Linear momentum conservation - rocket

In summary: No, those aren't the results I get.Hint: You need two positive speeds (one left, one right) which add up to 3, and one is 4 times the other. :smile:
  • #1
Ready2GoXtr
75
0

Homework Statement


A two-ended "rocket" that is initally stationary on a frictionless floor, with its center a the origin of an x axis. The rocket consists of a central block C(of mass M = 6.00kg) and blocks L and R (each of mass m = 2.00kg) on the left and right sides. Small explosions can shoot either of teh side blocks away fro mblock C and along the x axis. Here is the sequence: (1) At time t = 0, block L is shot to the left with a speed of 3.00 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next at the time t = .80 s, block R is shot to the right with a speed of 3m/s relative to the velocity that block C then has. At t = 2.80s, what are the (a) velocity of block C and (b) the position of its center?

http://ready2goxtr.googlepages.com/Problem943.jpg

Homework Equations



Pi = Pf




The Attempt at a Solution



Pi = m1v1 + m2v2 + m3v3 Pf = m1v1f + m2v2f + m3v3f

I don't really understand what's happening in the problem
 
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  • #2
Ready2GoXtr said:
Small explosions can shoot either of teh side blocks away fro mblock C and along the x axis. Here is the sequence: (1) At time t = 0, block L is shot to the left with a speed of 3.00 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next at the time t = .80 s, block R is shot to the right with a speed of 3m/s relative to the velocity that block C then has.
…
I don't really understand what's happening in the problem

Hi Ready2GoXtr! smile:

The question is rather badly worded :frown:, but I think it means that the first explosion is between L and C, and the second between R and C, so that each explosion pushes both blocks away from each other. :smile:
 
  • #3
alright so should I set PI = to 0 seconds and PF = to .80? seconds
 
  • #4
Ready2GoXtr said:
alright so should I set PI = to 0 seconds and PF = to .80? seconds

Not following you. :confused:

Pi = 0, yes … but P is constant, so Pf = 0 also.

Hint: after the first explosion, v2 = v3, and they are both 3 m/s different from v1 :smile:
 
  • #5
tiny-tim said:
Not following you. :confused:

Pi = 0, yes … but P is constant, so Pf = 0 also.

Hint: after the first explosion, v2 = v3, and they are both 3 m/s different from v1 :smile:

You mean at t = .80 seconds v2 = v3?
 
  • #6
Ready2GoXtr said:
You mean at t = .80 seconds v2 = v3?

Assuming that v2 is the intial velocity of R block and v3 is the initial velocity of the C block, that statement is true. (Note: Initial means before the R block is shot off).
 
  • #7
Ready2GoXtr said:
You mean at t = .80 seconds v2 = v3?

No, I mean between t = 0 and .80 …

the explosion is between blocks 1 and 2, so block 2 pushes block 3, and therefore blocks 2 and 3 have the same velocity (v2 = v3), until the second explosion. :smile:
 
  • #8
so could i say that m1v1i = 2kg * 3m/s?
 
  • #9
Ready2GoXtr said:
so could i say that m1v1i = 2kg * 3m/s?

Nooo …
tiny-tim said:
after the first explosion, v2 = v3, and they are both 3 m/s different from v1 :smile:

Going to bed now … :zzz:​
 
  • #10
Okay i think ima just sskip it, don't know how to do the question
 
  • #11
I decided to give it another try. since the momentums have to equal each other, I set m1v1 =- (m2+m3)v2 v2 = .75 m/s i

then i just decided. well what would have it been if it was just m1 and m2

m1v1 = m2v2 v2 = -1 then i took -1 + .75 = equal final velocity which is -.15 m/s

For part b that was easy

vit = .6m then vft = -.42 added them .18 m
 
  • #12
Ready2GoXtr said:
I decided to give it another try. since the momentums have to equal each other, I set m1v1 =- (m2+m3)v2 v2 = .75 m/s i
…

Hi Ready2GoXtr! :smile:

No, those aren't the results I get.

It's difficult to tell unless you show your full working, but I think you subtracted 1 from 3 instead of adding.

Hint: You need two positive speeds (one left, one right) which add up to 3, and one is 4 times the other. :smile:
 

1. How does linear momentum conservation apply to rockets?

Linear momentum conservation is a fundamental principle in physics that states that the total momentum of a system remains constant as long as there are no external forces acting on it. In the case of a rocket, the system consists of the rocket and the exhaust gases expelled from its engines. This means that the total momentum of the rocket and the exhaust gases must remain constant throughout the rocket's flight.

2. Why is linear momentum conservation important in rocket propulsion?

Linear momentum conservation is important in rocket propulsion because it explains how rockets are able to move in the opposite direction of the expelled exhaust gases. As the rocket's engines expel the gases at high speeds, the rocket experiences an equal and opposite force, propelling it forward. This is possible due to the conservation of momentum, which ensures that the total momentum of the system remains constant.

3. What factors affect linear momentum conservation in rockets?

There are several factors that can affect linear momentum conservation in rockets. These include the mass of the rocket and the exhaust gases, the velocity of the exhaust gases, and the duration of the rocket's engine burn. Additionally, external forces such as air resistance and gravity can also impact the conservation of momentum in rockets.

4. How does linear momentum conservation impact rocket design?

Linear momentum conservation plays a significant role in rocket design. Engineers must carefully consider the mass of the rocket, the amount and velocity of the exhaust gases, and the duration of the engine burn in order to achieve the desired change in momentum. Additionally, the design must also account for external forces that may disrupt the linear momentum conservation, such as air resistance and gravitational pull.

5. Can linear momentum conservation be violated in rocket propulsion?

Linear momentum conservation is a fundamental law of physics and cannot be violated. However, it is important to note that the momentum of the rocket and the exhaust gases can change due to external forces. For example, if the rocket collides with an object, the conservation of momentum may be disrupted. However, this does not violate the principle itself, but rather showcases the impact of external forces on the system's momentum.

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