Answer to Question #256544 in Physics for megs

Question

A 41 kg skateboarder running at 4.5 m/s [N] jumps onto a 4.4 kg skateboard moving at 2.5 m/s [N]. How fast will the skateboarder and board be traveling once she jumps on the board?

Your Answer:

Answer 1

We can find the velocity of skateboarder and board from the law of conservation of momentum:

$$m_1v_1+m_2v_2=(m_1+m_2)v,$$ $$v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2},$$ $$v=\dfrac{41\ kg\times4.5\ \dfrac{m}{s}+4.4\ kg\times2.5\ \dfrac{m}{s}}{41\ kg+4.4\ kg}=4.3\ \dfrac{m}{s}.$$