21) A particle of mass $m$ moving at speed $v_0$ collides with a particle of mass $M$ which is originally at rest. The fractional momentum transfer $f$ is the absolute value of the final momentum of $M$ divided by the initial momentum of $m$.

If the collision is perfectly $elastic$, what is the maximum possible fractional momentum transfer, $f_{max}$?

A) $0 < f_{max} < \frac{1}{2}$

B) $f_{max} = \frac{1}{2}$

C) $\frac{1}{2} < f_{max} < \frac{3}{2}$

D) $f_{max} = 2$

E) $f_{max} \ge 3$
1. **Conservation of Momentum:**
   The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. For the given problem, we have:
   \[
   mv_0 = mv_0' + Mv_1
   \]
   where \(v_0\) is the initial velocity of the particle with mass \(m\), \(v_0'\) is the final velocity of the particle with mass \(m\), and \(v_1\) is the final velocity of the particle with mass \(M\).

2. **Conservation of Kinetic Energy:**
   Since the collision is perfectly elastic, the total kinetic energy before and after the collision is conserved. Therefore:
   \[
   \frac{1}{2}mv_0^2 = \frac{1}{2}mv_0'^2 + \frac{1}{2}Mv_1^2
   \]

3. **Relative Velocity in Elastic Collision:**
   For elastic collisions, the relative velocity of approach before the collision is equal to the relative velocity of separation after the collision:
   \[
   v_0 - 0 = v_1 - v_0'
   \]
   This simplifies to:
   \[
   v_0 = v_1 - v_0'
   \]
   Therefore:
   \[
   v_0' = v_1 - v_0
   \]

4. **Substitute \(v_0'\) into the Momentum Equation:**
   Substitute \(v_0' = v_1 - v_0\) into the momentum conservation equation:
   \[
   mv_0 = m(v_1 - v_0) + Mv_1
   \]
   Simplify and solve for \(v_1\):
   \[
   mv_0 = mv_1 - mv_0 + Mv_1
   \]
   \[
   mv_0 + mv_0 = mv_1 + Mv_1
   \]
   \[
   2mv_0 = (m + M)v_1
   \]
   \[
   v_1 = \frac{2mv_0}{m + M}
   \]

5. **Calculate the Fractional Momentum Transfer (FMT):**
   The fractional momentum transfer \(f\) is defined as the absolute value of the final momentum of \(M\) divided by the initial momentum of \(m\):
   \[
   f = \left| \frac{Mv_1}{mv_0} \right|
   \]
   Substitute \(v_1 = \frac{2mv_0}{m + M}\) into the equation:
   \[
   f = \left| \frac{M \cdot \frac{2mv_0}{m + M}}{mv_0} \right|
   \]
   Simplify:
   \[
   f = \left| \frac{2M}{m + M} \right|
   \]

6. **Determine the Maximum Value of \(f\):**
   To find the maximum value of \(f\), consider the behavior of the function \(\frac{2M}{m + M}\). As \(m\) approaches 0, the expression \(\frac{2M}{m + M}\) approaches 2. Therefore, the maximum possible value of \(f\) is:
   \[
   f_{max} = 2
   \]

The final answer is \( \boxed{ 2 } \)