Three objects \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are kept in a straight line on a frictionless horizontal surface. These have masses $\mathrm{m}, 2 \mathrm{~m}\( and \)\mathrm{m}\( respectively. The object \)\mathrm{A}$ moves towards \(\mathrm{B}\) with a speed \(9 \mathrm{~m} / \mathrm{s}\) and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with \(\mathrm{C}\). All motion occur on the same straight line. Find final speed of the object \(\mathrm{C}\) (A) \(3 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(1 \mathrm{~m} / \mathrm{s}\)

We'll start by considering the elastic collision between object A with mass m and object B with mass 2m. The initial velocity of A is 9 m/s, and B is initially stationary. We can use the conservation of momentum and kinetic energy to find the final velocities of A and B after the collision. The momentum conservation equation is given by: \(m_1v_1 + m_2v_2 = m_1v_3 + m_2v_4\) Where: - \(m_1 = m\), mass of object A - \(v_1 = 9 m/s\), initial velocity of object A - \(m_2 = 2m\), mass of object B - \(v_2 = 0 m/s\), initial velocity of object B (stationary) - \(v_3\), final velocity of object A - \(v_4\), final velocity of object B Since the collision is elastic, kinetic energy is conserved: \(0.5m_1v_1^2 + 0.5m_2v_2^2 = 0.5m_1v_3^2 + 0.5m_2v_4^2\) Substitute in the mass and initial velocities values into the conservation of momentum and kinetic energy equations.

With the momentum conservation equation and conservation of kinetic energy equation: \(mv_1 + (2m)v_2 = mv_3 + (2m)v_4\) \(0.5mv_1^2 = 0.5mv_3^2 + 0.5(2m)v_4^2\) Substitute initial velocities: \(m(9) = mv_3 + 2m(v_4)\) \(0.5m(9^2) = 0.5m(v_3^2) + m(v_4^2)\) We can cancel `m` in both equations: \(9 = v_3 + 2v_4\) \(0.5(81) = 0.5(v_3^2) + v_4^2\) Solve the first equation for \(v_3\), and substitute it in the second equation: \(v_3 = 9 - 2v_4\) \(0.5(81) = 0.5((9 - 2v_4)^2) + v_4^2\) Solve for \(v_4\): \(v_4 = 3 m/s\) Find \(v_3\): \(v_3 = 9 - 2v_4 = 9 - 6 = 3 m/s\) After the elastic collision, object A has a final velocity of 3 m/s, and object B has a final velocity of 3 m/s.

Now we examine the completely inelastic collision between object B with mass 2m and object C with mass m. B has initial velocity 3 m/s, while C is initially stationary. As before, we'll use conservation of momentum to find the final velocity. The momentum conservation equation is: \(m_1v_1 + m_2v_2 = (m_1+m_2)v_f\) Where: - \(m_1 = 2m\), mass of object B - \(v_1 = 3 m/s\), initial velocity of object B - \(m_2 = m\), mass of object C - \(v_2 = 0 m/s\), initial velocity of object C (stationary) - \(v_f\), final velocity of both objects B and C after inelastic collision

With the momentum conservation equation: \((2m)v_1 + m(v_2) = (3m)v_f\) Substitute initial velocities: \((2m)(3) = (3m)v_f\) We can cancel `m`: \(6 = 3v_f\) Solve for \(v_f\): \(v_f = 2 m/s\) The final velocity of object C is 2 m/s. Therefore, the correct answer is: (D) \(2 \mathrm{~m} / \mathrm{s}\)