1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding final velocity and position of a two ended rocket

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M = 6.40 kg) and blocks L and R (each of massm = 1.90 kg) on the left and right sides. Small explosions can shoot either of the side blocks away from block 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 2.80 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.90 s, block R is shot to the right with a speed of 2.80 m/s relative to the velocity that block C then has (after the second explosion). At t = 2.80 s, what are (a) the velocity of block C (including sign) and (b) the position of its center?



    2. Relevant equations


    3. The attempt at a solution
    Okay so first I need to find the velocity of the center and right blocks after the explosion at t=0
    Mass of right block(relative velocity- velocity of center and right blocks)=combined mass of right and left blocks*velocity of right and left blocks
    1.90kg(2.80m/s-v)=8.30kg*
    v=0.523 m/s
    Taking that velocity and multiplying by 0.9 seconds gives the position of the center of the center block at the time that the second explosion occurs
    0.523m/s*0.9s= 0.4707m

    Next is to use the same method to find the velocity of the center block after the second explosion, I will skip the equation modeling as it is redundant and crude as I am unable to use notation.
    1.90kg(2.80m/s-v)=6.40kg*v
    v=0.6410 m/s but in the negative x direction so v=-0.6410 m/s
    Now this is where I became uncertain. Is this velocity of -0.6410 m/s the new velocity of the center block and as it is a frictonless surface the answer to part a or should it be added the the original velocity of 0.523 m/s?
    v=-0.6410m/s
    or
    v=-0.6410m/s+0.523m/s=-0.1703m/s
    That would also affect part b as it would be
    0.4707m+-0.6410(2.8-0.9)=-0.7472 m
    or
    0.4707m+-0.703(2.8-0.9)=0.14713m

    The problem being that I entered both sets of solutions and both proved to be incorrect. What is wrong with my approach of using the conservation of momentum?
     
  2. jcsd
  3. Oct 25, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I take it the 8.3 is a typo. I agree with the 0.52 (but 0.522, not 0.523).
    The idea is right, but check the arithmetic.
     
  4. Oct 25, 2016 #3
    Okay so my problem was that instead of adding the velocity from the first explosion to the velocity of the second, I added the position of the first explosion to the second. This was just me quickly glancing at the wrong number. I got the final velocity to be -0.119 m/s and the final position to be 0.244 m
     
  5. Oct 25, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks right.
     
  6. Oct 25, 2016 #5
    Haruspex comes to my rescue once again
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding final velocity and position of a two ended rocket
Loading...