A block of mass 2.0 kg travelling at 2.0 m/s undergoes a head-on...

The maximum loss in kinetic energy occurs when the collision is perfectly inelastic, meaning the two blocks stick together and move with a common velocity after the collision.

Let $m_1 = 2.0 \text{ kg}$, $u_1 = 2.0 \text{ m/s}$, $m_2 = 2.0 \text{ kg}$, and $u_2 = 0$.

By conservation of linear momentum:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$

$2.0 \times 2.0 + 0 = (2.0 + 2.0) v$

$4.0 = 4.0 v \Rightarrow v = 1.0 \text{ m/s}$

Initial kinetic energy of the system is:

$K_i = \dfrac{1}{2} m_1 u_1^2 = \dfrac{1}{2} \times 2.0 \times (2.0)^2 = 4.0 \text{ J}$

Final kinetic energy of the system is:

$K_f = \dfrac{1}{2} (m_1 + m_2) v^2 = \dfrac{1}{2} \times 4.0 \times (1.0)^2 = 2.0 \text{ J}$

The maximum possible loss in kinetic energy is:

$\Delta K = K_i - K_f = 4.0 - 2.0 = 2.0 \text{ J}$

Alternatively, using the direct formula for loss of kinetic energy in a perfectly inelastic collision:

$\Delta K = \dfrac{1}{2} \dfrac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2$

$\Delta K = \dfrac{1}{2} \times \dfrac{2.0 \times 2.0}{2.0 + 2.0} \times (2.0 - 0)^2 = \dfrac{1}{2} \times 1.0 \times 4.0 = 2.0 \text{ J}$