COM BYU 18

Physics Solution Q18

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: $v_B = \frac{m_A(1+e)}{m_A+m_B} u$