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

Let $m_A = 40$ kg be the mass of Ankit, $m_B = 40$ kg be the mass of Bharat, and $m = 4$ kg be the mass of the ball.

The speed of the ball relative to the platform is $5$ m/s. Let the direction from Ankit to Bharat be positive.

When Ankit rolls the ball towards Bharat, the ball's velocity is $v_1 = 5$ m/s. By conservation of linear momentum for Ankit and the ball:

$m_A v_{A1} + m v_1 = 0$

$40 v_{A1} + 4(5) = 0 \Rightarrow v_{A1} = -0.5$ m/s

When Bharat catches the ball, applying conservation of momentum for Bharat and the ball:

$m v_1 = (m_B + m) v_{B1}$

$4(5) = (40 + 4) v_{B1} \Rightarrow v_{B1} = \dfrac{20}{44} = \dfrac{5}{11}$ m/s

When Bharat rolls the ball back to Ankit, the ball's velocity is $v_2 = -5$ m/s. Applying conservation of momentum for Bharat and the ball:

$(m_B + m) v_{B1} = m_B v_{B2} + m v_2$

$44 \left(\dfrac{5}{11}\right) = 40 v_{B2} + 4(-5)$

$20 = 40 v_{B2} - 20 \Rightarrow 40 v_{B2} = 40 \Rightarrow v_{B2} = 1$ m/s

When Ankit catches the ball for the first time, applying conservation of momentum for Ankit and the ball:

$m_A v_{A1} + m v_2 = (m_A + m) v_{A2}$

$40(-0.5) + 4(-5) = (40 + 4) v_{A2}$

$-20 - 20 = 44 v_{A2} \Rightarrow 44 v_{A2} = -40$

$v_{A2} = -\dfrac{40}{44} = -\dfrac{10}{11}$ m/s

The speed of Ankit is the magnitude of his velocity, which is $\dfrac{10}{11}$ m/s.