We know the following

${\mathrm{m}}_{\mathrm{A}}=10 \mathrm{kg}$ .

${\mathrm{V}}_{\mathrm{Ai}}=2 \mathrm{m}/\mathrm{s}$

${\mathrm{m}}_{\mathrm{B}}=2 \mathrm{kg}$

${\mathrm{V}}_{\mathrm{Bi}}=-4 \mathrm{m}/\mathrm{s}$ . The negative sign is because the velocity is in the negative direction.

Now we need to find $V_{Af}\text{and}{V}_{Bf}$ . Use the equations from above. Let’s start with $V_{Af}$ .

${\mathrm{V}}_{\mathrm{Af}}=\frac{{\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}}{\mathrm{V}}_{\mathrm{Af}}+\frac{({\mathrm{m}}_{\mathrm{B}}-{\mathrm{m}}_{\mathrm{A}})}{({\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}})}{\mathrm{V}}_{\mathrm{Bi}}$

Plug in our known values.

${\mathrm{V}}_{\mathrm{Af}}=\frac{(10-2)\mathrm{kg}}{(10+2)\mathrm{kg}}.(2 \mathrm{m}/\mathrm{s})+\frac{2\left(2\mathrm{kg}\right)}{(10+2\mathrm{kg})}.(-4 \mathrm{m}/\mathrm{s})$

${\mathrm{V}}_{\mathrm{Af}}=\frac{8}{12}.(2 \mathrm{m}/\mathrm{s})+\frac{4}{12}.(-4 \mathrm{m}/\mathrm{s})$

${\mathrm{V}}_{\mathrm{Af}}=\frac{16}{12} \mathrm{m}/\mathrm{s}+\frac{-16}{12} \mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{Af}}=0 \mathrm{m}/\mathrm{s}$

The final velocity of the larger mass is zero. The collision completely stopped this mass.

Now for $V_{Bf}$

${\mathrm{V}}_{\mathrm{Af}}=\frac{2{\mathrm{m}}_{\mathrm{A}}}{({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}})}{\mathrm{V}}_{\mathrm{Af}}+\frac{({\mathrm{m}}_{\mathrm{B}}-{\mathrm{m}}_{\mathrm{A}})}{({\mathrm{m}}_{\mathrm{A}}-{\mathrm{m}}_{\mathrm{B}})}{\mathrm{V}}_{\mathrm{Bi}}$

Plug in our known values

${\mathrm{V}}_{\mathrm{Bf}}=\frac{2\left(10\mathrm{kg}\right)}{(10+2)\mathrm{kg}}.2 \mathrm{m}/\mathrm{s}+\frac{(2-10)\mathrm{kg}}{(10+2\mathrm{kg})}.-4 \mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{Bf}}=\frac{40}{12} \mathrm{m}/\mathrm{s}+\frac{32}{12} \mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{Bf}}=\frac{72 \mathrm{kg}}{12 \mathrm{kg}} \mathrm{m}/\mathrm{s}$

${\mathrm{V}}_{\mathrm{Bf}}=6 \mathrm{m}/\mathrm{s}$