Question

The mass of the skateboard is 1.8kg and the mass of the skateboarder is 42kg.

Answer 1

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:
$P = m_1v_1 + m_2v_2 = (1.8kg)(0) + (42kg)(0) = 0$

Momentum after the jump:
$P = m_1v_1 + m_2v_2 = (1.8kg)(v_1) + (42kg)(0.3 \frac{m}{s} ) = 0$

Solving for the velocity v:
$v_1 = - \frac{m_2v_2}{m_1} = - \frac{(42kg)(0.3 \frac{m}{s} )}{1.8kg} = -7 \frac{m}{s}$