Answer to Question #330112 in Physics for Jane

Question #330112

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?

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Expert's answer
2022-04-19T15:27:45-0400

a) According to the momentum conservation law:


m1v1=m2v2m_1 v_1 = m_2 v_2

where m1=65.0kg,m2=2.50kg,v2=3.50m/sm_1 = 65.0kg, m_2 = 2.50kg, v_2 = 3.50m/s and v1v_1 is the speed of the astronaut. Thus, obtain:


v1=m2v2m1=2.53.5650.135m/sv_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=m1v122+m2v222K=12(65(0.135)2+2.5(3.5)2)31.8JK = \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

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