Total momentum must be conserved since there are no external forces in the horizontal direction on the skateboarder or the skateboard.
Momentum before the jump:
[tex]P = m_1v_1 + m_2v_2 = (1.8kg)(0) + (42kg)(0) = 0[/tex]
Momentum after the jump:
[tex]P = m_1v_1 + m_2v_2 = (1.8kg)(v_1) + (42kg)(0.3 \frac{m}{s} ) = 0[/tex]
Solving for the velocity v:
[tex]v_1 = - \frac{m_2v_2}{m_1} = - \frac{(42kg)(0.3 \frac{m}{s} )}{1.8kg} = -7 \frac{m}{s} [/tex]
