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Distance Travelled in Perfectly Elastic Collision with Friction

  1. Jan 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A 5.0 kg block with a speed of 8.0 m/s travels 2.0 m along a horizontal surface where it makes a head-on, perfectly
    elastic collision with a 15.0 kg block which is at rest. The coecient of kinetic friction between both blocks and
    the surface is 0.35. How far does the 15.0 kg block travel before coming to rest?

    ANS. 1.79m
    2. Relevant equations
    Elastic Collision:
    K_0 = K
    m1iv1i = m1v1 + m2v2
    Ff = μmg

    3. The attempt at a solution
    First it travels 2.0 m with friction:
    K_o - Wf = K
    .5mv^2 - 2μmg = .5mv^2
    v_0^2 - 4μg = v^2
    v^2 = 50.28
    v = 7.1 m/s

    Now cons. of momentum:
    5(7.1) + 0 = 5(v1) + 15(v2)
    7.1 = v1 + 3v2

    KE:
    1/2mv_0^2 + 0 = 1/2m1v1^2 + 1/2m2v2^2
    160 = 2.5v1^2 + 7.5v2^2
    64 = v1^2 + 3v2^2
    Solving for v1 by substituting then quad formula gives:
    v2 = -.29 m/s
    This does not make any sense?
     
  2. jcsd
  3. Jan 28, 2013 #2

    PeterO

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    I have not checked your calculations - just your last line - but during an elastic collision, when a small mass collides with a large mass, the small mass bounces back while the large mass lumbers forward.
    Could it be that your negative velocity is the small mass bouncing back, and you have not yet calculated the speed of the large mass.
     
  4. Jan 28, 2013 #3
    The reason it does not make sense is that object 2 is the heavy object at rest so it should have a forward velocity.
     
  5. Jan 28, 2013 #4

    PeterO

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    Your figure of 160 is wrong.
    That is the energy of the 5kg block at the beginning - when travelling at 8 m/s. You already established that it had slowed to 7.1 by collision time [it will certainly have slowed, I assume that figure is correct]
     
  6. Jan 28, 2013 #5
    Ahh lets work with that.
    It becomes 126 then, given that it is 7.1
    126 = 2.5v1^2 + 7.5v2^2
    50.4 = v1^2 + 3v2^2
    Quad Form again yields:
    v = .000235 m/s - Something is still wrong.
     
  7. Jan 28, 2013 #6

    PeterO

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    I meant the last line of calculations, when you got the negative velocity.
    If 7.1 is correct, and you don't get the large mass traveling at 7.1/4, something has gone wrong.

    I get that answer by looking at the elastic collision from a different perspective. I can explain later after you have solved the problem.
     
  8. Jan 28, 2013 #7

    PeterO

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    Clearly you are doing something wrong in the simultaneous equation step. can you type that step out so I can comment.
     
  9. Jan 28, 2013 #8
    Yes here it is:
    sqroot(50.4 - 3v2^2) = v1

    Substitute the above into this equation:
    7.1 = v1 + 3v2
    7.1 = sqroot(50.4-3v2^2) + 3v2
    isolate sqroot
    7.1 - 3v2 = root(50.4-4v2^2)
    Sq both sides:
    50.4 - 42.6v2 + 9v2^2 = 50.4 - 4v2^2
    13v2^2 - 42.6v^2 = 0
    v≈0
    The error must be from earlier on.

    Gahh I can't find that pesky error.
    On another note, what is the alternate suggestion you had before, if you're willing to share?
     
  10. Jan 28, 2013 #9

    PeterO

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    A 4 became a 3, and a subscript became a square

    I see you are substituting the opposite way to the usual.

    you should substitute from linear into quadratic.

    You have 7.1 = v1 + 3v2

    so v1 = 7.1 - 3v2

    substitute that into the energy equation:
    50.4 = v1^2 + 3v2^2

    50.4 = (7.1 - 3v2)2 +3v22

    expand and solve.

    remember - one answer is 0 [the large mass was stationary before collision] so this quadratic will simplify very easily.
     
  11. Jan 28, 2013 #10
    Ok, thanks. So solving your quadratic yields v2 ≈ 0 and v2 ≈ 3.55 m/s. Only 3.55 m/s makes sense in the context of the problem.
    Now the problem is simple:
    K = Wf
    1/2mv^2 = μmgd
    v^2/2 = μgd
    d = v^2/2gμ
    d = 1.83m
    This is very close to the correct answer 1.79 m (probably due to rounding/sig fig errors).

    So this is the correct answer.

    I guess what to learn from this problem is to make organized, tidy work that doesn't make itself more complicated than it needs to be.
     
  12. Jan 28, 2013 #11

    PeterO

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    The other method [I made an error it was 14.2/4 rather than 7.1/4]

    for simplicity I will describe the frictionless situation, so that the incoming velocity is 8.0 m/s

    In the modern day of small cameras - like a GoPro, you would be able to film the collision from several points of view.

    If you mounted the camera on the small mass, the video would show a large mass approach, collide then rebound at the same speed as it arrived earlier [negative velocity f you like.

    If you mounted the camera on the large mass, the video would show a small mass approach, collide then rebound at the same speed as it arrived earlier.

    If you mounted the camera on a third mass, travelling a parallel path, at the speed of the centre of mass of the system, you would see a small mass approach from one direction while a large mass approached from the other direction, and then after collision both would rebound at their individual, initial speeds.

    The speed of the centre of mass is easy to calculate: total momentum/total mass

    Total momentum at first [and always] is 5 x 8 = 40 Ns
    Total mass = 20 kg

    velocity of Centre of Mass is thus 2 m/s.

    The small mass, travelling at 8 m/s, will be "catching" the C of M at a rate of 6 m/s. ie approaching at 6 m/s,
    After collision, it will be departing the CofM at 6 m/s - so travelling at -4 m/s. rebounding at 4 m/s.

    The CofM is approaching the large mass at 2 m/s - so the large mass is approaching the CofM at 2m/s
    After collision it will be departing the CofM at 2m/s, and thus travelling at 4 m/s.

    ie (8 / 2) m/s

    In this problem, the small mass was finally approaching at 7.1, so the large mass moved off at (7.1 / 2) m/s.
     
  13. Jan 28, 2013 #12
    Hmm could you explain that CoM bit again?
     
  14. Jan 28, 2013 #13

    PeterO

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    if you throw a ball at a brick wall, it bounces off at the same speed it approached.

    This is because the wall is attached to the Earth, and given the ratio of masses of the Earth and a ball, the Centre of Mass of the Earth-Ball system is effectively the Earth.

    The Earth is thgus "stationary" with respect to itself, and remains so.
    The ball approached the earth [or at least the brick wall appendage] at some speed, then rebounded from that centre of mass at the same speed.

    When the masses in a collision are more equal in mass, the centre of mass is not tagged to one of the masses.

    BUT, when viewed from the point of view of the centre of mass, the two bodies seem to just bounce off the centre of mass.


    Suppose you were standing at the end of a fixed, free-standing, wall in a playground, watching two people throw balls at each side of that wall.

    You would see the balls bounce back from each side of the wall.

    If you watched form a large distance, you might not be able to see the wall, and mistakenly think the balls were bouncing off each other. That would be rather confronting if one ball was a basket ball while the other was a table tennis ball!!
    What you are actually doing is watching the collisions form the frame of the Centre of Mass, since you are standing on, and the wall is attached to, the Earth.

    You can consider any collision from the from the point of view of the centre of mass, and you see the same thing.

    Another familiar example:

    Suppose a 2 kg cart, travelling at 12 m/s, approaches a 6kg cart travelling at 4 m/s in the same direction and have an elastic collision.
    If you carry out the usual calculations, you will find that after the collision, the 2kg mass will actually stop, while the 6 kg mass will be travelling at 8m/s
    [You can do the usual calculations to prove that]

    Initial momentum: 2x12 + 6x4 = 48 Ns
    Final momentum: 2x0 + 6x8 = 48Ns
    Initial Energy 1x144 + 3x16 = 192 J
    Final Energy 1x0 + 3x64 = 192 J

    Now for the velocity of the centre of mass.

    Total momentum 48 Ns [see above]
    Total Mass 2 + 6 = 8 kg

    VCofM = 48 / 8 = 6 m/s

    2kg mass:
    Initial vel: 12 m/s - so approaching the centre of mass at 6m/s
    Final vel: 0 m/s - so the centre of mass is moving away at 6 m/s

    6 kg mass
    Initial vel: 4 m/s - so the centre of mass is approaching at 2m/s
    Final vel: 8 m/s - so moving away from the centre of mass at 2 m/s

    Given how easy it is to calculate the velocity of the Centre of Mass, this second method is a really easy way to find the final velocities after an elastic collision.

    NOTE: Some people don't like you calculating it this way, but at least it is an easy way to confirm the answers you have calculated. There was never an easier answer to calculate, than one you already know!
     
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