Question

2 identical billiard balls are in contact on a smooth table. a third identical ball strikes them symmetrically and comes to rest after impact. the coefficient of restitution is_?

Answer 1

coefficient of restitution is = velocity after collision / velocity before collision \[C.O.R=\frac{ V _{b} -V_{a}}{ U _{a} -U _{b}}\] where Vb is final velocity of second object after collision and Va is final velocity of first object before collision , Ub is initial velocity of second object and Ua is initial velocity of first object
Now, Vb = 0 and Ua=0, if we consider two balls placed on a friction less table as a system then here it is a case of perfectly elastic collision and Va=Ub , therefore C.O.R=1 , the total momentum of ball that is colliding with the system of two balls is transferred to them and there is no losses also velocity of a single ball is = velocity of balls after collision , hence coefficient of restitution= 1

Answer 2

\[ \bf\text{Cruciall to the whole question is the meaning , given here, to the words} \\ \ \\ \color{red}{\text{"strikes them symmetrically"}}\] is it this configuration ?

Answer 3

If so, the whole computation may be 2-D. Otherwise ghazi is right