After perfectly inelastic collision between two identical particles moving with same speed in different directions, the speed of the particles become half the initial speed. The angle between the velocities of the two before collision is

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 ${120}^{°}.$