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At which part of the height of the building the balls collided?

  1. Nov 9, 2005 #1
    Help Needed Asap!!! Please!

    Hi!

    I need to solve the following problem before tomorrow.. otherwise I'll be in big trouble. :uhh: I've already discussed it with classmates, but they also don't know how to handle it. Please help me!
    Ok this is the question:

    A ball A is thrown off a building and at the SAME time a ball B is thrown up from the ground. When both balls hit eachother, they move in OPPOSITE direction and the velocity of ball A is twice the size of B. At which part of the height of the building the balls collided?
     
  2. jcsd
  3. Nov 9, 2005 #2

    Astronuc

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    Staff: Mentor

    So VA= 2*VB when they hit.

    Both balls traverse some distance in the same time T (to be determined) because they are released at the same time, and they collided at the same time.

    Now ball A is thrown (or dropped ?) and is accelerating down, while ball B is thrown up and decelerating, and the rate of acceleration g and rate of deceleration is -g.

    Can you write an equation for each ball, remembering v(t) = at + vo, and x = 1/2 a t2, assuming constant acceleration, where a = g or -g depending on acceleration or deceleration.
     
  4. Nov 10, 2005 #3
    Hi!

    Thank you for your help! I found the fraction where the collision took place to be 2/3 of the buildings height, assuming ball A had no initial velocity. Is that correct? (by the way: that wasn't mentioned in the problem, but if I use a v0A I can't find a proper fraction (without an initial velocity in it).
     
  5. Nov 10, 2005 #4

    Astronuc

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    I haven't had time to work through it, but I believe that is correct.

    I was going to look at this problem this evening.
     
  6. Nov 10, 2005 #5
    Thanks a hell of a lot! You are doing a great job helping people out with their homework! Keep up the good job ;) !
     
  7. Nov 10, 2005 #6

    Astronuc

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    Just curious, what equations of motion did you use?

    There are 4 equations/relationships here.

    Two equations for position or height of ball A and ball B, and two equations for velocity.

    There are two constraints. VA(T) = 2*VB(T), where T is the time of contact assuming a reference time of t=0.

    And height of building H = hA + hB at t=T, they meet at approximately the same location - actually their CM's are separated by rA and rB. hA is the distance A travels to meet B at hB. You can find either h and the proportion hB/H.
     
  8. Nov 11, 2005 #7
    Well I'll show you what I did to get to the answer of 2/3 and could you please tell me if this is the correct way ? :wink: o:) :

    First I formulated two equations for the velocity of ball A and ball B at a certain time t, so:

    v(t)A= gt
    v(t)B= v0B-gt

    The balls collide when v(t)A=2 v(t)B so when gt= 2v0B-2gt .
    From this equation the time t when the collision occured is obtained:

    gt = 2v0B - 2gt
    3gt= 2v0B
    t= 2v0B/3g

    The height ball A has travelled from the top of the building is given by
    x(t)A= 1/2 gt^2

    And the height ball B travelled from the ground is:
    x(t)B= v0B t - 1/2 gt^2

    Since the collision occured at a time t= 2v0B/3g the height from the ground where it happend is obtained by substituting t= 2v0B/3g in the x(t)B formula:

    x(2v0B/3g)B= v0B (2v0B/3g) - 1/2 g(2v0B/3g)^2=
    ((2v0B^2)/3g)- (1/2g((4v0b^2)/9g^2))= ((2v0B^2)/3g)- ((4v0b^2)/18g)= ((12v0B^2)/18g)- ((4v0b^2)/18g)= ((8v0b^2)/18g) =((4v0b^2)/9g)

    Ok so now I only need to know the height of the building, which is given by the height ball B travelled from the ground till the collision occured (x(2v0B/3g)B) + the height ball A travelled from the top of the building till the collision (x(2v0B/3g)A) occured. So h= x(2v0B/3g)B + x(2v0B/3g)A. x(2v0B/3g)B was already calculated, so:

    x(2v0B/3g)A= 1/2 g(2v0B/3g)^2 = (1/2g((4v0b^2)/9g^2))= ((4v0b^2)/18g)=((2v0b^2)/9g).

    The height of the building h= ((2v0b^2)/9g) + ((4v0b^2)/9g)= ((6v0b^2)/9g)

    The fraction of the height of the building where the collision occured is given by x(2v0B/3g)B / h = ((4v0b^2)/9g)/((6v0b^2)/9g) = 2/3
    (seen from the ground!, because x(2v0B/3g)B = the height where the collision occured measured from the ground)

    Now this last thing was actually my biggest doubt. I've talked to many fellow students who thought 1/3 was the height where the collision occured seen from a ground perspective, because they thought: ball A has twice the velocity of ball B when they meet, so ball B had travelled 1/3 from the ground and ball A 2/3 from the top of the building, therefore the collision occured at 1/3 of the height of the building above the ground.... I'm not very sure what's the correct answer.... What do you think?
     
  9. Nov 11, 2005 #8
    well fration works out to be 2/3 h from the ground
    lets assume the distance ball be travels befor collision be D which is equal to
    D= (4(v0b^2)/9g), ...........1
    and then the distance travelled in same time by ball a is
    H-D = (2(v0b^2)/9g)...............2
    from 1 and 2
    H = (6(v0b^2)/9g)
    fraction.....would be D/H which works out be 2/3, that mean ball B has to climb 2/3 of the height of the building.

    that also means that ball A has to drop to 1/3rd of the hieght of the building befor both ball collide.(doesn't matter where u are lookin from)
     
  10. Nov 11, 2005 #9
    Thank you! I already thought so! :D
     
  11. Nov 11, 2005 #10

    Astronuc

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    The equations of motion of the balls are correct, assuming ball is dropped, and the rest appears to be OK. :smile:
     
  12. Nov 11, 2005 #11
    Btw concerning the usage of an initial velocity for ball A: today I heard from a few fellow students that the problems we had to solve were translated from English to Dutch (I live in the Netherlands) and my teacher translated the phrase 'the ball was dropped' as 'de bal is gegooid', which literally means 'the ball was thrown', because 'gegooid' mean thrown and 'gevallen' means dropped. :surprised
    Ohhh I hate my teacher for doing that! It caused me to overwork till dawn! Anyways thanks again for your help!!!
     
  13. Nov 11, 2005 #12

    Astronuc

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    Yeah, but it made you think! :biggrin:

    Actually that is a nice problem. It's a bit like two trains coming at one another in opposite directions, possibly starting at different times, perhaps with differences in initial velocity and acceleration.

    Here the key was knowing about the acceleration and the time.
     
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