Answer to Question #330112 in Physics for Jane

Question

A 65.0 kg astronaut is doing a repair in space on the orbiting space station. The astronaut throws a what 2.50 kg toolbox away from him at 3.50 m/s to the space station. (a) With what speed and in direction will he begin to move? (b) How much kinetic energy does the astronaut create by this maneuver?

Answer 1

a) According to the momentum conservation law:

$$m_1 v_1 = m_2 v_2$$

where $m_1 = 65.0kg, m_2 = 2.50kg, v_2 = 3.50m/s$ and $v_1$ is the speed of the astronaut. Thus, obtain:

$$v_1 = \dfrac{m_2v_2}{m_1} =\dfrac{2.5\cdot 3.5}{65} \approx 0.135m/s$$

b) The resultant total kinetic energy of the astronaut with the tool is:

$$K = \dfrac{m_1v_1^2}{2} + \dfrac{m_2v_2^2}{2}\\ K = \dfrac{1}{2}(65\cdot (0.135)^2 + 2.5\cdot (3.5)^2) \approx 31.8J$$