How Do You Solve for tanB in an Elastic Collision Problem?

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The discussion revolves around solving for tanB in an elastic collision problem involving two spheres with different masses. The user has derived equations based on conservation of momentum and energy but is struggling to isolate tanB in terms of the masses and angle A. Suggestions from other participants include using the energy equation and considering the scenario where angle A equals angle B. The user expresses frustration with the algebra involved and the complexity of the equations. Ultimately, the conversation highlights the challenges of applying conservation laws in collision problems while seeking a clear expression for the angles involved.
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Homework Statement


A sphere of mass m1 traveling at velocity u1 collides elastically with a stationary sphere of m2. The particles move away with v1 and v2 respectively.

I'm at part 4, of 5. It asks us to resolve the conservation of momentum equations normal and tangential to the post-collision velocity of m2. I obtained these equations using the diagram that's provided. The part I'm stuck on is where they ask us to obtain an expression for tanB involving only the masses and angle A.

A = angle of incidence
B = angle of reflection

The 5th part asks "Hence when does the angle of incidence equal angle of reflection?". So I know the resultant equation should be something along the lines of:

tanB = ((m1 + m2)/m2)*tanA

as m1 should be << m2

Homework Equations


m1u1cosA = -m1v1cosB + m2v2

m1u1sinA = -m1v1sinB

m1u12 = m1v12 + m2v22

The Attempt at a Solution



I found terms for v1 and equated them eventually getting:

tanB = ((m1u1cosA)/(m1u1cosA - m2v2))*tanA

But now I'm stuck. Any pointers?
 
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Welcome to PF!

Welshy said:
m1u1cosA = -m1v1cosB + m2v2

m1u1sinA = -m1v1sinB

Hi Welshy! Welcome to PF! :smile:

What about the energy equation?
 


tiny-tim said:
Hi Welshy! Welcome to PF! :smile:

What about the energy equation?
I used that in an earlier part of the question and when I tried to use it here it only complicated things by giving me more velocity terms which I couldn't cancel.

Any pointers on how it could be used - or even if my existing methods are on the right track?
 
Hint:

i] put A = B

ii] use (u1 + v1) and (u1 - v1), and the energy equation. :smile:
 
tiny-tim said:
Hint:

i] put A = B

Sorry, I don't think I worded the question right. The bit I'm stuck on needs B = [mass terms]*A so I can't do that yet. But I'll try the other hint. Thanks.
 
tiny-tim said:
Hint:

i] put A = B

ii] use (u1 + v1) and (u1 - v1), and the energy equation. :smile:

I'm having no joy with this at all. Each of u1 and v1 can have 2 different equations. That gives a total of 8 combinations for (u1 - v1)(u1 + v1) and I'm sure I've tried them all.

I don't think I can make them cancel that way when A doesn't = B . :/
 
Any more hints that anyone can give? Is tiny-tim's method valid for A not equal to B? And, if so, where will I pull my speed equations from?

The sheer algebra of this just has me stumped. :/
 
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