Let the masses be $m_1 = 3m$ and $m_2 = m$. Let the initial velocities be $v_1 = 2$ m/s and $v_2 = -2$ m/s. The collision is perfectly inelastic, so momentum is conserved.

[Q1] The final velocity $v_f$ is found by conservation of momentum:

$$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$$ $$v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{(3m)(2 \text{ m/s}) + m(-2 \text{ m/s})}{3m + m} = \frac{4m}{4m} \text{ m/s}$$

[Q2] The percentage energy loss is $\frac{K_i - K_f}{K_i}$. Initial kinetic energy: $K_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(3m)(2)^2 + \frac{1}{2}m(-2)^2 = 8m$ Final kinetic energy: $K_f = \frac{1}{2}(m_1+m_2)v_f^2 = \frac{1}{2}(4m)(1)^2 = 2m$ Percentage loss:

$$\frac{K_i - K_f}{K_i} = \frac{8m - 2m}{8m} = \frac{6}{8}$$