy converted to heat. What is the maximum deflection angle (alpha or beta) at which one of the balls is observed?

I have come up with the following relationships using conservation of momentum and energy:

1. Vo = V1 cos a + V2 cos B

2. 0 = V1 sin a - V2 sin B

and 12 = m (V1f^2 + V2f^2) which i believe can be written

3. 0.75Vo = V1^2+V2^2

by rearranging initial kinetic energy of ball one to solve for m.

How do i go about comparing alpha and beta? Are my equations correct
Help please! thanks

 

[Andrew Mason]

I have come up with the following relationships using conservation of momentum and energy:

1. Vo = V1f cos a + V2f cos B

2. 0 = V1f sin a + V2f sin B

and 12 = m (V1f^2 + V2f^2) which i believe can be written

3. 0.75Vo = V1f^2+V2f^2

1. and 2. are fine. 3. should have Vo^2:

This is not a trivial problem to solve. Use 1 and 2. to find $cos\beta \text{ in terms of } sin\alpha \text{ and } cos\alpha$

Then look at the ranges of values that $cos\beta$ can have. There should be a $\sqrt{.75}$ term in there somewhere.

AM

 

[thenewbosco]

Well i have attempted this another way and have got the following:

$2 = mv_{1}v_{2} (cos \alpha }+ \beta)$

now i do not know how to compare $\alpha$ and $\beta$

any help

 

[thenewbosco]

?

 

[Andrew Mason]

Well i have attempted this another way and have got the following:

$2 = mv_{1}v_{2} (cos \alpha }+ \beta)$

now i do not know how to compare $\alpha$ and $\beta$

any help

It has to have $\beta$ in terms of $\alpha$ ONLY. You can do this with these equations: determine v2f in terms of v1f using 2. and substitute into 1. to find v1f in terms of $\alpha \text{ and } \beta$. Substitute also into 3. to find v1f. $(v_0^2 = 16/m)$. Then combine the two to find $beta$ in terms of $\alpha$ (the m falls out).

AM

 

[thenewbosco]

I don't know why this is not working out but after trying what you said i got:

[tex]3sin^2\beta (cos^2\alpha + 2cos\alpha + cos^2\beta) = 4(sin^2\beta+sin^2\alpha)[/itex]<br /> <br /> It should probably be simpler shouldn't it?<br /> <br /> I really want to get this one done but it just isn't working out<br /> Thanks for all your help on this one[/tex]

 

[thenewbosco]

??

 

[Andrew Mason]

I don't know why this is not working out but after trying what you said i got:

[tex]3sin^2\beta (cos^2\alpha + 2cos\alpha + cos^2\beta) = 4(sin^2\beta+sin^2\alpha)[/itex]<br /> <br /> It should probably be simpler shouldn't it?<br /> <br /> I really want to get this one done but it just isn't working out<br /> Thanks for all your help on this one[/tex]

$$I told you it was non-trivial. Have a look at the solution for an elastic collision at: <a href="http://rustam.uwp.edu/201/L12/lec12_w.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://rustam.uwp.edu/201/L12/lec12_w.html</a> (scroll down to the collision in 2 dimensions). For elastic collisions the two angles add to 90 degrees. The question is asking how the sum of the two angles in an inelastic collision where 25% of the energy is lost compares to 90 degrees.<br /> <br /> AM$$