Respuesta :
Given information:
Mass of Emily,
[tex]m_1=50\text{ kg}[/tex]Mass of the wagon;
[tex]m_2=14.2\text{ kg}[/tex]Initial velocity;
[tex]u=4.27\text{ m/s}[/tex]Final velocity of Emily;
[tex]v_1=5.41\text{ m/s}[/tex]According to the conservation of momentum the initial momentum of the system equals the final momentum of the system.
[tex](m_1+m_2)u=m_1v_1+m_2v_2[/tex]Here, v_2 is the velocity of the wagon after Emily jumps off.
The velocity of the wagon after Emily jumps off is given as,
[tex]\begin{gathered} m_2v_2=(m_1+m_2)u-m_1v_1 \\ v_2=\frac{(m_1+m_2)u-m_1v_1}{m_2} \end{gathered}[/tex]Substituting all known values,
[tex]\begin{gathered} v_2=\frac{\lbrack(50\text{ kg})+(14.2\text{ kg})\rbrack\times(4.27\text{ m/s})-(50\text{ kg})\times(5.41\text{ m/s})}{(14.2\text{ kg})} \\ \approx0.26\text{ m/s} \end{gathered}[/tex]Therefore, the velocity of the wagon after Emily jumps off is 0.26 m/s.
