Rocket questions, Fundamentals of Physics

[Volcano]

Hi,

I am confused with a solution of momentum in Rocket problems which published and solved in "Fundamentals of Physics, (Halliday-Resnick)".

There are two different approaches to solving two similar questions. As i noticed, only difference is, the thing which flings. Fuel or tank(or any part of rocket). This is really confusing.

Say, the rocket and fuel total mass is "M", fired fuel mass "m" and "v" is first velocity of rocket plus fuel, relative to earth. The fuel is firing backward with "u" relative to rocket body. For Earth final velocity of rocket body is "v'". The equation is;

M v = (M - m) v' + m (v - u) ... (1)


But if the thing which throws backward is the empty fuel tank beside of fired fuel then equation is,


M v = (M - m) v' + m (v' - u) ... (2)

highly similar. But notice the last parentheses. This questions are at the end of chapter 9. And the solutions are in the solutions book.

Chapter 9 - 6 : Preservation of Linear Momentum, exercises, question 15

and,

Chapter 9 - 7 : Practice of Momentum Principle, exercises, question 16.

these two questions are friend but equations not similar.

I am trying to guess about this difference but don't have any reasonable idea. Meanwhile, the questions,

9 - 7 - 18 (exercises): equation looks like (2) above
9 - 6 - 12 (problems) : equation looks like (2) above

are related too. Tired of thinking. Can you see the reason? Or any idea.



Note: This language is not my native for me. So, i tried my best for translating the name of chapters. BTW, i suppose the Chapter numbers are enough. If these explanations not enough then i will try to find them in English somewhere.

Have nice day

 

[Volcano]

Hi,

This question type is confusing me.

Initially A and B is staying quiet together. Suddenly, after a blast, both far away from other. Say, B is going to positive direction, A is negative. Relative to B, start velocity of A is "Vba". Thus what is the velocity of B? (Ma, Mb : mass)

Pi = Pf -> (Ma+Mb)0 = Ma Va + Mb Vb and Vba = Va - Vb then Va = Vb + Vba,

0 = Ma (Vb + Vba) + Mb Vb then Ma (Vb + Vba) = - Mb Vb,

After this equation, if we put the data in places, can solve the Vb.

I know question is easy. And my solution is true. However hard to explain without pen-paper.

Now say; A: Rocket Fuel, B: Rocket Body. Then above solution is wrong. It must,

Vb = Ma/Mb x Vba

for the last equation, look like "Vba" is not relative to rocket body but relative to quiet observer. But why? Why approaches are so different? I get this equation from Serway and "Fundamentals of Physics".

Honesty, was not easy to explain for me with my poor English. But, this is my first try to ask a Physics question in internet. If my statements needs more explanations feel free to say.

BTW, if you think is will be better to understanding then show me a link or maybe a name of book.

Greats

 

[thepatientmental]

!

Hi there,

I can understand your confusion about the different equations used in solving rocket problems. It can be frustrating when the equations seem similar but have slight differences that make them difficult to understand.

From what I can gather, the main difference between equations (1) and (2) is the reference frame being used. In equation (1), the velocities are relative to the Earth, while in equation (2), the velocities are relative to the rocket body. This means that in equation (1), the velocity of the fired fuel is (v-u) relative to the Earth, while in equation (2), it is (v'-u) relative to the rocket body.

The reason for this difference is because the equations are being applied in different scenarios. In the exercises and problems you mentioned, the rocket is being fired in a vacuum, where there is no external force acting on it. In this case, the reference frame of the Earth is more useful because it is fixed and does not change. However, in real-life situations, the rocket is often fired in the presence of air resistance, which can affect its velocity. In these cases, it is more useful to use the reference frame of the rocket body, as it is constantly changing with the motion of the rocket.

I hope this helps clarify the difference between the two equations and why they are used in different scenarios. Don't be discouraged by the slight variations in equations, as long as you understand the concept and can apply it correctly in different situations, you are on the right track. Keep practicing and you will become more comfortable with solving rocket problems. Best of luck to you!