# A problem regarding momentum(from an entrance exam to Uni in Japan)

1. Sep 2, 2008

### fantasy

a problem regarding momentum(from an entrance exam to Uni in Japan)(editted)

1. The problem statement, all variables and given/known data

2. Relevant equations

In the space ,A rocket with mass M moving with constant velocity of V with no external force acting on it.

1) after sometime, the rocket release the fuel mass M/2 in an instant to increase the velocity .the velocity of the fuel relative to the rocket = V
1.1) using the law of conservation of momentum find the velocity of the rocket after the increase of the velocity
1.2) after the increase of the velocity , what is the sum of the kinetic energy of the fuel gas and the rocket?

2) (2nd situation . not the time after question 1) after sometimes the rocket release the fuel gas of mass M/4 to increase velocity .after that the rocket realease another fuel gas of mass M/4 to increase velocity. both of the velocity of the fuel relative to the rocket are V.

2.3) after increases of the velocity 2 times what is the velocity of the rocket
2.4)the first time of release of fuel gas to increase the velocity of the rocket,what is the velocity of the fuel relative to the space
2.5)the second time of release of fuel gas to increase the velocity of the rocket,what is the velocity of the fuel relative to the space
2.6) after the increase of the velocity 2 times, what is the sum of the kinetic energy of the fuel gas and the rocket?

3) (3rd situation ) After sometimes the rocket release aa amount of fuel to increase its velocity.after that, it release the same amount of fuel as the first time to increase its velocity again.both of the velocity of the fuel relative to the rocket are V.
3.7) after the second time of the release of the fuel,the velocity of rocket kV ,find the value of k (k is more than 1)
3.8) from question 3.7, if the maximum amount of the fuel released is equal to M ,what is the maximum value of k

3. The attempt at a solution
this is what i thought of using this formulae

mv = − dm (−u +v + dv ) + (m + dm )(v + dv )

and i got the answer for 1.1 = 3/2M

is it correct?

for 1.2) should sum of the kinetic enegy equal to 2(1/2)(M/2)(3V/2)^2 ?

2.3) using the same formulae as 1.1 i got 5/4V and 1/5/6V

2.4,2.5,2.6,2.7 dun understand the concept= =

2.8) is it 0? because the mass of the rocket would also be reduced to 0?

sorry for not adding my own attempt at first I am new to this forum and hadnt yet to read the rule then
..................

this is the question in the entrance exam to a University in Japan.....so if the question seem confusing please do ask me....i might have mistranslated somethings.....thank you very much for helps

Last edited: Sep 2, 2008
2. Sep 2, 2008

### LowlyPion

It would help if you showed what work you have done and where you are having a problem.

3. Sep 2, 2008

### ZapperZ

Staff Emeritus
You have to show what you have attempted to do. That's why there is a section called "3. The attempt at a solution". Members here are prohibited from giving outright solutions.

Zz.

4. Sep 2, 2008

### fantasy

i stucked at the first question....= =.....so i didnt

this is what i thought of using this formulae

mv = − dm (−u +v + dv ) + (m + dm )(v + dv )

but when i substitutued all the variable into the formulae

MV = -M/2(dV) + M/2(V+dV)

but the dV i want to know is gone after sovling....

I really do not know how to continue....

5. Sep 2, 2008

### fantasy

hmmm...
is the formulae correct by the way?.....

i tried substituting again after relising that dm must be negative...

and i got the answer for 1.1 = 3/2M

is it correct?

for 1.2) should sum of the kinetic enegy equal to 2(1/2)(M/2)(3V/2)^2 ?

6. Sep 2, 2008

### tiny-tim

Welcome to PF!

Hi fantasy! Welcome to PF!
I don't understand this …

(i suspect your first bracket should just be dv)

you must write clearly if you want to pass the exam.

Start by saying "Let the new velocity of the rocket be w" (or v + dv, or whatever you want to call it).

Continue "So the velocity of the fuel is … "

(and why do you use dm when the mass of fuel is given?)

7. Sep 2, 2008

### fantasy

thank you very much for your advice....I use the mass given ..i just want to show the formulae i use =)...

mv = − dm (−u +v + dv ) + (m + dm )(v + dv )

i got this from some website
m- mass
v- velocity of rocket relative to observer
u- velocity of fuel relative to the rocket

ummm...so is the formulae usable and is 3V/2 the answer =)?

and the rest of the question?

8. Sep 2, 2008

### tiny-tim

Why use u when v is velocity of fuel relative to the rocket?

It makes you look as if you are using formulas without understanding them.

That is not the way to get into university.
Yes, 3V/2 is right.

But it would be much better to say "So the velocity of the fuel is dv, so mv = m dv/2 + m (v + dv)/2 …"
How do you get that?

9. Sep 2, 2008

### fantasy

i use the given V as the velocity relative to rocket to substituted into u in the formulae

i just want to quote the how the formulae i use looks like.....

oooo....i sum up the kinetic energy of the rocket and fuel gas after the increase of the velocity.....1/2 * M* (3V/2)^2 then multiplied by 2 ......is my concept wrong?....

10. Sep 2, 2008

### tiny-tim

Why did you multiply by 2?

The fuel has a different speed.

11. Sep 2, 2008

### fantasy

hmmmmm..... because my understanding was the speed of the fuel and the rocket must be the same becuase the mass is the same ......according to newton`s 3rd law...

but after using that formulae....my new answer is 1/2 * M/2* (V/2)^2 +1/2 * M* (3V/2)^2 = 19/16*MV^2 ? is it correct?

12. Sep 2, 2008

### tiny-tim

Yes … except they should both have M/2, shouldn't they?

13. Sep 2, 2008

### fantasy

yes.....typo sorry......1/2 * M/2* (V/2)^2 +1/2 * M/2* (3V/2)^2 =10/16*MV^2

right?......and thank you very much again tiny-tim for helping....

what abt next question...what concept or how should i continue?