After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

$$\begin{gathered} \theta =\varphi \\ V=v/2 \end{gathered}$$

We can find the momentum along x and y direction before the collision in terms of mass, velocity and unknown angles. Using the law of conservation of momentum, we can equate the initial momentum to the final momentum to find the angles made by the object’s velocity with each other before the collision.

Formula:

Initial momentum = Final momentum.

Let’s consider that the angle made by 1^st particle with x axis is$\theta$and angle made by the 2^nd particle with x axis is$\varphi$.

The momentum along x and y axis will be conserved. It means momentum before the collision is equal to momentum after the collision along x and y axis.

Therefore we can write, for y axis

$\mathrm{mv}\sin \hspace{0.33em}\mathrm{\theta }-\mathrm{mv}\sin \hspace{0.33em}\mathrm{\varphi }=0\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$ That gives us, $\theta =\varphi \cdot \cdot \cdot \cdot \cdot \left(2\right)$

For x axis, we can write,

$\mathrm{mv}\hspace{0.33em}\cos \hspace{0.33em}\mathrm{\theta }+\mathrm{mv}\hspace{0.33em}\cos \hspace{0.33em}\mathrm{\varphi }=2\mathrm{mV}$

Using equation (2), we have

$2\mathrm{mv}\hspace{0.33em}\cos \hspace{0.33em}\mathrm{\theta }=2\mathrm{mV}$

As per given information,$V=v/2$

Therefore equation (2) will be,

$\cos \theta =1l2$

It gives angle as$\theta =60°$

This angle is the angle made by each object with x axis. Therefore, the total angle between the objects’ velocities is$120°$

Angle between two objects after inelastic collision can be found by using the law of conservation of momentum.