Let $R_M$ and $R_m$ be the constant resistance forces (buoyancy plus drag) on the metal and wood blocks, respectively. This is implied by the constant acceleration $a$. For the combined system (M+m) during time $t$:

$$(M+m)g - (R_M + R_m) = (M+m)a$$

The velocity at time $t$ when the string breaks is $v_t = at$.

For the wood block (m) during the interval $t'$, its velocity changes from $at$ to 0. Its acceleration is $a_m = (0 - at)/t' = -at/t'$. The net force on the wood block is $mg - R_m = ma_m$.

$$R_m = mg - ma_m = mg - m(-at/t') = m(g + at/t')$$

For the metal block (M) after separation, its acceleration is $a_M$. The net force is $Mg - R_M = Ma_M$. From the combined system equation, $R_M = (M+m)(g-a) - R_m$.

$$Ma_M = Mg - R_M = Mg - [(M+m)(g-a) - R_m]$$ $$Ma_M = Mg - (M+m)(g-a) + R_m$$

Substitute the expression for $R_m$:

$$Ma_M = Mg - (M+m)(g-a) + m(g + at/t') = -mg + (M+m)a + mg + mat/t'$$ $$Ma_M = (M+m)a + mat/t'$$ $$a_M = \frac{M+m}{M}a + \frac{mat}{Mt'}$$

The final velocity of the metal block is $v_M = v_t + a_M t'$.

$$v_M = at + \left(\frac{M+m}{M}a + \frac{mat}{Mt'}\right)t' = at + \frac{M+m}{M}at' + \frac{mat}{M}$$ $$v_M = a(t+t') + \frac{m}{M}a(t+t') = \frac{M+m}{M}a(t+t')$$