I need some direction in solving this simple pulley problem.
Two masses of $m$ and $2m$ kg are connected by a light inextensible string passing over a smooth pulley. Find the acceleration of the system. If the mass $2m$ kg hits the ground (without rebounding) after the masses have been moving 3 seconds, find how much time elapses from the instant this happens until the system is instantaneously at rest with the string taught.
Finding the acceleration is straight forward using Newton's Second Law $F = ma$. The acceleration is $a=\dfrac{g}{3} m/s^2$. This is the book's answer.
My answer to the second part is $2\,s$. However, the books answer is $3\,s$.
My logic is thus: the velocity of the $m$ mass after $3\,s$ is
$$ v = u + at $$
$$ v = 0 + \frac{g}{3} \cdot 3 $$
$$ v = g $$
The $m$ kg mass moves upwards with an initial velocity of $g\,m/s$. It temporarily comes to rest at its highest point when $v = 0$.
$$ 0 = g - gt $$
$$ t = 1 $$
Total time is $2 \,s$, $1 \,s$ up and $1 \,s$ down. At which the string is taught instantaneously.
I seem to doing something wrong...
Update on this problem.
Thanks to ABC the 'second' part of this problem (finding this illusive extra second) involves Impulsive String Tension. (https://mathspanda.com/A2FM/Lessons/Impulsive_tension_in_strings_LESSON.pdf). As the mass $m$ returns to its position when the $2m$ hits the ground, impulsive string tension is now involved.
Using the conservation of linear momentum $mg = (m + 2m)v$, where $v$ is the velocity at which $m$ and $2m$ will move and $g$ is the velocity of $m$. When the taut string starts moving both $m$ and $2m$ have an initial velocity of $\dfrac{g}{3}$. The acceleration of the system is $\dfrac{-g}{3}$ (I found this using Newton's 2nd Law - just the reverse of the acceleration initially found). Then, using $v = u + at$ for the $2m$ mass:
$$ v = u + at $$ The $2m$ mass will come to instantaneous rest when $v = 0$.
$$ 0 = \frac{g}{3} - \frac{g}{3} t $$
$$ t = 1 $$
$ \therefore \Sigma t = 2 + 1 = 3\,sec$.
Thanks for all your help.
The other possibility is that the answer in the book is wrong. I get the same answer using the fact that the string is taught when the object of mass $m$ is at the same position as when $2m$ stops. $$0=\Delta y=vt-\frac12gt^2\\t=\frac{2v}g=2s$$