Two friends, Ankit and Bharat, each weighing 40 kg, sit at a separation of d on a frictionless platform. Ankit rolls a 4 kg ball along the platform towards Bharat, who catches it. Bharat subsequently rolls the ball back to Ankit, who catches it. This back-and-forth motion of the ball between the two friends continues. The ball travels at a constant speed of 5 m/s relative to the platform. Find Ankit's speed after he

Step-by-step solution

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 v₁ = 5 m/s. By conservation of linear momentum for Ankit and the ball: m_A v_ A1 + m v₁ = 0 40 v_ A1 + 4(5) = 0 v_ A1 = -0.5 m/s When Bharat catches the ball, applying conservation of momentum for Bharat and the ball: m v₁ = (m_B + m)