In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.
The equation that defines the linear moment is given by
[tex]mV_i = (m-m_O)V_f - m_OV_O[/tex]
where,
m=Total mass
[tex]m_O =[/tex] Mass of Object
[tex]V_i =[/tex] Velocity before throwing
[tex]V_f =[/tex] Final Velocity
[tex]V_O =[/tex] Velocity of Object
Our values are:
[tex]m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg[/tex]
Solving to find the final speed, after throwing the object we have
[tex]V_f=\frac{mV_0+m_TV_O}{m-m_O}[/tex]
We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.
That way during each section the equations should be modified depending on the previous one, let's start:
A) [tex]5.3Kg\rightarrow 15m/s[/tex]
[tex]V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}[/tex]
[tex]V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}[/tex]
[tex]V_{f1}=0.8511m/s[/tex]
B) [tex]7.9Kg\rightarrow 11.2m/s[/tex]
[tex]V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}[/tex]
[tex]V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}[/tex]
[tex]V_{f2} = 2.0173m/s[/tex]
C) [tex]10.5Kg\rightarrow 7m/s[/tex]
[tex]V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}[/tex]
[tex]V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}[/tex]
[tex]V_{f3} = 3.63478m/s[/tex]
Therefore the final velocity of astronaut is 3.63m/s
