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Problem regarding linear momentum

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    See the image: http://postimage.org/image/xnim2zumv/ [Broken]

    If you want to enlarge the image, just left click on it...

    2. Relevant equations

    Conservation of linear momentum , when no net external force acts in a particular direction.

    3. The attempt at a solution

    I have already done parts (a) and (b).. So please I want assistance in parts (c) and (d). Kindly start from part (c)..

    In part (a) I got u= mv/(M+m) which is the speed of the small block, when it was travelling on the vertical part of the larger mass.

    In part (b) I got v1= [{(M2+Mm+m2)v2/(M+m)2} - 2gh]1/2 , which is the speed of smaller block when it breaks off the larger mass at height h.

    Ok, so here is my attempt for part (c),

    Extra height h' which smaller block ascends above larger mass is,

    h'=v12/2g = {(M2+Mm+m2)v2/(M+m)22g} - h

    Thus total height above ground which smaller block ascends is :

    H=h+h' = {(M2+Mm+m2)v2/(M+m)22g}

    But this is the wrong answer as per the book. Where did I go wrong ?

    Please help !!

    Thanks in advance...:smile:
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 13, 2013 #2


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    That looks like the correct equation for part a's answer. But it's the speed of the large block, not the small block. It is however the horizontal component of the small block's velocity (if I'm not mistaken).
    That looks about right to me. :approve:

    Just keep in mind that the above is the small block's total speed: A component of it is in the horizontal direction, and another component is in the vertical direction.
    Wait, something is going wrong here I think. You're treating the small block as though it's shooting straight up, but only part of the small block's velocity is in the vertical direction. You need to find the vertical component of the small block's velocity. :wink:
    Last edited by a moderator: May 6, 2017
  4. Mar 14, 2013 #3
    Thanks for reply collinsmark !!

    Yes, sorry..


    Ahhh yes...

    Ok, so

    Vertical component is given by,


    I successfully found vertical component using above.

    Then treating upper part to be symmetric projectile,

    H=vy2/2g + h
    H= Mv2/2g(M+m)

    Which is the correct answer!! Thanks collinsmark !! :smile:

    Ok, now to part d.

    Velocity of large mass is u due right. Then shall I find the total time of flight ?
  5. Mar 14, 2013 #4


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    Or left, I think.
    That's the next thing I would do, yes.
  6. Mar 15, 2013 #5
    Sorry... The second thing the question asks is ---> "Show that the smaller block will again land on the bigger mass."

    If this were true, then how can large has velocity due left ? In starting, small one had velocity due left, so by conservation of linear momentum due horizontal, large mass should move due right.

    The smaller block can again land on the bigger mass, if horizontal velocity of the smaller one, i.e. horizontal component of its velocity is same as the velocity of larger mass and is directed parallel to it. Correct ?
  7. Mar 15, 2013 #6


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    I agree that the the original problem statement was poorly worded when it comes to this. I think it could have cleared up some potential confusion by elaborating was it meant by "The smaller mass is pushed on the longer one at speed v and the system is left to itself."

    When I first read that, I didn't know what it was trying to say. Was it pushed? Or was it left to itself? You can't have both!

    So after thinking about it, here is what it's trying to say: When the smaller mass is on the horizontal section of the longer mass, it is pushed to the left until it reaches speed v. Once the smaller mass reaches speed v the pushing stops, and the system is left to itself. The entire pushing process is completed before the smaller mass leaves the horizontal section of the longer mass.

    I'm pretty sure that's the correct way to interpret it, since if the correct interpretation was any other way, the answers we've been getting would not have been correct so far.

    Okay, now lets talk conservation of linear momentum.

    Conservation of linear momentum doesn't apply while the pushing is taking place, since that's an outside, external force. (It does apply after the pushing stops though.)

    Conservation of linear momentum doesn't apply to the y-direction, because we're treating the normal force from the ground to the longer mass as an external force. (We could instead apply conservation of momentum and take into account the change of momentum of the entire Earth, but that would be getting silly. It's just easier to model it as an external force.)

    So when conservation of linear momentum starts, in the x-direction, the small mass is moving to the left at speed v. The longer mass is at rest.

    When the smaller mass "breaks off" the larger mass as height h, it has a x-component [of velocity] that is still to the left (although less than the original v), and the longer mass is also moving a little to the left.

    Since the top part of the longer mass is vertical, and the longer mass is sloped such that the smaller mass slopes up to it gradually; there is no true "collision" (so to speak)*. When the smaller mass reaches the vertical section of the larger mass there is no longer a force between the two, and the velocity of the larger mass, and the x-component of the smaller mass are identical -- both moving to the left. And this speed is what you have calculated as u.

    Yes, that is correct.

    (Regarding the vector directions: the larger mass is only moving along the x-direction. The x-component of the smaller block is along the x-direction by definition. So those two vectors are parallel, yes. [Edit: well, technically both vectors are actually pointing in the negative x direction, but you know what I mean.])

    *(Edit: Regarding what I said about there not being a "collision": This is not entirely true. The entire process, if extended out longer, after the smaller mass lands back on the larger mass and then slides all the way back down, can be treated as one, big, elastic collision. Don't concern yourself with this now though. First solve the problem and come back to this later if you are curious. But if you are curious [after you solve all the parts of the original problem statement], continue to calculate the velocities of both masses all the way to the very end, after the smaller mass slides back down the larger mass, and then off the larger mass altogether. You should find that the final velocities are exactly as if a perfectly elastic collision had occurred. [This of course ignores all friction, including air resistance.])
    Last edited: Mar 15, 2013
  8. Mar 15, 2013 #7
    Yes, thank you. I actually understood it before. I don't know what happened. I just ignored the sign conventions while using conservation of linear momentum.:smile:

    However we know very special and very unique case in which conservation of linear momentum does hold even in the presence external force. And that is when time interval is infinitesimally small.

    By impulse-momentum conservation :

    Pf=Pi+ FΔt

    As Δt→0, and F is not very large, we get very nearly Pi almost equal to Pf, and that small difference can be very well neglected, as per my textbook.

    Ok, I am very grateful to you for providing such a wonderful explanation. If we take earth + 2 masses as a system, then we can apply conservation of linear momentum in y direction, but for that we require earth's upward velocity, which we do not know, so that will not work.

    Ok, so this already proves that small block will land on the large mass again. Now I'm going to find the distance traversed by large mass during this interval, but tomorrow. Its already 1:45 am here. I am off to bed. Edit : Since its "am" , its tomorrow only..

    Thanks once again. :smile:
    Last edited: Mar 15, 2013
  9. Mar 15, 2013 #8


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    Yes, impulse (force times time) equates to change in momentum. And as a matter of fact, Δt doesn't even need to be small for this to apply [for a uniform force]. But let's not use that here, since it's not necessary.

    But as long as we're speaking hypothetically, one could model the "pushing" by acknowledging that the person pushing is attached to the earth, and then modeling that as the small block is pushed to the left, the person and the entire earth move to the right a little bit. That way conservation of linear momentum even apply during the pushing. But again, let's not do that. That's just getting silly. It's much easier to model the pushing as an external force.
    Actually conservation of momentum will work in that case, but it's just overkill for what we want to do.

    Suppose you are standing at rest on the Earth. We define our frame of reference such that the total momentum at this point is 0. Now suppose you jump up in the air with some initial velocity. What actually happens is when you do that, the Earth moves down a little with some initial velocity of its own. The Earth's initial velocity is much, much smaller in magnitude than yours only because the Earth has such a larger mass. But if you sum your mass times your velocity, together with the Earth's mass times the Earth's velocity (keep in mind the velocities are in opposite directions), they sum together as 0. Momentum is conserved. (In the process of doing this you must first calculate the barycenter of you and the earth, and the velocities and positions are with respect to this barycenter.)

    Of course we don't want to do that for this problem. That's complete overkill. This is the sort of thing that is necessary when calculating the orbits of celestial bodies around each other, when their masses are comparable. But that's way to much for what we want to do here. I'm just saying that we could apply conservation of momentum in the y direction, in principle. Not that we should.
    Goodnight, and good luck!
    Last edited: Mar 15, 2013
  10. Mar 17, 2013 #9
    Thank you Collin.

    Ok, so, time of flight is T= 2vy/g

    I calculated that successfully.

    Then I calculated distance traversed by large mass as,

    D= uT, u being the velocity of large mass left, which I found in part (a).

    And I got the correct answer !! Thanks collinsmark.

    I did all the calculations in copy. It would be really messy and time consuming if I posted them here.:smile:
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