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Kinematics Rocket Question

  1. Jun 18, 2014 #1
    1. The problem statement, all variables and given/known data

    A rocket is moving in a gravity free space with a constant acceleration of 2 m/s2 along + x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in + x direction with a speed of 0.3 m/s relative to the rocket. At the same time, another ball is thrown in -x direction with a speed of 0.2 m/s from its right end relative to the rocket. The time in seconds when the two balls hit each other is ?

    Image: http://postimg.org/image/fh9trs9r5/

    The answer ranges from 0 to 9.

    2. Relevant equations

    http://en.wikipedia.org/wiki/Relative_velocity

    3. The attempt at a solution

    Here is what I did.

    If rocket accelerates by 2 m/s2 in +x direction, then balls inside it also accelerates with it with same value and in same direction in ground frame. Thus when we switch to frame of the rocket there is no acceleration on balls. Inside that frame I impose the concept of relative velocity and convert this two body problem to one body.

    Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

    Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.

    This is the correct answer.

    However exam in which this question was asked says that both 8 seconds and 2 seconds answers are acceptable.

    One book offers solution as :

    Maximum displacement of the left ball from the left wall of the chamber is 2.25 cm, so the right ball has to travel almost the whole length of the chamber (4m) to hit the left ball. So the time taken by the right ball is 1.9 sec (approximately 2 sec).

    Now my doubt is as follows :

    How is 2 seconds a correct answer ? Also how can there be "two" values of time of collision ?

    Please help !!

    Thanks in advance... :)
     
    Last edited: Jun 18, 2014
  2. jcsd
  3. Jun 18, 2014 #2

    adjacent

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    How do you know that? If the ball is thrown from one end, then why should it accelerate along with the rocket?
     
  4. Jun 18, 2014 #3
    Observe from ground frame yourself.

    See, if you are sitting in an accelerated train suppose, and you throw a ball vertically, it comes back to your hand right ? This is because it is accelerated with the train.

    Waiting for someone else to reply though.
     
  5. Jun 18, 2014 #4

    adjacent

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    I know that. I actually forgot about the "relative" word in your question
     
  6. Jun 18, 2014 #5
    No problem.

    Just waiting for someone to reply and help me. xD

    It's confusing question.. :confused:
     
  7. Jun 18, 2014 #6
    I think Im going to have to refute that. As long as the ball is in your hand while you are on the accelerated train, then yes it is accelerating, but as soon as you throw it what forces are acting on it? None. The train keeps accelerating due to some force, but the ball being in air is subject only to its initial velocity and g with respect to the ground, no? If the train is going at a constant velocity, then the relative velocity is zero, yet if the trian is accelerating with acceleration [itex] \hat{a} [/itex], then the balls acceleration relative to the train is [itex] -\hat{a}[/itex].

    EDIT: Remember: [itex] \hat{a}_{\text{1,2}} = \hat{a}_{\text{1,g}} - \hat{a}_{\text{2,g}} [/itex]
     
    Last edited: Jun 18, 2014
  8. Jun 18, 2014 #7
    This problem can be solved for either the rocket frame of reference or for the fixed frame of reference. Both approaches will give the same answer. Which frame do you prefer?

    Chet
     
  9. Jun 18, 2014 #8
    Hello Chet ! :)

    I would like to solve the problem with both approaches. First I will do rocket frame approach. But do you not think I already solved it in my attempt at solution using rocket frame ? :confused:
     
    Last edited: Jun 18, 2014
  10. Jun 18, 2014 #9
    In the rocket frame of reference, the system behaves as if there is a gravitational acceleration of 2 m/s2 in the negative x-direction. So, for the ball thrown from the left end at x = 0, the ball's location at time t is xl = 0.3t - t2. For the ball thrown from the right end at x = 4, its location at time t is xr=4 -0.2t - t^2. The balls meet when xl=xr.

    time left ball right ball

    0 0 4
    1 -0.7 2.8
    2 -3.4 -0.4
    3 -8.1 -5.6
    4 -14.8 -12.8
    etc.

    These results make no sense. Are you sure about that acceleration? Could it be 0.2 m/s2? With this acceleration, the left ball falls back to the left wall in a very short time.

    Chet
     
  11. Jun 18, 2014 #10
    The left ball falls back to the left wall every 0.3 seconds, and it never gets further from the wall than 0.0225 m, even if it bounces elastically. So the question really boils down to how long it takes for the right ball to fall to ~ 0.

    Chet
     
  12. Jun 18, 2014 #11
    Hi Sankalp

    Nice question :)

    This approach is not correct because as a result of collision of the left ball with the left wall of the rocket ,the velocity of the left ball changes after every 0.3 seconds (assuming elastic collision) .

    I think 8 seconds cannot be the correct answer .OTOH approx. 2 seconds looks alright.
     
  13. Jun 19, 2014 #12

    ehild

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    NO.

    It is true if the train travels with constant velocity.

    In an accelerating train, the released ball moves backward. Just like you if you stand and do not grab something fixed.

    ehild
     
  14. Jun 19, 2014 #13

    ehild

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    Very good!

    ehild
     
  15. Jun 19, 2014 #14
    Thanks to all. :)

    I think I figured it out.

    From the rocket frame we have acceleration of 2 m/s2 in -x direction on each ball.

    So maximum displacement traveled by left ball is u2/2a = 0.025 m.

    Note that acceleration is still acting on left ball in -x direction.

    Time taken :

    0=0.3-2t
    t=0.15 s.

    In this time:

    Distance by right ball :

    -s=-0.2t-t2

    s= 0.05 m.

    Now 0.08 + 0.05 <<4

    The right ball has to traverse a distance of 4-x, where x<0.02..

    Hence approximating 4-x≈4 we have

    -4=-0.2t-t2

    This gives t=1.9 s ≈ 2 s...

    Thanks to Chet.. His third post explains the reason for this theoretically.

    Now why exam accepted 8 s ?

    That's my reasoning :

    If the balls are "kept on the floor of rocket cabin", then they have the same acceleration as rocket. Hence from rocket frame there is no acceleration on balls. Inside that frame I impose the concept of relative velocity and convert this two body problem to one body.

    Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

    Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.


    So both answers are correct.

    Correct ?
     
    Last edited: Jun 19, 2014
  16. Jun 19, 2014 #15
    Thanks :smile:
     
  17. Jun 19, 2014 #16
    I haven't been able to follow your reasoning.

    That is still the same situation as in OP .Keeping the balls on the floor cabin do not give the balls same acceleration as that of rocket.

    I think there is only one correct answer and that is 2 seconds.

    What is 0.02 on the LHS ?
     
  18. Jun 19, 2014 #17
    Why ? When you are sitting in the train, you are accelerated with the train. When you throw the ball up, then that acceleration doesn't count with it.


    I can't say that 8 s cannot be the answer.

    0.02 is the distance traversed by the left ball when it velocity becomes zero due to deceleration.
     
  19. Jun 19, 2014 #18
    OK.

    Assuming frictionless surface ,balls motion would be independent of the motion of the rocket .

    In the train example ,when the person throws the ball up ,the acceleration of the ball will still be 9.8ms-2 . But if you are working from the train reference then surely you need to take into account the pseudo force .

    But accounting for pseudo force is not the same thing as if the ball is having horizontal acceleration of the train in addition to its vertical gravitational acceleration.

    If 2 sec is the correct answer ,what does 8 seconds signify ?
     
  20. Jun 19, 2014 #19
    More to the point.

    8 s should signify that the right ball is taking approx. 6 seconds to move from left to right and then again left.

    It has already taken 2 s to move from right to left initially.

    Assume that left ball is fixed at x=0 for the sake of approximation.

    Correct ?
     
    Last edited: Jun 19, 2014
  21. Jun 19, 2014 #20
    Have you calculated the time it takes for the right ball to move from left to right end ,and left again ?

    Does it come out to be 6 sec ?

    I am curious to know what makes you so sure that 8 seconds is the correct answer :smile: .The answer key may be wrong .Which book are you referring to ?
     
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