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

Let the mass of Ankit be $M = 40$ kg and the mass of the ball be $m = 4$ kg.

Initially, Ankit and the ball are at rest on the frictionless platform, so the initial momentum of the system is zero.

When Ankit rolls the ball, it travels at a constant speed of $v = 5$ m/s relative to the platform. Let Ankit's velocity after rolling the ball be $V$.

According to the law of conservation of linear momentum:

$M V + m v = 0$

Substituting the given values:

$40 V + 4 \times 5 = 0$

$40 V + 20 = 0$

$V = -\dfrac{20}{40} = -0.5$ m/s

The negative sign indicates that Ankit moves in the direction opposite to the ball. Therefore, Ankit's speed is $0.5$ m/s.