A small particle travelling with a velocity v undergoes an elastic...

Let the angle between the initial velocity of the small particle and the line of impact (normal to the sphere at the point of contact) be $\theta$.

From the geometry of the collision, $\sin\theta = \dfrac{\rho}{r}$ and $\cos\theta = \dfrac{\sqrt{r^2 - \rho^2}}{r}$.

The initial velocity of the small particle can be resolved into two components:
Normal component: $v_n = v \cos\theta$
Tangential component: $v_t = v \sin\theta$

The spherical body is initially at rest. Since the collision is elastic and the masses of the two bodies are equal, they exchange their velocities along the line of impact (normal direction).

Final normal velocity of the small particle $= 0$
Final normal velocity of the spherical body $= v \cos\theta$

The tangential components of the velocities remain unchanged because no force acts in the tangential direction.

Final tangential velocity of the small particle $= v \sin\theta$
Final tangential velocity of the spherical body $= 0$

Therefore, the small particle moves along the tangent with speed $v \sin\theta = v\dfrac{\rho}{r}$.

The spherical body moves along the normal with speed $v \cos\theta = \dfrac{v}{r}\sqrt{r^2 - \rho^2}$.

Since the normal and tangent are perpendicular to each other, the spherical body moves perpendicular to the small particle.