A sphere collides with another sphere of identical mass. After collision, the two spheres move. The collision is inelastic. Then the angle between the directions of the two spheres is
A. Different from ${{90}^{\circ }}$
B. ${{90}^{\circ }}$
C. ${{0}^{\circ }}$
D. ${{45}^{\circ }}$

Hint: While the other body is at rest, the other body is in motion. As a result, the first body will have momentum, and the second body will have zero momentum since, as was already mentioned, it is at rest. We can now determine the answer using momentum and energy conservation.

Complete step by step solution:
An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In an elastic collision, both momentum and kinetic energy are conserved.
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_1},{v_2} = \]Final velocity

In order for two bodies to collide exactly elastically, their paths must be at an angle of \[{90^ \circ }\]. When some of the kinetic energy of a colliding object or system is wasted, the collision is said to be inelastic.
\[{m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right){v_f}\]
Where,
\[{m_1},{m_2} = \] Object’s mass
\[{u_1},{u_2} = \]Initial velocity
\[{v_f} = \]Final velocity

As a result, the angle for an inelastic collision should be different from \[{90^ \circ }\]. A sphere with identical mass collides with each other. The two spheres move once they have collided. There is no elastic collision. If the collision is perfectly elastic, the angle between the directions of the two spheres will be \[90\] degrees.

Hence, option A is correct.

Note: We are aware that whenever the body moves, the body should also experience momentum, and vice versa. Mass and velocity are multiplied to create momentum. We now understand that the second sphere has no momentum prior to impact. Following the impact, the second will feel momentum as a result of the movement it receives. Therefore, we can simply answer this question using momentum conservation.