Questions: A bullet with mass 15 g is traveling at 350 m / s when it strikes a stationary wooden object with mass 500 g. The bullet emerges on the other side of the object but has slowed to 75 m / s. What is the kinetic energy of the wooden object?

Solution Steps

First, we need to convert the masses from grams to kilograms.

\[ m_{\text{bullet}} = 15 \, \text{g} = 0.015 \, \text{kg} \] \[ m_{\text{wood}} = 500 \, \text{g} = 0.5 \, \text{kg} \]

Next, we calculate the initial and final momentum of the bullet.

Initial momentum of the bullet: \[ p_{\text{initial, bullet}} = m_{\text{bullet}} \times v_{\text{initial, bullet}} = 0.015 \, \text{kg} \times 350 \, \text{m/s} = 5.25 \, \text{kg} \cdot \text{m/s} \]

Final momentum of the bullet: \[ p_{\text{final, bullet}} = m_{\text{bullet}} \times v_{\text{final, bullet}} = 0.015 \, \text{kg} \times 75 \, \text{m/s} = 1.125 \, \text{kg} \cdot \text{m/s} \]

The change in momentum of the bullet is:

\[ \Delta p_{\text{bullet}} = p_{\text{initial, bullet}} - p_{\text{final, bullet}} = 5.25 \, \text{kg} \cdot \text{m/s} - 1.125 \, \text{kg} \cdot \text{m/s} = 4.125 \, \text{kg} \cdot \text{m/s} \]

By conservation of momentum, the change in momentum of the bullet is equal to the momentum gained by the wooden object.

\[ \Delta p_{\text{bullet}} = p_{\text{wood}} \]

\[ 4.125 \, \text{kg} \cdot \text{m/s} = m_{\text{wood}} \times v_{\text{wood}} \]

Solving for \( v_{\text{wood}} \):

\[ v_{\text{wood}} = \frac{4.125 \, \text{kg} \cdot \text{m/s}}{0.5 \, \text{kg}} = 8.25 \, \text{m/s} \]

Finally, we calculate the kinetic energy of the wooden object using the formula:

\[ KE_{\text{wood}} = \frac{1}{2} m_{\text{wood}} v_{\text{wood}}^2 \]

\[ KE_{\text{wood}} = \frac{1}{2} \times 0.5 \, \text{kg} \times (8.25 \, \text{m/s})^2 = 0.25 \, \text{kg} \times 68.0625 \, \text{m}^2/\text{s}^2 = 17.0156 \, \text{J} \]

Final Answer