Ball on String Elastic Collision

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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. :(
 
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.