Ball on String Elastic Collision

AI Thread Summary
In the discussion about the elastic collision of two steel balls, the problem involves a 250 g ball and a 500 g ball, both hanging from strings and released after being pulled to a 22-degree angle. The participant calculated the velocities before the collision using conservation of energy and momentum principles but encountered difficulties in deriving the correct rebound angles. The equations for final velocities were discussed, with corrections made to the momentum equations to ensure proper directionality. Ultimately, the participant sought clarification on their approach and the derivation of the formulas used. The conversation highlights the importance of accurately applying conservation laws in collision problems.
jzwiep
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Homework Statement



A 250 g steel ball and a 500 g steel ball each hang from 3.5-m-long strings. At rest, the balls hang side by side, barely touching. The 250 g ball is pulled to the left until the angle between its string and vertical is 22 degrees The 500 g ball is pulled to a 22 degree angle on the right. The balls are released so as to collide at the very bottom of their swings.

What angle does each ball rebound?

Homework Equations



Conservation of Energy
U = K' ----> K = U'
Conservation of Momentum (at collision)
p1 + p2 = p1' +p2'

h=L(1-cos(22))

The Attempt at a Solution



I found the v for both immediately before the collision by:

U=K'
v=sqrt(2g(L-Lcos(22))

Then used that value to find v1 and v2 after the collision with:

-v=2v2' - v1'

and

3v2=v1'2 + 2v2'2

and got two quadratics:

v2
and
v1

I plugged those v values back into the K = U' formula and got:

Theta 1: 14.6636
Theta 2: 8.45

Where did I go wrong?
 
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Show the equations you solved.
 
For substitution into the second double variable formula:

v2'=(-v+v1')/2

v1'=(2v2'-v)

For turning the v' values back into angles:

h=v2/2g

theta=arccos((l-h)/l) (where l is the length of string)

Anything else? I'm not really sure which equations you meant.
 
Anyone? I tried working through it again, and ended up with the same wrong answer. It's due tonight. :(
 
What did you get for v1' and v2' in terms of v?

ehild
 
jzwiep said:
Then used that value to find v1 and v2 after the collision with:

-v=2v2' - v1'

-v=2v2' + v1'

What do you get for v1' and v2' in terms of v?

ehild
 
Last edited:
ehild said:
-v=2v2' + v1'

What do you get for v1' and v2' in terms of v?

ehild

Thanks, that did it. Just out of curiosity, where did I go wrong deriving the original formula?

mv -2mv = 2mv2' - mv1'
-mv=m(2v2' - v1')
-v=2v2' - v1'

Mass 1 is going in the positive direction first, then negative. Mass 2 is vice-versa.
 
I see, you used the speeds instead of velocities. Go ahead. What are the final results for v1' and v2'?

ehild
 
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