Answer to Question #242901 in Physics for rshiii

Question

7. A 68.5-kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.25-kg tool away from her at 3.20 m/s relative to the space station. With what speed and in what direction will she begin to move?

Answer 1

Using the momentum conservation law, obtain:

$$0 = m_tv_t + m_av_a$$

where represents the momentum before she throws the tool, $m_t = 2.25kg, m_a = 68.5kg$ are the masses of the tool and astronaut respectively, $v_t = 3.20m/s,\space v_a$ are the velocities of the tool and astronaut after the throw. Expressing $v_a$, obtain:

$$v_a = -\dfrac{m_tv_t}{m_a}\\ v_a = -\dfrac{2.25kg\cdot 3.20m/s}{68.5kg} \approx -0.105m/s$$

Sign "-" means that the direction of this velocity is opposite to the direction of tool's velocity.

Answer. 0.105 m/s away from the tool.