- #1

Father_Ing

- 33

- 3

- Homework Statement:
- -

- Relevant Equations:
- momentum conservation

Consider a rocket with mass ##m## in space is going to move forward. In order to do so, it needs to eject mass backwards. Let the mass that is ejected has velocity ##u## relative to the rocket. What is the equation for the final velocity?

It is said that after ##dt## second, the rocket will have mass ##m-dm##, and velocity ##v+dv##.But, isn't it also possible for the speed to increase in high sum after a very small amount of time, or even, not changing at all?

And I tried to find this out by using conservation of momentum.

Let ##v'## be the rocket's speed after ##dt## second, and ##v## is the initial speed.

$$mv = dm(v-u)+(m-dm)v'$$

$$v'=\frac {(m-dm)v +udm}{m-dm}$$

$$v'=v+\frac {udm}{m-dm}$$

Since ##dm## is small, we can take the limit of dm->0. Therefore,$$v'= v$$

It can be concluded that the velocity neither increase nor decrease.

But, I searched about this matter in the internet, and they said that ##v'## is ##v+dv##. Are there any mistakes in my method?

It is said that after ##dt## second, the rocket will have mass ##m-dm##, and velocity ##v+dv##.But, isn't it also possible for the speed to increase in high sum after a very small amount of time, or even, not changing at all?

And I tried to find this out by using conservation of momentum.

Let ##v'## be the rocket's speed after ##dt## second, and ##v## is the initial speed.

$$mv = dm(v-u)+(m-dm)v'$$

$$v'=\frac {(m-dm)v +udm}{m-dm}$$

$$v'=v+\frac {udm}{m-dm}$$

Since ##dm## is small, we can take the limit of dm->0. Therefore,$$v'= v$$

It can be concluded that the velocity neither increase nor decrease.

But, I searched about this matter in the internet, and they said that ##v'## is ##v+dv##. Are there any mistakes in my method?