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Glancing elastic collision, and center of mass

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    So this is just something i came up with to check if i am understanding everything allright, it all comes down to collisions in regard to the center of mass

    So two "masses" experience a glancing collision
    M1:
    vi - 3m/s (refered to as v1i) ;
    m - 2Kg (m1)
    M2
    vi - 0 (v2i) ;
    m - 6Kg (m2)

    2. Relevant equations
    what i did was figure out the speed of the center of mass (Vcm) using:
    Eq(1) (v1i*m1 +v2i*m2)/(m1+m2)
    and then since the Vcm is constant before and after the collision (right?) i figured out v1f and v2f because:
    eq(2) v2f=2vcm-v2i (and same for v1f)

    this is were i get stuck, using the system:
    Sys(1)
    -v1f*m1*sin(theta) + v2f*m2*sin(theta2)=0
    m1*v1i=v1f*m1*cosin(theta) + v2f*m2*cosin(theta2)
    3. The attempt at a solution
    so using eq 1 and 2 i get Vcm=3/4 and v1f=-3/2 and v2f=3/2
    and i cant solve the system, all the other glancing colision problems i have seen give one of the angles.
    plus the masses are different so the angle bettwen theta and theta2 isnt 90º, i don't know what to do.
    I am probably doing something wrong as everything i have seen says i cant solve this kind of problem if i am not given one of the 4 unknown quantities (the two angles and the two final velocities) so probably my mistake is regarding the two final velocities, i dont know.
     
  2. jcsd
  3. Jan 7, 2016 #2

    Merlin3189

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    It seems to me that there is a solution where the Ms continue to move in the same direction (with the small M going in the opposite sense of course). Therefore, if some other specific solution is required, there needs to be some piece of information which precludes the simple one - such as a given final speed which does not match or an angle of contact. Otherwise there will be many other solutions depending on the angle.
    Since the other extreme case is grazing incidence, with M1 continuing undeviated and M2 remaining stationary, I think any angle of deviation for M1 is possible.

    Edit: I don't think you have done anything wrong. If you have shown all the given data, nothing you can work out can add anything.
     
  4. Jan 7, 2016 #3

    mfb

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    This is true in one dimension only, where "glancing collision" does not make sense. The magnitude of the velocity stays the same, but the direction can change.
     
  5. Jan 7, 2016 #4
    Ohh i see, thanks, so there is no way to solve 2-d exercices using the centre of mass?
    so, new questions:
    1. there is no way of solving a problem like this using only the information provided, correct?
    2. provided 1 is correct, imagine say v1f is a given value by the exercice, i should figure out v2f using conservation of kinectic energy, what should i do next? i still dont think i have enough information to solve the system...

    i've set v1f to be 3/2, so v2f is +/- sqrt(2), how do i figure out the angles? i dont know any of the components of the two velocity vectors, only the magnitude...

    Im failing to see how to relate the angles of the system.
    The book clearly states as long as no more than 2 of the (seven) quantities of the equations in the system are unknown it is solvable, so there is a way of solving the system i just can't figure out how
     
  6. Jan 7, 2016 #5

    haruspex

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    You can use centre of mass, but whatever method you use you need to know the angle of impact. That is, the line that the initial relative velocity makes to the normal where the surfaces contact. (This is ignoring friction.). If that normal does not pass through the mass centres then it gets nastier, since torques will be involved and the objects will acquire spin. So assume the objects are spheres.

    Whether it is a matter of solving for two unknowns, or three, or four, depends on how you count them. You could argue that there are four unknowns: the final speeds and angles. Counting that way, you will need four equations. You have conservation of momentum overall in the direction normal to the surface. It is only conserved 'overall' because there is an unknown impulse between the surfaces. In the direction parallel to the surfaces, with no friction, there is no impulse, so momentum is conserved separately for each object. That gets you to three equations. The fourth is conservation of energy, but that requires you to know the coefficient of restitution. Note, the elasticity operates only normal to the surfaces, so only apply it to those components of velocity.
     
  7. Jan 7, 2016 #6
    thanks @haruspex, alllow me to clarify, this is a simple case (i think)
    this is the sort of collision im talking about:
    51.jpg

    So, imagine i know, v1i (v2i is 0) and i know the masses (which are different), and im also given v1f

    with this info, knowing its elastic, i can use the conservation of kinectic energy equation to figure out v2f.

    all that is left to find out is theta and phi,
    with:
    m1*v1f*cos(theta)+m2*v2f*cos(phi)=m1*v1i
    and
    -m1*v1f*sin(theta)+m2*v2f*sin(phi)=0
    because there are 2 unknowns and 2 equations. im supposed to find out theta and phi but i dont know how (algebra failure?)
     
  8. Jan 7, 2016 #7

    haruspex

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    What is the relationship between the sin of an angle and its cosine?
     
  9. Jan 7, 2016 #8
    tan(x)=sin(x)/cos(x) i think thats what you mean
    I'm not seeing how that will help me o_O but is that it? i gotta write it downand try to solve it, but i think ive tried going with the tangent and it didnt seem to help, but il try again tommorow
    its almost 4 am, gotta get some sleep, il get back on it tommorow
    thank you :)
     
  10. Jan 7, 2016 #9

    haruspex

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    No, a relationship involving only the sine and the cosine and constants.
     
  11. Jan 8, 2016 #10
    sine (x)=cosin (90-x) ?? dont thinks thats it as it doesnt seem to help much
    im running through all the identitied and relatioships i can think of, i cant get there :/
     
  12. Jan 8, 2016 #11

    haruspex

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    Pythagoras!
     
  13. Jan 8, 2016 #12
    the basic? sin(x)= opp/hyp ? Ok im doing this and i think its working.
    i get something like
    m1*v1fx+m2*v2fx=m1*v1i
    -m1*v1fy+m2*v2fy=0

    and then i write v1fy in respect to v2fy (and v1fx to v2fx), and use the pythagoras hypotnuse formula of the both triangles as a system and solve (since i know the hypotenuses, and i wrote v1x and v1y "as" b*v2x and a*v2y)
    Am i correct?
     
  14. Jan 8, 2016 #13
    everything seemed to work out.
    One last question, i got both positive angles, and both positive values for Vy, i imagine this is because it doesnt matter which particle goes up or down, i just have to pick which one i set..
    but the angles are both positive also? that confused me a bit, is it also because "i get to choose" which particle goes up and which goes down?
    the angles i might have messed up/confused somehow with the signs in the equations, but im pretty sure i didnt mess up with the V signs
     
  15. Jan 8, 2016 #14

    mfb

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    The way you defined the signs, everything should be positive.
     
  16. Jan 8, 2016 #15

    haruspex

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    I meant sin2+cos2=1, which is equivalent to the Pythagorean formula. Looks like you got there anyway.
     
  17. Jan 8, 2016 #16
    ah i get the values i have now, im evaluati g each triangle "outside" the reference frame, after figuring out the trianglesnl i have to apply them in a way that makes sense to the problem.

    hm i thought about that but it seemed to me i wouldn't be able to "line up" same angle cosines and sines with a sum, il give it a go on paper when i get home.

    Thanks for all the help
     
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