As per the given figure, two blocks each of mass 250 g are connected to a spring of spring constant 2 Nm^–1. If both are given velocity v in opposite directions, then maximum elongation of the spring is:

Given: Two blocks, each with a mass of 250 g, are connected to a spring with a spring constant of 2 N/m. Both blocks are given velocity "v" in opposite directions. The task is to calculate the maximum elongation of the spring.

Solution:

When both blocks are given velocities in opposite directions, the system undergoes a situation where the kinetic energy of the blocks is converted into potential energy stored in the spring. Let's use the principle of conservation of mechanical energy to solve the problem.

Step 1: Loss in Kinetic Energy

When the blocks are moving, they have kinetic energy. The total kinetic energy (KE) of the system is the sum of the kinetic energies of both blocks:

KE_total = 2 × ½ mv²

Here, each block has a mass of 250 g = 0.25 kg, and they both move with the same velocity "v". Hence, the total kinetic energy becomes:

KE_total = 2 × ½ × 0.25 × v² = 0.25 × v²

Step 2: Spring Potential Energy

The potential energy stored in the spring at maximum elongation is given by the formula for elastic potential energy:

PE_spring = ½ k x²

Where k is the spring constant and x is the elongation of the spring. Here, k = 2 N/m, so the spring potential energy becomes:

PE_spring = ½ × 2 × x² = x²

Step 3: Applying Conservation of Energy

According to the law of conservation of mechanical energy, the total kinetic energy of the system is equal to the total potential energy stored in the spring at maximum elongation:

KE_total = PE_spring

Substituting the expressions for the total kinetic energy and spring potential energy:

0.25 × v² = x²

Step 4: Solving for the Elongation "x"

Solving the above equation for x:

x = √(0.25 × v²) = 0.5 × v

Therefore, the maximum elongation of the spring is:

x = 0.5 × v

Thus, the correct answer is B.