after first collision MvA=MvA′+Mv′B vA=vA′+vB′ ⇒ vA′+vB′=9 .......(i) We know that e=
vB′−vA′
vB−vA
−1=
VB′−vA′
0−vA
[ for elastic collision e=−1] vA=vB′−vA′ 9=vB′−vA′ .........(ii) From Eqs. (i) and (ii), we get 2vB′=18 vB′=9m∕s After second collision MvB′=(2M+M)vC′ MvB′=3MvC′ vB′=3vC′ vC′=