A body of mass m undergoes an elastic collision with another...

Let the initial velocity of the first body be $\vec{u}$ and the second body be at rest.

Let their final velocities be $\vec{v}_1$ and $\vec{v}_2$.

By conservation of linear momentum:

$m\vec{u} = m\vec{v}_1 + m\vec{v}_2$

$\vec{u} = \vec{v}_1 + \vec{v}_2$

Taking the dot product of this equation with itself:

$u^2 = v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2$

Since the collision is elastic, kinetic energy is conserved:

$\dfrac{1}{2}mu^2 = \dfrac{1}{2}mv_1^2 + \dfrac{1}{2}mv_2^2$

$u^2 = v_1^2 + v_2^2$

Comparing the two equations for $u^2$:

$v_1^2 + v_2^2 = v_1^2 + v_2^2 + 2\vec{v}_1 \cdot \vec{v}_2$

$2\vec{v}_1 \cdot \vec{v}_2 = 0$

$\vec{v}_1 \cdot \vec{v}_2 = 0$

Since the collision is not head-on, neither $v_1$ nor $v_2$ is zero. Thus, the angle $\theta$ between $\vec{v}_1$ and $\vec{v}_2$ must satisfy $\cos \theta = 0$.

\theta = 90^{}^{{}^{\circ}}