Question

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 $\alpha ={45}^{\circ }$. Assuming the balls to be smooth, the fraction $\eta$ of the kinetic energy of the striking ball that turned into potential energy at the moment of the maximum deformation .

Answer 1

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 struck 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 $u\sin \alpha =v_{1y}$, $v_{2y}=0$

$mu\cos \alpha =m{v}_{1x}+m{v}_{2x}$

The impulsive force of mutual interaction satisfies

$\frac{d}{dt}\left(v_{1x}\right)=\frac{F}{m}=-\frac{d}{dt}\left(v_{2x}\right)$

($F$ 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

$D=\frac{1}{2}m{u}^2-\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_2^2$

$=\frac{1}{2}m{u}^2{\cos }^2\alpha -\frac{1}{2}m{v_{1x}}^2-\frac{1}{2}m{v_{2x}}^2$

$=\frac{1}{2m}[m^2{u}^2{\cos }^2\alpha -m^2{v_{1x}}^2-{(mu\cos \alpha -m{v}_{1x})}^2]$

$=\frac{1}{2m}[2m^{2}u\cos \alpha v_{1x}-2m^2{v_{1x}}^2]=m(v_{1x}u\cos \alpha -{v_{1x}}^2)$

$=m[\frac{u^2{\cos }^2\alpha }{4}-{(\frac{u\cos \alpha }{2}-v_{1x})}^2]$

We see that $D$ is maximum when

$\frac{u\cos \alpha }{2}=v_{1x}$

and $D_{max}=\frac{m{u}^2{\cos }^2\alpha }{4}$

Then $\eta =\frac{D_{max}}{\frac{1}{2}m{u}^2}=\frac{1}{2}{\cos }^2\alpha =\frac{1}{4}$

On substituting $\alpha ={45}^{\circ }$