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Conservation of linear momentum of rocket

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    The figure below 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 = 6.00 kg) and blocks L and R (each of mass m = 2.50 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 2.60 m/s relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time t = 0.80 s, block R is shot to the right with a speed of 2.60 m/s relative to the velocity that the explosion gives to block C?

    (a) What is the velocity of block C at t = 2.80 s?

    (b) What is the position of block C's center at that time?

    2. Relevant equations

    3. The attempt at a solution

    I know how to conserve momentum for the different pieces of the problem; however, if someone could help me approach the problem it would be greatly appreciated. I believe I need to find the velocity of each piece after the explosion but am unsure of how to go about finding it. Are the individual velocities required, or do simply use the relative velocity between the two? Any hints on how to approach would be greatly appreciated!
  2. jcsd
  3. Nov 29, 2008 #2
    Dear mbisCool,

    I take it you do know the equation that describes the general case of the conservation of linear momentum. Perhaps this question may help you start: what is the total momentum of the system before L is shot? Remember, this is an explosion.

  4. Nov 29, 2008 #3

    Might it help to consider the TOTAL MOMENTUM before the explosions and the TOTAL at the end bearing in mind the 'explosions' are internal to the system.
  5. Nov 29, 2008 #4
    I know the total momentum for the system in conserved but i am not sure where to go from there as the individual pieces of the system do not conserve momentum. Sorry if its painfully obvious too :)
  6. Nov 29, 2008 #5
    It's OK... Let's approach it step-by-step. Since no other forces act on the system, overall momentum is conserved. Hence, it would right to say that the total momentum of C and the launched object will be equal to the momentum before the launch.

    That brings us to the question, what is the total momentum of the system? We know that the system of C,R and L were at rest at the beginning (ie before the launches), what does that tell you about the total momentum of the system?

    In each launch, we know the mass of the launcher and the object, right? We know the total momentum, and we also know the speed of the object. So, you could find the speed of the launcher after each launch. Is it clearer now?

  7. Nov 29, 2008 #6
    Well then my question is regarding the velocity. I know the relative velocity between the two is 2.6 however this is ambiguous. My question was can you simply use 2.6 m/s? this could be anywhere a range of different velocites with a difference of 2.6. I would need to solve for 8.5kg * V1 = 2.5kg * V2 and the relative velocity would be 2.6
  8. Nov 29, 2008 #7
    Ok, it may seem a little ambiguous that the relative velocity is given. But don't panic. Let the velocity of the launcher be v (relative to the ground), and we know for certain that the 2 bodies are moving in opposite directions. If that is so, it is apparent that the velocity of the launched object is -(2.6 - v), since the magnitude of the relative velocity must be higher than the "absolute" velocity. Substituting these 2 values into the equation will give you an equation in 1 unknown, so it's solvable.

    Last edited: Nov 29, 2008
  9. Nov 29, 2008 #8
    Much appreciated ^^
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