Answers:
a) 0.053 kg
b) 5.22 N
Explanation:
We have the following data:
[tex]m_{A}= 70 kg[/tex] is the mass of the astronaut
[tex]M= 110 kg[/tex] is the mass of the MMU
[tex]a=0.029 m/s^{2}[/tex] is the acceleration of the system astronaut-MMU
[tex]V_{N_2}=490 m/s[/tex] is the velocity of the ejected fuel
Now, we have the following two equations to find the amount of gas (mass)[tex]dm[/tex] ejected by the thruster and the thrust [tex]F[/tex]:
[tex]a=-\frac{V_{N_2}}{m} \frac{dm}{dt}[/tex] (1)
[tex]F=-V_{N_2} \frac{dm}{dt}[/tex] (2)
Where:
[tex]m=m_{A} + M=70 kg + 110 kg=180 kg[/tex] is the total mass of the system astronaut-MMU
[tex]dm[/tex] is the mass of the ejected fuel
[tex]dt=5 s[/tex] is the time interval in which this occurs
Knowing this, let's begin with the answers:
a) How much gas is used by the thruster in 5s?
Here we have to isolate [tex]dm[/tex] from (1) considering [tex]m[/tex] as constant, since the mass of the fuel is negligible compared to the total mass [tex]m[/tex]:
[tex]dm=-\frac{m a}{V_{N_2}} dt[/tex] (3)
[tex]dm=-\frac{(180 kg)(0.029 m/s^{2})}{490 m/s} 5 s[/tex] (4)
[tex]dm=-0.053 kg[/tex] (5) Note the negative sign is because the direction of the ejected mass is opposite to the direction of the system astronaut-MMU. However, the value of [tex]dm[/tex] is 0.053 kg.
b) What is the thrust of the thruster?
Let's use equation (2) with the [tex]dm[/tex] calculated in (5):
[tex]F=(-490 m/s) \frac{-0.053 kg}{5s}[/tex] (6)
Finally:
[tex]F=5.22 N[/tex] (7)
