Answer to Question #141317 in Physics for Sydney

Question

A longboarder (45.0 kg) is travelling on her 2.00 kg skateboard at 5.35 m/s. She jumps off the board, continuing in the same direction at 4.25 m/s. What is the velocity of the skateboard after she jumps off?

Answer 1

The initial momentum of the system was

$$p_i =m_Sv_i + m_lv_i = (m_s+m_l)v_i$$

where $m_s = 2kg$ is the mass of the skateboard, $m_l = 45kg$ is the mass of the longboarder and $v_i = 5.35m/s$ is their initial speed. Thus, we have:

$$p_i = (45kg+2kg)\times 5.35m/s = 251.45\space kg\cdot m/s$$

After the jump the longboarder took away the following momentum:

$$p_{l}' = m_lv_l'$$

where $v_l' = 4.25m/s$ is the speed of the longboarder after the jump. Thus, obtain:

$$p_{l}' = m_lv_l' = 45kg\times 4.25m/s = 191.25\space kg\cdot m/s$$

Thus, the momentum of the skateboard after the jump is:

$$p_s' = p_i - p_l' = 251.45\space kg\cdot m/s-191.25\space kg\cdot m/s = 60.2\space kg\cdot m/s$$

The speed of the skateboard then is:

$$v_s' = \dfrac{p_s'}{m_s} = \dfrac{60.2\space kg\cdot m/s}{2kg} = 30.1m/s$$

Answer. 30.1 m/s.