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

Since the platform is frictionless, there is no net external force acting on the system consisting of Ankit, Bharat, and the ball in the horizontal direction.

Therefore, the center of mass of the system remains stationary throughout the motion, regardless of the internal movements.

Let the initial position of Ankit be at the origin, $x_A = 0$.

The initial position of Bharat is $x_B = d$.

The ball is initially with Ankit, so its initial position is $x_{ball} = 0$.

The given masses are $m_A = 40$ kg, $m_B = 40$ kg, and $m_{ball} = 4$ kg.

The position of the center of mass of the system is calculated as:

$X_{cm} = \dfrac{m_A x_A + m_B x_B + m_{ball} x_{ball}}{m_A + m_B + m_{ball}}$

Substituting the values:

$X_{cm} = \dfrac{40(0) + 40(d) + 4(0)}{40 + 40 + 4}$

$X_{cm} = \dfrac{40d}{84}$

$X_{cm} = \dfrac{10}{21}d$

Thus, the center of mass of the system is permanently located at a distance of $\dfrac{10}{21}d$ away from the initial position of Ankit towards Bharat.