91Ó°ÊÓ

Calculate the initial momentum of the system. Since there is no external force acting on the system, the overall linear momentum of the system will be conserved. Initially, 3 kg block is moving with \(2\,\text{m/s}\) towards the left (positive direction), and 6 kg block is at rest. Total initial momentum of the system, P_initial\(=m_1v_1 + m_2v_2 =3(2) + 6(0) = 6\,\text{kg m/s}\).

Using the total initial momentum, we can determine the velocity of the center of mass, V_CM. Given that the linear momentum is conserved, the velocity of the center of mass is a constant throughout the motion. V_CM\(=\frac{\text{total initial momentum}}{\text{total mass}} = \frac{6}{3+6} =\frac{2}{3}\,\text{m/s}\).

(A) When the velocity of \(3\,\text{kg}\) block is \(\frac{2}{3}\,\text{m/s}\). (B) When the speed of \(3\,\text{kg}\) block is \(\frac{2}{3}\,\text{m/s}\). (C) When the speed of \(3\,\text{kg}\) block is minimum. This occurs when the springs' deformation is at its maximum. (D) When the velocity of \(6\,\text{kg}\) block is maximum. This also occurs when the springs' deformation is at its maximum.

1. Velocity of the center of mass is \(\frac{2}{3}\,\text{m/s}\). This is a constant throughout the motion, so it is true for all the cases in Column-I. Therefore, (A1), (B1), (C1), and (D1) are correct matches. 2. Deformation of the spring is zero. This occurs at the initial state, but none of the cases in Column-I indicate this situation. 3. Deformation of the spring is maximum. This is true for (C) and (D) as discussed in Step 3. Therefore, (C3) and (D3) are correct matches. 4. Both the blocks are at rest with respect to each other. This condition is met at the instant when the deformation of the spring is maximum since the velocity of both blocks is equal (either positive or negative). As mentioned earlier, this occurs for cases (C) and (D). Therefore, (C4) and (D4) are correct matches. So, the final column matches are: (A1), (B1), (C1), (D1), (C3), (D3), (C4), and (D4).