Questions: Lee is rolling along on her 4-kg skateboard at a speed of 3 m/s when she jumps off the back and continues forward with a velocity of 2 m/s relative to the ground. This causes the skateboard to go flying forward with a speed of 15.5 m/s relative to the ground. What is Lee's mass?

Determine the initial momentum of the system.

Calculate the initial momentum of Lee and the skateboard.

The initial momentum of the system is the sum of the momentum of Lee and the skateboard. Let \( m_L \) be Lee's mass. The initial momentum is given by:

\[ p_{\text{initial}} = (m_L + 4 \, \text{kg}) \times 3 \, \text{m/s} \]

Express the initial momentum in terms of \( m_L \).

\[ p_{\text{initial}} = 3m_L + 12 \, \text{kg} \cdot \text{m/s} \]

\(\boxed{p_{\text{initial}} = 3m_L + 12 \, \text{kg} \cdot \text{m/s}}\)

Determine the final momentum of the system.

Calculate the final momentum of Lee and the skateboard.

The final momentum of the system is the sum of the momentum of Lee and the skateboard after she jumps off. The final momentum is given by:

\[ p_{\text{final}} = m_L \times 2 \, \text{m/s} + 4 \, \text{kg} \times 15.5 \, \text{m/s} \]

Express the final momentum in terms of \( m_L \).

\[ p_{\text{final}} = 2m_L + 62 \, \text{kg} \cdot \text{m/s} \]

\(\boxed{p_{\text{final}} = 2m_L + 62 \, \text{kg} \cdot \text{m/s}}\)

Apply the conservation of momentum to solve for Lee's mass.

Set the initial momentum equal to the final momentum.

According to the conservation of momentum:

\[ 3m_L + 12 \, \text{kg} \cdot \text{m/s} = 2m_L + 62 \, \text{kg} \cdot \text{m/s} \]

Solve for \( m_L \).

Subtract \( 2m_L \) from both sides:

\[ m_L + 12 \, \text{kg} \cdot \text{m/s} = 62 \, \text{kg} \cdot \text{m/s} \]

Subtract 12 from both sides:

\[ m_L = 50 \, \text{kg} \]

\(\boxed{m_L = 50 \, \text{kg}}\)

\(\boxed{p_{\text{initial}} = 3m_L + 12 \, \text{kg} \cdot \text{m/s}}\)

\(\boxed{p_{\text{final}} = 2m_L + 62 \, \text{kg} \cdot \text{m/s}}\)

\(\boxed{m_L = 50 \, \text{kg}}\)