The bullet in the previous problem strikes a \( 2.5-\mathrm{kg} \) steel ball that is at rest. The bullet bounces backward after its collision at a speed of \( 5.0 \mathrm{~m} / \mathrm{s} \). How fast is the ball moving when the bullet bounces backward?

To solve this problem, we can use the principle of conservation of momentum. Before the collision, only the bullet has momentum since the steel ball is at rest. After the collision, both the bullet and the ball have momentum. Let: - \( m_b \) = mass of the bullet (not given explicitly, but we can calculate using its speed change), - \( v_b \) = initial speed of the bullet, - \( m_{s} = 2.5 \, \text{kg} \) (mass of the steel ball), - \( v_{s} \) = speed of the steel ball after the collision, - Bullet bounces backward with a speed of \( 5.0 \, \text{m/s} \) which we can treat as negative after the direction change. Initially: \[ \text{Initial Momentum} = m_b v_b \] After the collision: \[ \text{Final Momentum} = m_b(-5.0) + m_{s} v_{s} \] Since momentum is conserved, we can write: \[ m_b v_b = m_b(-5.0) + m_{s} v_{s} \] We can rearrange this to find the velocity of the ball \( v_s \): \[ m_b v_b + 5.0 m_b = m_s v_s \] \[ v_s = \frac{m_b v_b + 5.0 m_b}{m_s} \] \[ v_s = \frac{m_b (v_b + 5.0)}{m_s} \] However, to find a numerical answer, we need the initial speed of the bullet prior to the collision. Assuming a hypothetical initial bullet speed of \( v_b \) which we would need to set, we’ll eventually require it to determine the final speed of the steel ball in a real situation. The answer for \( v_s \) thus is more dependent on specific values you may have for \( v_b \).