8. A 35.0 g bullet strikes a 5.0 kg stationary piece of lumber and embeds itself in the wood. The piece of lumber and bullet fly off together at \( 8.6 \mathrm{~m} / \mathrm{s} \). What was the original speed of the bullet?

Let the mass of the bullet be \( m_b = 0.035 \, \text{kg} \) and the mass of the wood be \( m_w = 5.0 \, \text{kg} \). The bullet embeds into the wood, so by the conservation of momentum the initial momentum of the bullet is equal to the combined momentum of the bullet and wood after the collision. The momentum conservation equation is \[ m_b v_b = (m_b + m_w)v_f, \] where \( v_f = 8.6 \, \text{m/s} \) is the final velocity and \( v_b \) is the original speed of the bullet. Solving for \( v_b \): \[ v_b = \frac{(m_b + m_w)v_f}{m_b}. \] Substitute the given values: \[ v_b = \frac{(0.035 \, \text{kg} + 5.0 \, \text{kg}) \times 8.6 \, \text{m/s}}{0.035 \, \text{kg}}. \] First, calculate the total mass: \[ m_b + m_w = 0.035 \, \text{kg} + 5.0 \, \text{kg} = 5.035 \, \text{kg}. \] Then the bullet speed is: \[ v_b = \frac{5.035 \, \text{kg} \times 8.6 \, \text{m/s}}{0.035 \, \text{kg}}. \] Calculate the numerator: \[ 5.035 \times 8.6 \approx 43.301 \, \text{kg} \cdot \text{m/s}. \] Now, divide by \(0.035 \, \text{kg}\): \[ v_b \approx \frac{43.301}{0.035} \approx 1237.17 \, \text{m/s}. \] Thus, the original speed of the bullet was approximately \( 1.24 \times 10^3 \, \text{m/s} \).