# Collison betwen two balls

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?

## Answers and Replies

radou
Homework Helper
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?

radou
Homework Helper
would it be 36sin56 instead?

I suggest you draw a diagram on this one.

well i want to find the lenght in the y direction and i see 2 ight triangles i could get y from 36cos36 or take 90-36=56 and do 36sin56