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

Let the mass of Ankit and Bharat be $M = 40$ kg each, and the mass of the ball be $m = 4$ kg.
The speed of the ball relative to the platform is $v = 5$ m/s.
Since the platform is frictionless, the net external force on the system (Ankit + Bharat + ball) in the horizontal direction is zero. Thus, the total momentum of the system remains conserved at zero.

Let us analyze the change in momentum of Bharat during one round trip:
When Bharat catches the ball, he receives a momentum of $mv = 4 \times 5 = 20$ kg m/s.
When Bharat throws the ball back to Ankit at a speed of $5$ m/s relative to the platform, he imparts a momentum of $-20$ kg m/s to the ball. By conservation of momentum, Bharat gains an additional momentum of $+20$ kg m/s.
Therefore, the total momentum gained by Bharat in one round trip is $20 + 20 = 40$ kg m/s.

After 5 round trips, Bharat has caught and thrown the ball 5 times.
Total momentum of Bharat, $P_B = 5 \times 40 = 200$ kg m/s.
Speed of Bharat, $v_B = \dfrac{P_B}{M} = \dfrac{200}{40} = 5$ m/s.

After 5 round trips, the ball is held by Ankit. Let the velocity of Ankit and the ball together be $v_A$.
Using the conservation of total momentum for the system:
$P_{Ankit+ball} + P_B = 0$
$(M + m)v_A + 200 = 0$
$(40 + 4)v_A = -200$
$44v_A = -200 \Rightarrow v_A = -\dfrac{200}{44} = -\dfrac{50}{11}$ m/s.

The speed of Ankit is $\dfrac{50}{11}$ m/s.

The respective speeds of Ankit and Bharat are $\dfrac{50}{11}$ m/s and $5$ m/s.

Answer: $\dfrac{50}{11}$ m/s, 5 m/s