Answer :
a) [tex]\Delta m = 1300\Delta t[/tex] [kg]
b) [tex]p=vR\Delta t[/tex]
c) F = vR, [tex]5.85\cdot 10^7 N[/tex]
Explanation:
a)
In this problem, we are given the rate at which the mass is expelled by the rocket:
[tex]R=\frac{\Delta m}{\Delta t}[/tex]
Therefore, the total mass expelled by the rocket in a time [tex]\Delta t[/tex] is the given by the equation
[tex]\Delta m = R \Delta t[/tex]
where
[tex]\Delta m[/tex] is the mass expelled
R is the rate at which it is expelled
[tex]\Delta t[/tex] is the time interval
Here we have
R = 1300 kg/s
Therefore, substituting into the equation,
[tex]\Delta m = 1300\Delta t[/tex] [kg]
b)
The momentum of an object is given by
[tex]p=mv[/tex]
where
p is the momentum
m is the mass of the object
v is the velocity of the object
In this problem, we want to find the momentum carried by the mass of the gas, [tex]\Delta m[/tex], that is expelled by the rocket. This will be given by
[tex]p=v\Delta m[/tex]
where
[tex]\Delta m = 1300 \Delta t[/tex] is the mass of gas expelled in a time of [tex]\Delta t[/tex]
[tex]v=4.5\cdot 10^4 m/s[/tex] is the velocity of the gas
Substituting,
[tex]p=vR\Delta t[/tex]
c)
According to Newton's second law of motion, the average force exerted on an object is equal to its rate of change of momentum:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]
where
F is the average force
[tex]\Delta p[/tex] is the change in momentum
[tex]\Delta t[/tex] is the time interval
In this problem, the change in momentum of the rocket is equal to the momentum of the gas expelled, so:
[tex]\Delta p = vR \Delta t[/tex]
Substituting into the equation, we can therefore find the average force on the rocket:
[tex]F=\frac{vR\Delta t}{\Delta t}=vR = (4.5\cdot 10^4)(1300)=5.85\cdot 10^7 N[/tex]