A sphere collides with another sphere of identical mass. After collision, the two sphere 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) \(45^{\circ}\)

Let's denote the initial velocities of the two spheres as \(\vec{v}_{1i}\) and \(\vec{v}_{2i}\), and their final velocities as \(\vec{v}_{1f}\) and \(\vec{v}_{2f}\), respectively. The mass of both spheres is \(m\). Since the goal of the problem is to find the angle between the directions of the two spheres after the collision, let the angle be denoted by \(\theta\).

The total linear momentum of the system before the collision should be equal to the total linear momentum after the collision. This can be written as: \[ m\vec{v}_{1i} + m\vec{v}_{2i} = m\vec{v}_{1f} + m\vec{v}_{2f} \] We can simplify the equation by dividing both sides of the equation by \(m\): \[ \vec{v}_{1i} + \vec{v}_{2i} = \vec{v}_{1f} + \vec{v}_{2f} \]

Since the collision is inelastic, both spheres stick together after the collision. Therefore, their final velocities are the same. We can denote this common velocity as \(\vec{v}_{f}\). This allows us to rewrite the conservation of momentum equation as: \[ \vec{v}_{1i} + \vec{v}_{2i} = 2\vec{v}_{f} \]

To find the angle between the directions of the two spheres, we will use the dot product between their final velocity vectors. The dot product between \(\vec{v}_{1f}\) and \(\vec{v}_{2f}\) is given by: \[ \vec{v}_{1f} \cdot \vec{v}_{2f} = |v_{1f}||v_{2f}|\cos{\theta} \] Since \(v_{1f} = v_{2f}\), we can rewrite the dot product as: \[ \vec{v}_{f} \cdot \vec{v}_{f} = |v_{f}|^{2}\cos{\theta} \] Since \(\vec{v}_{1f} = \vec{v}_{2f} = \vec{v}_{f}\), the angle between the two final velocity vectors is necessarily zero: \[ \theta = \arccos\frac{\vec{v}_{1f} \cdot \vec{v}_{2f}}{|v_{1f}||v_{2f}|} = \arccos\frac{|v_{f}|^{2}}{|v_{f}|^{2}} = \arccos{1} = 0^{\circ} \] Therefore, the correct answer is: (C) \(0^{\circ}\)