A moving particle of mass m collides head-on and elastically with a...

Let the initial velocity of the moving particle be $u$.

For a perfectly elastic head-on collision with a stationary target, the fraction of kinetic energy transferred from the projectile to the target is given by:
$f = \dfrac{4mM}{(m + M)^2}$

Given that $64\%$ of the initial kinetic energy is transferred:
$\dfrac{4mM}{(m + M)^2} = \dfrac{64}{100} = \dfrac{16}{25}$

Dividing numerator and denominator by $m^2$ and letting the ratio $\dfrac{M}{m} = x$:
$\dfrac{4x}{(1 + x)^2} = \dfrac{16}{25}$

$\Rightarrow \dfrac{x}{1 + 2x + x^2} = \dfrac{4}{25}$

Cross-multiplying gives:
$25x = 4 + 8x + 4x^2$

$4x^2 - 17x + 4 = 0$

Solving this quadratic equation:
$4x^2 - 16x - x + 4 = 0$
$4x(x - 4) - 1(x - 4) = 0$
$(4x - 1)(x - 4) = 0$

Thus, $x = 4$ or $x = \dfrac{1}{4}$.

Since it is given that $M > m$, the ratio $\dfrac{M}{m}$ must be greater than $1$. Therefore, the correct ratio is $4$.

Answer: $4$