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Elastic collision between ball and block

  1. Apr 25, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A block with large mass M slides with speed V0 on a frictionless table towards a wall. It collides elastically with a ball with small mass m, which is initially at rest at a distance L from the wall. The ball slides towards the wall, bounces elastically, and then proceeds to bounce back and forth between the block and the wall.

    a) How close does the block come to the wall ?
    b) How many times does the ball bounces off the block, by the time the block makes its closest approach to the wall?
    Assume that M >> m, and give your answers to leading order in m/M.

    2. Relevant equations

    http://www.luiseduardo.com.br/mechanics/conservationlaws/conservationlawsproblems_arquivos/image023.jpg [Broken]

    3. The attempt at a solution

    Let the velocity of block and ball be v1 and v2 after collision. Applying conservation of momentum and law of restitution,

    [itex]v_1 = \dfrac{M-m}{M+m} v_0 \\
    v_2 = \dfrac{2M}{M+m} v_0 [/itex]

    Using approximations
    [itex]v_1 = v_0 \\ v_2 = 2v_0 [/itex]

    At distance of closest approach the velocity of bigger block should become zero. But I can't think how to find that distance as the velocity of bigger block is not constant and changes after every collision.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 25, 2014 #2
    You're gonna have to be a bit cleverer that that aren't you. Start by making a better choice of notation. Don't call the speeds after the first collision v1 and v2. Call them V1 and v1. In general Call the speeds after the ith collision Vi and vi. Your job is, given Vi and vi, find Vi+1 and vi+1. Off course, you also need to keep track of Li - the distance from the location of the ith collision and the wall. Also, don't be so fast at making hasty approximations.

    Come back after you've done that.
     
  4. Apr 25, 2014 #3

    utkarshakash

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    [itex]V_{i+1} = \dfrac{M-m}{M+m} V_i - \dfrac{2m}{M+m} v_i [/itex]

    Let Li be the distance from the point of ith collision to wall. By the time i+1 th collision occurs, let the block travel a distance Δx.
    ∴Li+1 = Li - Δx

    In terms of Li
    [itex]L_{i+1} = L_i \left( 1- \dfrac{2V_i}{V_i+v_i} \right) [/itex]

    Am I correct so far?
     
    Last edited: Apr 26, 2014
  5. Apr 26, 2014 #4

    utkarshakash

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    Please help me.
     
  6. Apr 26, 2014 #5

    SammyS

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    What is your result for ##\ v_{1+1} \ ?##
     
  7. Apr 26, 2014 #6

    utkarshakash

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    [itex]V_{1+1} = \dfrac{M-m}{M+m} v_0 - \dfrac{2m}{M+m} 2v_0[/itex]
     
  8. Apr 26, 2014 #7

    SammyS

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    I asked about ## \ v_{1+1} \ , ##

    not about ## \ V_{1+1} \ . ##

    Or are you not taking dauto's suggestion for notation ?
     
  9. Apr 26, 2014 #8

    utkarshakash

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    Sorry about that.

    [itex]v_{i+1} = \left( \dfrac{2M}{M+m} \right) V_i + \left( \dfrac{M-m}{M+m} \right) v_i [/itex]

    If I put i=1

    [itex]v_2 = \left( \dfrac{2M}{M+m} \right) V_1 + \left( \dfrac{M-m}{M+m} \right) v_1 [/itex]
     
  10. Apr 27, 2014 #9
    Now you can start making approximations. You should try to simplify those expressions. may be define the parameter ## \mu = m/M ## and expand those expressions in power series of that parameter keeping only the independent term and the linear term. Any quadratic or higher term may be dropped. That will help clear things up quite a bit while still preserving the essence of the problem. You should be able to get the correct order of magnitude for the solution but probably not much more than that, but it seems that's all the problem text asked you to do.
     
  11. Apr 27, 2014 #10

    utkarshakash

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    [itex]V_{i+1} = (1-2 \mu) V_i - 2 \mu (1- \mu) v_i \\
    v_{i+1} = 2(1- \mu ) V_i + (1-2 \mu) v_i [/itex]

    But I really don't see how does this help me. The problem asks me to find the closest distance.
     
  12. Apr 27, 2014 #11
    One step at a time. The expession for the distance depends on the velocities, so you should solve the velocities first because their expressions do not depend on the distances. There is still one quadratic term in your expressions that may be dropped. Try writing the pair of equations in matrix form by defining the column vectors Xi = (Vi v_i)T and express Xi+1 as a matrix product

    Xi+1 = M Xi,

    where the 2 by 2 matrix may be written as M = A + μB.

    A and B are independent of μ.

    Once you've done that calculate M2, M3, and M4, and look for patterns that will allow you to find the general expression for Mn.

    Remember to drop any quadratic terms that arise as you do your calculation.

    Once you have Mn you can easily find Xn = Mn X0.

    That should get you busy for a while.
     
  13. Apr 27, 2014 #12
    Alternatively you might find the expression for the momenta change δPi = M(Vi+1-Vi) and δpi = μM(vi+1-vi) and approximate

    δV = dV/dn and δv = dv/dn,

    and solve the differential equations.
     
  14. Apr 27, 2014 #13
    OK, I solved the problem using both methods described above, The second method turned out to be easier and surprisingly accurate (I compared my answers with a numerical solution produced with a spread-sheet using M = 100000 m). You might want to go that way.
     
  15. Apr 27, 2014 #14

    utkarshakash

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    Transforming the equation in matrix notation, I get

    [itex]\left( \begin{array}{c} V_{i+1} \\ v_{i+1} \end{array} \right) = \left( \begin{array}{cc} 1-2\mu & -2\mu \\ 2(1-\mu) & 1-3\mu \end{array} \right) \left( \begin{array}{c} V_{i} \\ v_{i} \end{array} \right) [/itex]

    [itex]M^2 = \left( \begin{array}{cc} 1-8\mu & -4\mu \\ 2(2-6\mu) & 1-8\mu \end{array} \right) \\
    M^3 = \left( \begin{array}{cc} 1-18\mu & -6\mu \\ 2(3-19\mu) & 1-18\mu \end{array} \right) [/itex]

    But I can't see a definite pattern except that the matrix is symmetric and M12 = 2n (-μ) .

    [EDIT] Oops, I'm sorry. The matrix is not symmetric. Only the principal diagonal elements are equal.
     
    Last edited: Apr 27, 2014
  16. Apr 27, 2014 #15

    utkarshakash

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    Your second method is not clear to me. Can you please elaborate?
     
  17. Apr 27, 2014 #16
    There is a mistake in your matrix. The right bottom term is not correct, I don't think.
     
  18. Apr 27, 2014 #17
    You already have the expressions for Vi+1 and vi+1. Find the momentum change δPi = M δ Vi = M (Vi+1 - Vi) and similarly for δpi.

    Its possible to express
    $$\delta P = \frac{\delta P}{1} = \frac{\delta P}{\delta n} \approx \frac{d P}{d n}$$
    where n counts the collision number.

    That way you find a pair of differential equations for dP/dn and dp/dn. This differential equations turn out to be easily solvable if you keep in mind That P >> p >> μP.
     
  19. Apr 27, 2014 #18

    utkarshakash

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    That's a typo. It should be 1-2μ
     
  20. Apr 27, 2014 #19
    OK.

    Don't you see the pattern? Only the bottom left term is difficult (and you won't need the to get a zeroth order solution). all the other ones either grow linearly or quadratically.
     
  21. Apr 27, 2014 #20

    utkarshakash

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    I get these two equations.

    [itex] \dfrac{dP}{dn} = -2(\mu P + p) \\
    \dfrac{dp}{dn} = 2\mu (P-p) [/itex]

    If I follow your approximations, the first equation changes to
    [itex] \dfrac{dP}{dn} = -2p [/itex]

    and second becomes

    [itex]\dfrac{dp}{dn} = 2\mu P [/itex]

    If I divide both equations I get something like this

    [itex]\mu P dP + pdp = 0 [/itex]

    Integrating both sides

    [itex]\mu \dfrac{P^2}{2} + \dfrac{p^2}{2} + c=0[/itex]

    How to find the constant of integration?
     
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