 Solution

With $M\gg m$ and $Em\gg M^2$, one also has $E\gg M^2/m\gg M$. The denominator in a is then dominated by $2Em$, and the numerator by $E{M}^2$:This residual total energy is still much larger than the incident particle's rest energy $M$, but is a small fraction of its incident energy $E$. The target acquires most of the available energy in this high-energy, heavy-projectile limit.