Question

2. Consider a situation where change in momentum come from a change in mass rather than that

Answer 1

a) $\Delta m = 1300\Delta t$ [kg]

b) $p=vR\Delta t$

c) F = vR, $5.85\cdot 10^7 N$

Explanation:

a)

In this problem, we are given the rate at which the mass is expelled by the rocket:

$R=\frac{\Delta m}{\Delta t}$

Therefore, the total mass expelled by the rocket in a time $\Delta t$ is the given by the equation

$\Delta m = R \Delta t$

where

$\Delta m$ is the mass expelled

R is the rate at which it is expelled

$\Delta t$ is the time interval

Here we have

R = 1300 kg/s

Therefore, substituting into the equation,

$\Delta m = 1300\Delta t$ [kg]

b)

The momentum of an object is given by

$p=mv$

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, $\Delta m$, that is expelled by the rocket.  This will be given by

$p=v\Delta m$

where

$\Delta m = 1300 \Delta t$ is the mass of gas expelled in a time of $\Delta t$

$v=4.5\cdot 10^4 m/s$ is the velocity of the gas

Substituting,

$p=vR\Delta t$

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:

$F=\frac{\Delta p}{\Delta t}$

where

F is the average force

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

In this problem, the change in momentum of the rocket is equal to the momentum of the gas expelled, so:

$\Delta p = vR \Delta t$

Substituting into the equation, we can therefore find the average force on the rocket:

$F=\frac{vR\Delta t}{\Delta t}=vR = (4.5\cdot 10^4)(1300)=5.85\cdot 10^7 N$