Conservation of momentum and energy, but missing mass Steps:

• System initially at rest, so total momentum is zero: $m_1 v_1 + m_2 v_2 = 0$, with $m_1 = 2$ kg, $v_1 = 2$ m/s (left), so $m_2 v_2 = 4$ kg m/s (right).
• Elastic potential energy in spring equals total kinetic energy: $E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = 4 + \frac{8}{m_2}$ J.
• Mass $m_2$ of second trolley not given, so $E$ cannot be calculated numerically.
• Not enough information to determine exact energy.

Why D is correct:

• Not applicable; insufficient data prevents selecting any option definitively.

Why the others are wrong:

• A: Assumes total energy 4 J (only one trolley's KE), ignores second trolley.
• B: No basis; arbitrary value not matching partial KE of 4 J.
• C: Assumes equal masses (both 2 kg, $v_2 = 2$ m/s), total 8 J, but other mass unknown.

Final answer: Not enough information.