Three balls A, B and C of masses 2 kg, 4 kg and 8 kg, respectively, move along the same straight line and in the same direction, with velocities 4 m/s, 1 m/s and 3/4 m/s. The velocities of ball A and ball B after collision are (e = 1)

In this case there will be two collisions, 

first between balls A and B and second is between B and C

For first collision ${\mathrm{V}}_{\mathrm{A}}=\frac{\left({\mathrm{m}}_{\mathrm{A}}-{\mathrm{em}}_{\mathrm{B}}\right){\mathrm{u}}_{\mathrm{A}}+(1+\mathrm{e}){\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}$

${\mathrm{V}}_{\mathrm{A}}=\frac{(2-1\times 4)4+(1+1)4\times 1}{2+4}=0 \mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{B}}^{\text{'}}=\frac{\left({\mathrm{m}}_{\mathrm{B}}-{\mathrm{em}}_{\mathrm{A}}\right){\mathrm{u}}_{\mathrm{B}}+(1+\mathrm{e}){\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}_{\mathrm{A}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}$

${\mathrm{V}}_{\mathrm{B}}^{\text{'}}=\frac{(4-1\times 2)1+(1+1)2\times 4}{2+4}=3 \mathrm{m}/\mathrm{s}$

For second collision

${\mathrm{V}}_{\mathrm{B}}=\frac{\left({\mathrm{m}}_{\mathrm{B}}-{\mathrm{em}}_{\mathrm{c}}\right){\mathrm{V}}_{\mathrm{B}}^{\text{'}}+(1+\mathrm{e}){\mathrm{m}}_{\mathrm{c}}{\mathrm{u}}_{\mathrm{c}}}{{\mathrm{m}}_{\mathrm{c}}+{\mathrm{m}}_{\mathrm{B}}}$

${\mathrm{V}}_{\mathrm{B}}=\frac{(4-1\times 8)3+(1+1)8\times \frac{3}{4}}{8+4}=0 \mathrm{m}/\mathrm{s}$