Solution for Question 18
1. Impulse Method
Block A ($m_A$) pulls B ($m_B$) via a cord with coefficient of restitution $e$.
During deformation (cord tightening), common velocity $v$ is reached:
$$v = \frac{m_A u}{m_A + m_B}$$Impulse of deformation on B: $J_D = m_B v$.
2. Restitution
Total impulse on B: $J = J_D(1 + e)$.
$$m_B v_B = m_B v (1 + e)$$ $$v_B = \frac{m_A u}{m_A + m_B} (1 + e)$$
Answer: $v_B = \frac{m_A(1+e)}{m_A+m_B} u$
