Two bodies of the same mass are moving with the same speed, but in different directions in a plane. They have a completely inelastic collision and move together thereafter with a final speed which is half of their initial velocities of the two bodies (in degree) is ___

Let us denote the mass of each body by $$m$$ and the common magnitude of their initial velocities by $$v$$. Thus, at the very beginning each body possesses a momentum vector of magnitude

$$p_0 = m\,v$$

but the two momentum vectors point in different directions within the same plane. Suppose the angle between these two directions is $$\theta$$ (this is precisely the angle we are required to find).

To combine the two individual momentum vectors into a single “net” or “resultant” initial momentum, we treat the vectors exactly the way we add two sides of a parallelogram. For two vectors of equal magnitude $$p_0$$ separated by an angle $$\theta$$, the magnitude of their resultant, by the law of cosines for vectors, is

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{p_0^{\,2} + p_0^{\,2} + 2\,p_0\,p_0\,\cos\theta}\;=\;\sqrt{2p_0^{\,2}\,(1+\cos\theta)}$$

Inside the square root we notice the standard trigonometric identity $$1 + \cos\theta = 2\cos^{2}\dfrac{\theta}{2}$$. Substituting this, we obtain

$$\lvert\vec P_{\text{initial}}\rvert \;=\;\sqrt{2p_0^{\,2}\,\bigl(2\cos^{2}\tfrac{\theta}{2}\bigr)} \;=\; \sqrt{4p_0^{\,2}\,\cos^{2}\tfrac{\theta}{2}} \;=\; 2p_0\,\cos\dfrac{\theta}{2}.$$

Replacing $$p_0$$ by $$m\,v$$ gives

$$\lvert\vec P_{\text{initial}}\rvert \;=\;2\,m\,v\,\cos\dfrac{\theta}{2}.$$

Now the collision is stated to be completely inelastic, meaning the two bodies stick together and thereafter behave like a single composite body. The mass of this composite body is simply the sum of the individual masses, namely

$$m_{\text{final}} \;=\; m + m \;=\; 2m.$$

The problem also tells us that the common final speed of this combined body is half of the original speed, that is

$$v_{\text{final}} \;=\; \dfrac{v}{2}.$$

Therefore the magnitude of the final momentum is

$$\lvert\vec P_{\text{final}}\rvert \;=\; m_{\text{final}}\,v_{\text{final}} \;=\; 2m \times \dfrac{v}{2} \;=\; m\,v.$$

According to the principle of conservation of linear momentum, the total momentum of an isolated system remains constant, so we can equate the magnitudes of the initial and final momenta:

$$2\,m\,v\,\cos\dfrac{\theta}{2} \;=\; m\,v.$$

Dividing both sides by the common positive factor $$m\,v$$ leaves us with a very simple trigonometric equation:

$$2\,\cos\dfrac{\theta}{2} \;=\; 1.$$

We now isolate the cosine term:

$$\cos\dfrac{\theta}{2} \;=\; \dfrac{1}{2}.$$

The standard value of the cosine function that gives one-half is $$60^\circ$$, so we can write

$$\dfrac{\theta}{2} \;=\; 60^\circ.$$

Multiplying both sides by 2, we finally obtain

$$\theta \;=\; 120^\circ.$$

So, the answer is $$120^\circ$$.