Inelastic collision: momentum conserved, kinetic energy not.

Steps:

• Initial momentum: $mv + 2m \cdot 0 = mv$.
• Post-collision velocity $V$: combined mass $3m$, so $3mV = mv$, thus $V = v/3$.
• Initial KE: $\frac{1}{2}mv^2$.
• Final KE: $\frac{1}{2}(3m)(v/3)^2 = \frac{1}{2}mv^2 \cdot \frac{1}{3}$.
• Fraction lost: $1 - \frac{1}{3} = \frac{2}{3}$.

Why C is correct:

• In perfectly inelastic collisions, KE loss fraction is $1 - \frac{m_1 m_2}{(m_1 + m_2)^2} \cdot \frac{(v_1 - v_2)^2}{(v_1^2 + v_2^2)}$, but here simplifies to $\frac{2}{3}$ via momentum conservation.

Why the others are wrong:

• A: $1/9$ assumes equal masses or elastic collision.
• B: $1/3$ is the retained fraction, not lost.
• D: $8/9$ overestimates loss, ignoring mass ratio.

Final answer: C