Linear momentum conservation - rocket

  • #1

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 dont really understand whats happening in the problem
 

Answers and Replies

  • #2
tiny-tim
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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 dont really understand whats 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
tiny-tim
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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
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
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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
tiny-tim
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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
tiny-tim
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so could i say that m1v1i = 2kg * 3m/s?

Nooo …
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, dunno 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
tiny-tim
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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:
 

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