Question

A ball strikes with angle 45 degree to the horizontal. The coefficient of restitution is1/2.then fractional loss in kinetic energy is?

Answer 1

$$\begin{gathered} lettheinitialspeedbeu \\ initialKE=\frac{m{u}^2}{2} \\ ycomponentofthevelocityaftercollision=\frac{u\cos 45}{2}=\frac{u}{2\sqrt{2}} \\ xcomponent=u\cos 45=\frac{u}{\sqrt{2}} \\ velocityaftercollisionv=\sqrt{{\left(\frac{u}{2\sqrt{2}}\right)}^2+{\left(\frac{u}{\sqrt{2}}\right)}^2}=u\sqrt{\frac{1}{8}+\frac{1}{2}}=\sqrt{\frac{5}{8}}u \\ lossinkE=\frac{m{u}^2}{2}-\left(\frac{5}{8}\right)\frac{m{u}^2}{2}=\left(\frac{3}{8}\right)\frac{m{u}^2}{2}=\frac{3}{8}ofinitialKE \\ Regards \end{gathered}$$