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Inelastic collision, please help

  1. Mar 31, 2007 #1
    Forgive the probable incorrect use of words of physics, I'm not used to writing them in english :) Hope you will understand anyway.

    Object number 1 is moving horizontally (left to right) with Uniform rectilinear motion at the speed of 10.5 m/s. Its mass is 7.5 Kg. The coefficient of dynamic friction is 0.15. A drawing force drags object number one from left to right (with an angle of 45) and its vertical component(projection) has the same value of its horizontal component, which is 9,6N.

    Evidently the frictional force is -9,6N, otherwise the motion would not be uniform rectilinear.

    Now Object number 2 (same mass of number 1, 7.5 Kg) falls on object number 1 vertically, causing a perfectly inelastic collision and it stays attached to object number 1.

    The drawing force stays constant (it doesn't change from 9.6N vertically and 9.6N horizontally).

    what is the speed right after the collision?

    what distance does the object number 1+2 cover before it stops?

    Thanks for the help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 31, 2007 #2


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    Nicely written, and welcome to Physics Forums! Are you familiar with the conservation of momentum law and its use in totally inelastic collisions? You'll have to start from there. Please show your attempt. Your dynamic friction force calculation looks good.
  4. Mar 31, 2007 #3
    thanks for the fast reply

    Not really, I tried to study it, but with no good results. And all the examples on the book are about collisions on the same axis.

    Doesn't that rule apply only when there are no external forces? Before the collision the frictional force compensates the drawing force, so the external forces are zero. But after that, the situation changes.

    And I'm also confused about the axis where the momentum is conserved.

    I would really appreciate an explanation :rolleyes:
  5. Mar 31, 2007 #4


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    During the collision, there are no net external force acting in the x direction, so momentum is conserved in that direction. Immediately after the collision, there is a net external force acting, so momentum is not conserved after the collision. Treat it as 2 separate problems. What is the initial speed of the dropped mass in the horizontal direction?
  6. Mar 31, 2007 #5
    er, simultaneous post here.

    If i understand the problem, its like dropping a block onto a moving cart. Momentum should be conserved from the standpoint of instantaneous velocity before and after the block strikes the "cart" and is carried with it. Note the block has no momentum along the x axis. only thje cart.

    At that time the external force will no longer be large enough to maintain motion as the frictional force will have increased due to the increase in mass, and will undergo linear deceleration.

    The normal force should be computed before and after the collision I think.

    Then you will have a new sum of forces along the x axis along with which to compute acceleration along with the new initial velocity (from conservation of momentum). At least thats how I interpret the problem.
  7. Mar 31, 2007 #6
    ... ok, then it would be:

    7.5 x 10.5 = 15v

    v=5.25 m/s


    is it so... I don't know, simple? That the speed just becomes half the initial one?

    :) well, good, anyway :) the rest I think it's quite easy (correct me if I am wrong):

    The normal force added is 7.5*9.8 = 73.57 N and the frictional force added on the objects is 73.575*.15 = -11.03625 N. This is also the net force on the 2 attached objects, so their acceleration is a = F/m = -11.03625/15 = -.73575 m/s²
    The distance covered is (V2)²/2a = 5.25²/(2*.73575) = 18.73 m.

    ... at first I thought I could only express the speed after collision with a function, since it is not constant... I didn't consider it to be the instantaneous speed after collision.... the function being v= v0-at= v0 - .735t

    ... thanks for the help, I'll take some more advantage of it with some other exercises I have ;)
    Last edited: Mar 31, 2007
  8. Mar 31, 2007 #7
    "banal", nice word, but never laugh of the gods of physics homework.

    well the interesting part imo is how the frictional force changes, remember part of the force is directed against the weight of the "cart".
  9. Mar 31, 2007 #8


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    Yes, the solution is not correct; the normal force is not correctly calculated.
  10. Apr 1, 2007 #9
    another try....

    gravitational force after collision: 15 Kg * 9.8 m/s2= 147N
    vertical drawing force (it stays constant) : 9.6N
    normal force + vert. force - grav. force =0
    normal force= grav.force - vert.force = 147N - 9.6N=137.4N

    frictional force= 137.4N * 0.15= 20.6N
    horizontal force (constant)= 9.6N

    Total horizontal force = horiz.force - frict.force= 9.6N - 20.6N = -11N

    F=ma, -11N= 15Kg * a, a=-0.73 m/s2

    The distance covered is (V2)²/2a = 5.25²/(2*0.73) = 18.73 m.
  11. Apr 1, 2007 #10
    Yea I did it both ways as well, and agreed with all results, must be overlooking something.
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