Who's In-volute?

Here, $e = \frac {v_{cylinder}-v_{ball}}{u_{ball}-u_{cylinder}}$

$=\frac {0-v_{ball}}{u_{ball}-0}$

Thus after each collision $v_{ball} = -eu_{ball}$ Thus, after $n$ collisions, $|v| = e^{n}u_0$ .

The speed of the ball is not affected by the string because the string always exerts a force which is normal to the ball's velocity.

It can also be found out that the length of the involute path is $\frac {{l_0}^{2}}{2R} = a$

and the length of the Quarter circle is $\frac {\pi l_0}{2} = b$

For the first collision the ball travels $a+b$ and for all the successive collisions it travels $2(a+b)$ .

Total time for the velocity to be reduced to $\frac {v_0}{n}$ = $\frac {a+b}{v_0} + \frac {2(a+b)}{ev_0} + \frac {2(a+b)}{e^{2}v_0} . . . . \frac {2(a+b)}{e^{n-1}v_0}$

$T = \frac {a+b}{v_0} + \frac {2(a+b)}{ev_0}(\frac {(\frac {1}{e})^{n-1} - 1}{\frac {1}{e} - 1})$

As final velocity has to be $\frac {v_0}{16}$ the ball must collide 8 times $in$ $all$ . Thus $n=8$ .

On substituting the values of $l_0, v_0, R$ , $n$ and $e$ we get $T=107.1s$ which can be rounded off to $\boxed{107s}$