Before collision

(V₁)_x = 10 cos 60° = 5 ms^–1    ;    (V₁)_y = 10 sin 60° =  ms^–1
After collision
(V₂)_x = V₀(say)    ;    (V₂)_y = 0

${\left({\mathrm{V}}_1\right)}_{\mathrm{y}}=5\sqrt{3}{ \mathrm{ms}}^{-1}; {\left({\mathrm{V}}_1\right)}_{\mathrm{x}}={\mathrm{V}}_1\text{ (say) }$

conservation of $\overrightarrow{\mathrm{p}}$ in x-direction will give
    1 × 5 = 1 × V₀    + 1 × V₁
    V₀ + V₁ = 5         ......(1)
Writting e-equation 

$\frac{1}{2}=\frac{{\mathrm{V}}_0-{\mathrm{V}}_1}{5}$

${\mathrm{V}}_0-{\mathrm{V}}_1=\frac{5}{2}$         .....(2)

$\therefore {\mathrm{V}}_1=\frac{5}{4}{ \mathrm{ms}}^{-1}$

∴   Speed of ball 1 after collision

$=\sqrt{(5\sqrt{3}{)}^2+(5/4{)}^2}=5\sqrt{3+\frac{1}{16}}=\frac{5\times 7}{4}=8.75 \mathrm{m}/\mathrm{s}$