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Old Feb 16th 2018, 11:46 PM   #1
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Momentum of Rocket

Greetings to all

I have two small questions about the Momentum of a Rocket and conservation of Momentum , the questios are :

1- Can the speed of a Rocket exceeds the exhaust speed of the Fuel ?

2- A bomb, initially at rest, explodes into several pieces.
(a) Is linear momentum conserved? (b) Is kinetic energy
conserved? Explain

In the attachement Photo , questions number 20 and 22

Thanks in Advance
Best regards
Razi
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Momentum of Rocket-1.jpg  
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Old Feb 17th 2018, 01:37 AM   #2
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Your list of questions are a good way of applying the basic concepts to a variety of situations.

Have you no thoughts on these?

For instance you have only asked about the second part of question 20, what was your answer to the first part?

Do you have a mathematical formula for momentum and kinetic energy?
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Old Feb 18th 2018, 05:18 AM   #3
Pmb
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Originally Posted by Razi View Post
Greetings to all

I have two small questions about the Momentum of a Rocket and conservation of Momentum , the questios are :

1- Can the speed of a Rocket exceeds the exhaust speed of the Fuel ?

2- A bomb, initially at rest, explodes into several pieces.
(a) Is linear momentum conserved? (b) Is kinetic energy
conserved? Explain

In the attachement Photo , questions number 20 and 22

Thanks in Advance
Best regards
Razi
We're required by forum rules not to simply give you the answers to your question but to help lead you to the answer. Can you take a shot at answering these questions yourself, please?.

I'll give you a two hints:

(1) Try thinking in terms of absolutes by which I mean is it possible for one object to be expelled (exhaust) by another object (rocket) in which the speed of the first object is less than the second object? What would happen if the first object was a bowling ball and the second object a flea?

(2) Conservation of a quantity means that the total before equals the total after.

(a) What is the total linear momentum before the explosion and what is it after the explosion? Find the answer by using the laws of conservation of linear momentum.

(b) What is the total kinetic energy before the explosion and what is it after the explosion? How do you find the answer?

If that doesn't lead you to your answer then please let me know. In such case turn to your text on the subject of conservation of momentum and kinetic energy first.
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Old Feb 20th 2018, 01:29 AM   #4
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Dear Studiot , Pmb

Thanks for replay
About your questions , Yes , I know well the answer of the first part of question 20 and about the second part I have derived some equations and I would like to ask you about them :
First let us assume that the movement of the rocket in free space ( no air resistance ) and there is no any other external forces affect on it .

a) if the engine of the rocket not working but the rocket has a constant initial velocity vi : ( please see the attached photos )

Now we can calculate the initial momentum of the rocket :

Pi = ( M + 𝜟m) v
Where M : Mass of the rocket
𝜟m : mass of the fuel
Pi : initial momentum according to its initial speed
V : speed of the rocket relative to the Earth

Over a short time interval 𝜟t , the rocket ejects fuel of mass 𝜟m , and so at
the end of this interval the rocket’s speed is v+𝜟v where 𝜟v is the change in
speed of the rocket

b) If the fuel is ejected with a speed ve relative to the
rocket (the subscript “e” stands for exhaust, and ve is usually called the exhaust
speed) , the velocity of the fuel relative to a stationary frame of reference is v-ve
Thus, if we equate the total initial momentum of the system to the total final momentum,
we obtain:
Pf = ( M ) (v+𝜟v)
Where Pf : final momentum according to its final velocity
Pi = Pf ( conservation of momentum )
( M+𝜟m)v = M (v+𝜟v) + 𝜟m(v-ve) , where ve is the exhaust speed of the burned fuel

( v-ve ) represent the speed of the exhaust relative to Earth

Mv + 𝜟m v = Mv + M𝜟v + 𝜟m v – 𝜟m ve
M𝜟v – 𝜟m ve = 0 ⇒ M𝜟v = 𝜟m ve
If we now take the limit as 𝜟t goes to zero , we get 𝜟v⇒ dv and 𝜟m ⇒ dm
M dv = ve dm
The increase in the exhaust mass dm corresponds to an equal decrease in the rocket mass
So that dm = - dM
Now


M dv = - ve dM
dv= -ve dM/M ⇒ ∫_vi^vf▒〖dv=ve∫_Mi^Mf▒〖-dM/M〗〗 ⇒ vf-vi=-ve [Ln(Mf)-Ln(Mi) ]
vf-vi=ve Ln(Mi/Mf)
vf=vi +ve Ln(Mi/Mf) ,Final Equation

My question is very clear ,can the speed of a rocket v exceeds the exhaust speed of the fuel ve ????
I have derived two equations according to this question but i am not sure if there are true or not
case number one∶ if the speed of the rocket became equal the speed of exhaust speed ,v=ve
the original equation was ∶ ( M+ Δm )v= M(v+Δv)+ Δm(v-ve)
the last term will equal to zero because we assumed that v=ve
now we have ∶ ( M+ Δm )v= M(v+Δv) ⇒Mv+ Δm v=Mv+MΔv ⇒ Δm v=MΔv
Δt⇒0 ⇒ Δm⇒dm and Δv⇒dv
v dm=M dv , dm= -dM
-v dM=M dv ⇒ -dM/M=dv/v ⇒-∫_Mi^Mf▒〖dM/M= ∫_vi^vf▒〖dv/v ⇒ -[Ln(Mf)-Ln(Mi) ]=Ln(vf)-Ln(vi)〗〗
Ln(Mi)-Ln(Mf)= Ln(vf)-Ln(vi)⇒ Ln(Mi/Mf)= Ln(vf/vi) ⇒ Mi/Mf=vf/vi
this result means that the initial momentum=finnal momentum ( i am not sure about this result )
case number two∶ if the speed of the rocket became greater than the speed of exhaust speed ,v>ve
Now the original equation will be∶(M+Δm)v=M(v+Δv)- Δm(v-ve)
Mv+Δm v=Mv+M Δv- Δm v+Δm ve ⇒2 Δm v=M Δv+ Δm ve ⇒2Δm v-Δm ve=M Δv
(2v-ve)Δm=MΔv
Δt⇒0 ⇒ Δm⇒dm and Δv⇒dv
dm= -dM
(2v-ve)dm=M dv ⇒ -(2v-ve)dM=M dv ⇒(ve-2v)dM=M dv
dM/M=dv/(ve-v)⇒ ∫_Mi^Mf▒〖dM/M= ∫_vi^vf▒〖dv/(ve-2v) ⇒Ln(Mf/Mi)= -1/2 [Ln(ve-2vf)- Ln(ve-vi) ] 〗〗
Ln(Mf/Mi)=1/2 [Ln(ve-2vi)-Ln(ve-2vf) ]
Ln(Mf/Mi)=1/2 Ln((ve-2vi)/(ve-2vf))⇒ Ln(Mf/Mi)=Ln(√((ve-2vi)/(ve-2vf))) ⇒Mf/Mi=√((ve-2vi)/(ve-2vf)) ( i am not sure about this result )



about the second question which was ∶ a bomb initially at rest ,explodes into several pieces .(a)is linear momentum conserved ?(b)is kinetic energy conserved ?
explane ?

My answer is∶ the momentum is not conserved ?but the kinetic energy is conserved because the kinetic energy of all picese qual
to the chemical potential enegry which was into the bomb before exploding . ( i am not sure about this answer )

can anybody just tell me if these equation correct or not ?

Thanks in advanve
Best regards
Razi
Attached Thumbnails
Momentum of Rocket-3.jpg   Momentum of Rocket-4.jpg   Momentum of Rocket-5.jpg   Momentum of Rocket-6.jpg   Momentum of Rocket-7.jpg  

Momentum of Rocket-8.jpg   Momentum of Rocket-1.jpg   Momentum of Rocket-2.jpg  
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Old Feb 20th 2018, 02:10 AM   #5
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about the second question which was ∶ a bomb initially at rest ,explodes into several pieces .(a)is linear momentum conserved ?(b)is kinetic energy conserved ?
explane ?

My answer is∶ the momentum is not conserved ?but the kinetic energy is conserved because the kinetic energy of all picese qual
to the chemical potential enegry which was into the bomb before exploding . ( i am not sure about this answer )
Let's get momentum cleared up first.

Momentum is conserved.

Initially the bomb is at rest so has zero momentum in any direction
Momentum is a vector quantity.
Hence direction is also important.

I see from your formulae that you understand momentum can be positive or negative.


So the momentum equation for the explosion event is:


Vector sum of momentum after explosion = Vector sum before explosion = 0 in this case.


You will find that the vector sum, taking signs into account, of the moementa of all the fragments adds up to zero.

Of course we add the components in three mutually perpendicular directions, where they will separately add to zero in each of those directions.



Now for the energy.

The question asks specifically for the kinetic energy.

Again initially the bomb is at rest so the kinetic energy is initially zero.

So yes the total of chemical and kinetic energies (and any others that may be involved) is constant ie preserved.

But the kinetic energy is one half mass (always positive) times the velocity squared (again always positive).

KE is always positive.

So the kinetic energies of all the fragments are all positive (unlike the momenta)

And they are moving so their sum is not zero.

So kinetic energy is not conserved



This is why it is so useful to have both momentum and KE available for calculation.


Would you like to revisit your other answers in the light of this?
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Old Feb 27th 2018, 03:40 AM   #6
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Dear Studiot

Thank you so much for your explanation and really you have explained it clearly , now this problem is very clear to me , thanks again and please accept my best regards .


Razi
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