What is the maximum height reached by the other ball after collision?

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    Balls Collison
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The discussion revolves around the collision of two identical steel balls, where one ball is pulled back and released, colliding elastically with the other. The initial calculations involve determining the velocity of the first ball just before impact using energy conservation principles, leading to a velocity of approximately 23.89 m/s. The momentum conservation equation is then applied to find the final velocity of the second ball after the collision, suggesting it may reach the same height as the first ball. There is a debate about the correct trigonometric function to use for calculating the height, with suggestions to consider both cosine and sine functions for the angle involved. The conversation emphasizes the importance of visualizing the problem with a diagram to clarify the calculations.
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Two identical steel balls, each of mass 1.4 kg, are suspended from strings of length 36 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle q = 36° with the vertical and let it go. It collides elastically with the other ball. How high will the other ball rise?

ok so what i am using is pi=pf and ei=ef

first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

then v=23.89

so pi=pf of the system so m1v1+m2v2=m1v1f+m2v2f

which is (1.4)(23.89)+0=0+1.4v2f

so v2f=23.89 and if that is true, wouldn't ball 2 go as high as ball 1?
 
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Rasine said:
first i calcuate the velocity of of th first ball when it is just about to hit the second which is
mgh=1/2mv^2
(1.4)(9.8)36cos36=1/2(1.4)v^2

Reconsider the red step.
 
would it be 36sin56 instead?
 
Rasine said:
would it be 36sin56 instead?

I suggest you draw a diagram on this one.
 
well i want to find the length in the y direction and i see 2 ight triangles i could get y from 36cos36 or take 90-36=56 and do 36sin56
 
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