1. Set up the collision

A proton of mass $m$ moves initially with speed $u$ and strikes a particle of unknown mass $M$ at rest.

After elastic collision:

• proton moves with speed $v_1$
• unknown particle moves with speed $v_2$
• the angle between their final velocities is $90^\circ$

We need to find $M$.

2. Use momentum conservation

Initial momentum is only along the initial direction: $\vec p_i = m u\,\hat{i}$

Final momentum: $\vec p_f = m\vec v_1 + M\vec v_2$

Since the two final velocities are perpendicular, we can use the magnitude relation: $|m\vec v_1 + M\vec v_2|^2 = (mu)^2$

Because $\vec v_1 \perp \vec v_2$, $m^2 v_1^2 + M^2 v_2^2 = m^2 u^2 \qquad (1)$

3. Use kinetic energy conservation

Since the collision is elastic, $\frac12 m u^2 = \frac12 m v_1^2 + \frac12 M v_2^2$

So, $m u^2 = m v_1^2 + M v_2^2 \qquad (2)$

4. Compare equations (1) and (2)

From (1): $m^2 u^2 = m^2 v_1^2 + M^2 v_2^2$

Multiply (2) by $m$: $m^2 u^2 = m^2 v_1^2 + mM v_2^2$

Comparing these two expressions: $M^2 v_2^2 = mM v_2^2$

Assuming $v_2 \neq 0$ and $M \neq 0$, $M = m$

5. Check options

Thus the unknown mass is: $M = m$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.