Along the x direction,

Initial momentum,

${\mathrm{P}}_1=2\mathrm{mvcos}\left(\mathrm{\theta }/2\right)$

final momentum,

${\mathrm{P}}_2=2\mathrm{mV}=2\mathrm{m}\times \frac{\mathrm{v}}{2}=\mathrm{mv}$

By conservation of momentum along x direction,

$$\begin{gathered} 2\mathrm{mvcos\theta }/2=\mathrm{mv} \\ \Rightarrow \mathrm{cos\theta }/2=\frac{1}{2} \\ \Rightarrow \mathrm{\theta }/2={60}^{°} \end{gathered}$$

Before a collision, there is an angle between the two velocities of

$\mathrm{\theta }={120}^{°}.$

Hence the correct answer is ${120}^{°}.$