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2 people throwing balls at each other on cart - cons. of momentum

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Emil and Kerry are standing on a stationary cart with frictionless wheels. Total mass of cart and riders is 130 kg. At the same instant, Emil throws a 1.0 kg ball at 4.5 m/s to Kerry while Kerry throws a 0.5 kg ball to Emil at 1.0 m/s. Emil's throw is to the right and Kerry's to the left. a) While balls are in the air, what are the speed and direction of the cart and its riders? b) After balls are caught, what are speed and direction of the cart and riders?


    2. Relevant equations
    mcartvcart initial+mEmil ball(Eb)vEb initial-mKerry ball(Kb)vKb initial = mcartvcart air+mEmil ball(Eb)vEb air-mKerry ball(Kb)vKb air
    Conservation of momentum essentially (mvinitial=mvfinal)

    3. The attempt at a solution
    So for (a) the entire left hand side goes to zero since the cart is stationary and the balls haven't been thrown. This allowed me to solve for vcart air, which meant I ended up with [(0.5*1)-(4.5*1)]/130, leaving me with a -0.031 m/s or essentially 0.031 leftwards.

    I'm having a bit of trouble with (b) though, provided that (a) is correct. Would I essentially solve mcartvcart air+mEmil ball(Eb)vEb air-mKerry ball(Kb)vKb air = mcartvcart final+0+0--zeroes since the balls are caught and not moving--for vcart final? I would get the answer of -2.31x10-4. Is that the correct way of thinking because what I was also thinking is that I could just set the initial left side equal to the final rather than use the balls in air equal to the final and that would get me an answer of 0 m/s in no particular direction.
     
  2. jcsd
  3. Mar 26, 2014 #2

    Simon Bridge

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    Welcome to PF;
    Lets call them E and K for short.
    ... it can help to keep the subscripts to a minimum so I have illustrated a way to do this below.
    What you've done is OK, just hard to read.

    M=130kg.
    Notice how I've eliminated a subscript?

    Note: I'm guessing they mean that the mass of each rider+cart is 130kg and not that they are 130kg together?

    so we know the momenta carried by the balls:
    ... tells us the signs:

    E's ball: ##p_E = +4.5 \text{ kgm/s}## ##m_E=1.0\text{ kg}##
    K's ball: ##p_K = -0.5 \text{ kgm/s}## ##m_K=0.5\text{ kg}##

    We may need the masses later.

    Conservation of momentum twice.

    Now to the nitty gritty:
    ... don't forget units. Why did you subtract one of the balls? Each guy is only holding the one to start with.

    I'll model one of the calculations for you - the trick is to be careful about what the initial and final situations are. To get this clear: draw a quick sketch.
    Then it is a matter of following the template below:

    (a) throwing the ball:

    Emile before (stationary holding ball):
    ##p_b=0##

    Emile after (both moving, ball in flight):
    ##p_a=Mu+p_E##

    conservation of momentum:
    ##P_b=p_a\implies Mu+p_E=0##

    solve for u:
    ##u= -p_e/M = -4.5/130 =0.035\text{ m/s}## (2 sig fig)

    conclude: Emile moves at 3.5cm/s to the left.

    Now you finish up this one:

    Kerry before (holding his ball, both stationary):
    ##p_b=0##

    Kerry after:
    ##p_a=Mu+\cdots##

    conservation of momentum:
    ##p_b=p_a\implies \cdots##

    solve for u:
    ##u= \cdots##

    conclude: Kerry moves at _____cm/s to the _____.

    ... same template as above, but this time the initial momentum is not zero.
     
  4. Mar 26, 2014 #3

    BvU

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    Hi doc,
    I want to help a little here, because I think you are doing fine but Simon interprets the exercise as if K&E are on separate carts. I (and you) tend to read "Total mass of cart and riders is 130 kg" to mean that there is only one cart involved.

    For the one-cart case I can see no error in your reasoning in (a). The minus sign appears because you interpret v as positive and add the sign to indicate that the direction is opposite. And then outcome < 0 means: to the left.

    Now for (b) you can use exactly the same equation, but you have to be aware that vEb initial is now the air speed (idem vKb initial) and on the righthand side all final speeds are the same. Is the result surprising ?

    Now for some nitpicking: what does the 4.5 m/s mean ? Is that wrt the thrower or wrt the air ?
    Your (a) answer shows you think the latter. Would it make a difference for (a) if the first was true? and for (b) ?


    In the case of two carts (and heavy throwers!) both throwing and catching increases the speed at which they separate. Nice difference!
     
  5. Mar 26, 2014 #4
    Yes, I think it should be only one cart rather than two. Simon's answer scared me for a moment until I realized how he worked through it.

    So then, I'd get something like -mcvc*air+mEBvEB-mKBvKB=mc,KB,EBvfinal, where vEB is still 4.5 m/s and vKB is 1.0 m/s and the negative means going leftwards. Solving that I get -2.28x10-4m/s, which means the cart will move leftward just a smidge.

    I guess I'm trying to figure out why the final velocity can't just be zero after the balls are caught (It's not a question in the problem; merely a question that came up in my mind).

    I think it would make a difference if the speed is with regards to the thrower, right...? I get too confused when it comes to frame references. :(
     
  6. Mar 26, 2014 #5

    Simon Bridge

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    I was worried about that, which is why I included the caveat at the start:
    ... and I guessed wrong :) (Could also have meant that there were two carts which, together, massed 130kg ... argh!)

    The template approach should still work ... only one cart answers my question:
    ... I'm glad it didn't confuse you too much.

    Note - It is usual to assume an inertial observer unless otherwise mentioned.
    Some observers make more sense than others.

    You could reason thusly: we have three steps:
    in step 1. cart (with E and K) and both balls stationary.
    in step 2. both balls in the air, cart has velocity u - probably not zero.
    in step 3. both balls and the cart at the same velocity v - possibly zero.

    step 1: total momentum = 0, because everyone stationary wrt inertial observer O.
    I'll do everything in O's reference frame. Lets fix O to the ground.

    step 2: total momentum = 0, because it was zero at step 1.
    If the cart moves at velocity u, then ##u=(p_K-p_E)/M## (what you did right?)

    step 3: total momentum = 0, because it was zero at step 2.
    If the cart+balls moves at velocity v, then doesn't that mean: ##(M+m_K+m_E)v=0\implies v=0##
    (this is basically your question right? To answer it: compare the above calculation with the one you used.)

    Note: In the above, M= mass of cart and occupants, mX = mass of ball thrown by person X, and pX = momentum of ball thrown by person X (according to inertial observer O.)

    Notice:
    We could fix observer O to the cart, or to the center of mass, or something else.
    The question does not specify so, technically, you are free to make any choice you like... provided you say what choice you made. I'd pick whichever makes the math easier me.
     
  7. Mar 27, 2014 #6

    BvU

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    Strange, since before you had
    -mcvc*air+mEBvEB-mKBvKB=0 exactly.

    Did you do ( -0.031 * 130 + 4 ) / 130 ? That gives me a smidge of -2.31x10-4m/s.

    But (-4/130 * 130 + 4) / 130 = really zero
     
  8. Mar 27, 2014 #7
    Yes. Although instead of dividing by 130, I divided by 131.5 kg, which comes from adding the weight of the balls on the cart.

    Hmm. I'm lost a bit now. So the velocity of the cart after the balls are caught is zero then? I'm not completely sure how you're getting -4/130*130+4...
     
  9. Mar 27, 2014 #8

    BvU

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    -4/130 was what you did to calculate v after throwing. Rounding off gave 0.031. If you round off, you can't dig up someething as small as the "smidge" any more... My eternal argument to avoid numbers if expressions can be used..!


    +4 is the net momentum the balls give back when caught.

    The 131.5 or 130 doesn't matter: (-4/130 * 130 + 4) / 131.5 = also really zero
     
  10. Mar 27, 2014 #9
    Ah I see. hah

    This is the part where I'm having a bit of trouble with right now. How would I end up with a +4 in momentum? Conceptually, it makes sense, but mathematically, I'm not sure how a +4 would come up, other than saying Emil and Kerry's momentum switched (mvEmil=-0.5 and mvK=4.5 on the right side of the conservation of momentum equation?)

    Nevermind. I figured out what you were saying all along. Wow. Staring at me right in the face...Thanks for the replies and the help BvU and Simon!
     
    Last edited: Mar 27, 2014
  11. Mar 27, 2014 #10

    BvU

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    Well done. If you want to torture yourself (or if you want to go on in physics, mechanics or the like, which is pretty equivalent :smile:), work it out once more for the case the throwing speed is wrt the thrower. (Small difference in a, no difference in b).
     
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