22) 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 $m$.

The fractional energy transfer is the absolute value of the final kinetic energy of $M$ divided by the initial kinetic energy of $m$.

If the collision is perfectly elastic, under what condition will the fractional energy transfer between the two objects be a maximum?

A) $\frac{m}{M} \ll 1$

B) $0.5 < \frac{m}{M} < 1$

C) $m = M$

D) $1 < \frac{m}{M} < 2$

E) $\frac{m}{M} \gg 1$
1. **Conservation of Momentum:**
   The initial momentum of the system is given by the momentum of the particle of mass \( m \) moving at speed \( v_0 \):
   \[
   p_{\text{initial}} = mv_0
   \]
   After the collision, let \( v_1 \) be the final velocity of the particle of mass \( m \) and \( v_2 \) be the final velocity of the particle of mass \( M \). By conservation of momentum:
   \[
   mv_0 = mv_1 + Mv_2
   \]

2. **Conservation of Kinetic Energy:**
   Since the collision is perfectly elastic, kinetic energy is also conserved. The initial kinetic energy is:
   \[
   E_{\text{initial}} = \frac{1}{2}mv_0^2
   \]
   The final kinetic energy is:
   \[
   E_{\text{final}} = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2
   \]
   By conservation of kinetic energy:
   \[
   \frac{1}{2}mv_0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}Mv_2^2
   \]

3. **Solving for \( v_1 \) and \( v_2 \):**
   From the conservation of momentum equation:
   \[
   v_1 = \frac{mv_0 - Mv_2}{m}
   \]
   Substituting \( v_1 \) into the conservation of kinetic energy equation:
   \[
   mv_0^2 = m\left(\frac{mv_0 - Mv_2}{m}\right)^2 + Mv_2^2
   \]
   Simplifying:
   \[
   mv_0^2 = \frac{(mv_0 - Mv_2)^2}{m} + Mv_2^2
   \]
   \[
   mv_0^2 = \frac{m^2v_0^2 + M^2v_2^2 - 2mMv_0v_2}{m} + Mv_2^2
   \]
   \[
   mv_0^2 = mv_0^2 + \frac{M^2v_2^2}{m} + Mv_2^2 - 2Mv_0v_2
   \]
   \[
   0 = \frac{M^2v_2^2}{m} + Mv_2^2 - 2Mv_0v_2
   \]
   Factoring out \( v_2 \):
   \[
   v_2 \left( \frac{M^2}{m} + M - 2Mv_0 \right) = 0
   \]
   Since \( v_2 \neq 0 \):
   \[
   \frac{M^2}{m} + M = 2Mv_0
   \]
   Solving for \( v_2 \):
   \[
   v_2 = \frac{2mv_0}{M + m}
   \]

4. **Fractional Energy Transfer:**
   The fractional energy transfer is the ratio of the final kinetic energy of \( M \) to the initial kinetic energy of \( m \):
   \[
   \frac{E_2}{E_0} = \frac{\frac{1}{2}Mv_2^2}{\frac{1}{2}mv_0^2} = \frac{Mv_2^2}{mv_0^2}
   \]
   Substituting \( v_2 \):
   \[
   \frac{E_2}{E_0} = \frac{M \left( \frac{2mv_0}{M + m} \right)^2}{mv_0^2}
   \]
   \[
   \frac{E_2}{E_0} = \frac{M \cdot \frac{4m^2v_0^2}{(M + m)^2}}{mv_0^2}
   \]
   \[
   \frac{E_2}{E_0} = \frac{4mM}{(M + m)^2}
   \]

5. **Maximizing the Fractional Energy Transfer:**
   To maximize \(\frac{4mM}{(M + m)^2}\), we need to find the condition under which this expression is maximized. Consider the function:
   \[
   f(x) = \frac{4x}{(1 + x)^2}
   \]
   where \( x = \frac{m}{M} \). Taking the derivative and setting it to zero to find the critical points:
   \[
   f'(x) = \frac{4(1 + x)^2 - 4x \cdot 2(1 + x)}{(1 + x)^4} = \frac{4(1 + x - 2x)}{(1 + x)^3} = \frac{4(1 - x)}{(1 + x)^3}
   \]
   Setting \( f'(x) = 0 \):
   \[
   1 - x = 0 \implies x = 1 \implies m = M
   \]
   Therefore, the fractional energy transfer is maximized when \( m = M \).

The final answer is \( \boxed{m = M} \)