A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90$$^\circ$$ with respect to each other. The mass of unknown particle is:

Let the proton of mass $$m$$ approach along the positive $$x$$-axis with speed $$u$$. The target particle of unknown mass $$M$$ is initially at rest, so its initial momentum is zero.

After the perfectly elastic collision the proton acquires speed $$v_1$$ and the unknown particle acquires speed $$v_2$$. The two velocities are given to be at right angles, so the angle between the vectors $$\mathbf v_1$$ and $$\mathbf v_2$$ is $$90^\circ$$, i.e. $$\mathbf v_1 \cdot \mathbf v_2 = 0$$.

Because the collision is elastic, both linear momentum and kinetic energy are conserved. We now write these two vector equations and then expand them algebraically.

Conservation of momentum (vector form): $$m\,\mathbf u = m\,\mathbf v_1 + M\,\mathbf v_2.$$

Taking the square (dot-product) of both sides and using $$\mathbf v_1 \cdot \mathbf v_2 = 0$$ (since the angle is $$90^\circ$$) we get $$\bigl(mu\bigr)^2 = (m v_1)^2 + (M v_2)^2 + 2\,m\,M\,\mathbf v_1 \cdot \mathbf v_2$$ $$\Rightarrow m^2 u^2 = m^2 v_1^2 + M^2 v_2^2.$$

Conservation of kinetic energy: $$\tfrac12 m u^2 = \tfrac12 m v_1^2 + \tfrac12 M v_2^2.$$

For convenience we multiply the energy equation by 2 to eliminate the fractions: $$m u^2 = m v_1^2 + M v_2^2.$$

Next we multiply this last equation by another factor of $$m$$ so that the left side becomes $$m^2 u^2$$, matching the momentum equation: $$m^2 u^2 = m^2 v_1^2 + m\,M\,v_2^2.$$

We now have two equations with identical left-hand sides:

From momentum: $$m^2 u^2 = m^2 v_1^2 + M^2 v_2^2,$$ From energy : $$m^2 u^2 = m^2 v_1^2 + m\,M\,v_2^2.$$

Since the left sides are equal, their right sides must also be equal: $$m^2 v_1^2 + M^2 v_2^2 = m^2 v_1^2 + m\,M\,v_2^2.$$

Subtracting the common term $$m^2 v_1^2$$ from both sides we obtain $$M^2 v_2^2 = m\,M\,v_2^2.$$

The speed $$v_2$$ cannot be zero (the second particle does move after collision), so we can cancel $$v_2^2$$. We then also divide by the non-zero factor $$M$$, giving

$$M = m.$$

Thus the unknown particle must have the same mass as the proton.

Hence, the correct answer is Option D.