Two friends, Ankit and Bharat, each weighing 40 kg, sit at a...

Let the mass of Ankit and Bharat be $M = 40$ kg, and the mass of the ball be $m = 4$ kg.
The speed of the ball relative to the platform is constant at $v = 5$ m/s. Thus, the momentum of the ball is always $p = mv = 4 \times 5 = 20$ kg m/s when moving towards Bharat, and $-20$ kg m/s when moving towards Ankit.

Each time Bharat catches the ball, he absorbs $20$ kg m/s of momentum. When he rolls it back, he imparts $-20$ kg m/s to the ball, gaining another $20$ kg m/s. Thus, in each catch-and-throw cycle, Bharat's momentum increases by $40$ kg m/s.

After $(n-1)$ such cycles, Bharat's momentum is $40(n-1)$ kg m/s. Since his mass is $40$ kg, his velocity is:
$v_B = \dfrac{40(n-1)}{40} = n - 1$ m/s.

For Ankit's $n$-th roll to reach Bharat, the velocity of the ball must be strictly greater than Bharat's velocity:
$v_{ball} > v_B \Rightarrow 5 > n - 1 \Rightarrow n < 6$.

This means Ankit's 6th roll will not reach Bharat. However, we must check if Ankit can receive the ball from Bharat's 5th roll to make this 6th roll.

Ankit's momentum after $n$ rolls is $-40n + 20$. For $n = 5$, his momentum is $-40(5) + 20 = -180$ kg m/s, so his velocity is $-180 / 40 = -4.5$ m/s.
Bharat's 5th roll sends the ball towards Ankit at $-5$ m/s. Since the ball's velocity ($-5$ m/s) is more negative than Ankit's velocity ($-4.5$ m/s), the ball will catch up to Ankit.

Ankit catches the ball and rolls it for the 6th time. After this 6th roll, the ball moves at $5$ m/s, but Bharat is also moving at $v_B = 6 - 1 = 5$ m/s. The ball will never reach Bharat, so the back-and-forth motion stops.

Therefore, the maximum number of times Ankit can roll the ball is 6.

Answer: 6