A ball moving translationally collides elastically with another, stationary, ball of the same mass. At the moment of impact the angle between the straight line passing through the centres of the balls and the direction of the initial motion of the striking ball is equal to `alpha=45^@`. Assuming the balls to be smooth, find the fraction `eta` of the kinetic energy of the striking ball that turned into potential energy at the moment of the maximum deformation.

Explanation

If `(v_(1x), v_(1y))` are the instantaneous velocity components of the incident ball and `(v_(2x), v_(2y))` are the velocity components of the stuck ball at the same moment, then since there are no external impulsive forces (i.e. other than the mutual interaction of the balls)
We have `u sin alpha=v_(1y)`, `v_(2y)=0`
`m u cos alpha=m v_(1x)+m v_(2x)`
The impulsive force of mutual interaction satisfies
`(d)/(dx)(v_(1x))=F/m=-(d)/(dt)(v_(2x))`
(F is along the x axis as the balls are smooth. Thus Y component of momentum is not transferred). Since loss of K.E. is stored as deformation energy D, we have
`D=1/2m u^2-1/2m v_1^2-1/2m v_2^2`
`=1/2m u^2cos^2alpha-1/2mv_(1x)^2-1/2mv_(2x)^2`
`=(1)/(2m)[m^2u^2cos^2alpha-m^2v_(1x)^2-(m u cos alpha-mv_(1x))^2]`
`=(1)/(2m)[2m^2u cos alpha v_(1x)-2m^2v_(1x)^2]=m(v_(1x)u cos alpha-v_(1x)^2)`
`=m[(u^2cos^2alpha)/(4)-((ucosalpha)/(2)-v_(1x))^2]`
We see that D is maximum when
`(ucos alpha)/(2)=v_(1x)`
and `D_(max)=(m u^2cos^2alpha)/(4)`
Then `eta=(D_(max))/(1/2m u^2)=1/2cos^2alpha=1/4`
On substituting `alpha=45^@`