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Momentum (velocity)

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data

    2 balls, both moving to the right - what are their velocity after collision?
    Elastic collision

    Ball 1
    M 60 kg
    u0= 18 m/s

    Ball 2
    m=220 kg
    v0= 7 m/s

    2. Relevant equations

    Law on momentum conservation
    m*v0+M*u0= m*v+M*u

    Momentum before = Momentum after


    3. The attempt at a solution

    Law on momentum conservation gives:
    220*7+60*18=220*v+60*u (1)

    As it is a elastic collision, it gives :
    1/2(220)(7)^2+1/2(60)(18)^2=1/2(220)(v)^2+1/2(60)(u)^2 (2)

    2620=220v*60u (1)

    15110=1/2(220)(v)^2+1/2(60)(u)^2 (2)

    v=(2620-60u)/220 (1)


    I need to simplify equation (2) and put in (1), but I can't get further with that
    Any help would be appreciated
    Would be great if someone could double-check as well, not used with this.

    Thanks in advance!
     
  2. jcsd
  3. May 27, 2012 #2

    Or.... you can simplify (1) and put it in (2) :wink:. It works out easier. Just substitute the v you got from equation (1) into (2), which will give you 'u' without much trouble.

    Your approach does look correct to me :smile:
     
  4. May 27, 2012 #3
    Oh, okay
    So you mean:

    v=12-0,27u (1)

    15110=110v^2+30u^2 (2) (further simplifed)

    15110=110(12-0,27u)^2+30u^2

    Now to get 'u' alone.. hmm.. any hint?
     
    Last edited: May 27, 2012
  5. May 27, 2012 #4
    Haven't checked the exact numbers, but it looks correct to me. Now comes the boring part. You expand the square term, get a quadratic in u, solve it to get your answer. It surely doesn't look pretty but what are calculators for...:rolleyes:
     
  6. May 27, 2012 #5


    Expand the square term, ok:
    15110=1320-29.7u^2+30u^2

    Have a slight problem understanding what is meant by get a quadratic in 'u'
     
  7. May 27, 2012 #6

    cepheid

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    A quadratic expression in u is a second order polynomial in which "u" is the variable that is being raised to increasing powers in each successive term. In this case, since it is a second order polynomial (also known as a quadratic), terms exist with powers of u up to and including the second power of u.


    Edit: some powers of u can be missing, that's fine. (In your case you are missing the term that has u raised to the first power). What defines it as second order / quadratic is that there is a u-squared term, and this is the highest power of u in the expression.
     
  8. May 27, 2012 #7
    Um, dont you know quadratic equations?

    http://en.wikipedia.org/wiki/Quadratic_equation

    Quadratic in u means the variable is u, instead of x.

    You need to find the roots of your quadratic and find the corresponding value for v from (1).

    ---------------------

    To simplify this for you(but still, learn quadratics, they're fundamental for most problems in physics), you can keep the terms in variable form, and you'll get,

    [itex]v_{f1} = ((m_1 - m_2)·v_{i1} + 2 ·m_2·v_{i2})/{(m_1 + m_2)}[/itex]

    Now you can find out vf2
     
  9. May 27, 2012 #8
    I will definitely look up quadratics.
    I was able to get the answers, which seems to be correct -thank you for all your help :)
     
  10. May 28, 2012 #9
    Another question while I'm at it - if the collision was inelastic, how do you think then? :smile:

    Does it have to be a "perfect" collision? - I'm intrested in a non-perfect collision if it is possible, meaning two velocities instead of one


    2 balls, both moving to the right - what are their velocity after collision?
    INelastic collision

    Ball 1
    M 60 kg
    u0= 18 m/s

    Ball 2
    m=220 kg
    v0= 7 m/s


    Law on momentum conservation is still true,
    m*v0+M*u0= m*v+M*u

    Kinetic energy is not conserved
     
  11. May 28, 2012 #10
    Is the collision perfectly inelastic? If yes, then the momentum conservation statement would be like

    [itex]mv_0 + Mu_0 = (M+m)v_f[/itex]

    If its a partially elastic collision, you would need extra information i.e the knowledge about the co-efficient of restitution of the impact.
     
  12. May 28, 2012 #11
    The question simply says "inelastic".
    Also, it says to find out velocities (in plural), so I assume it's not perfect.

    So I'll have to take a closer look at the formulas now, thank you.
     
  13. May 28, 2012 #12
    If it says inelastic, its most definitely perfectly inelastic.
     
  14. May 28, 2012 #13
    Oh, good. :smile:
    Should be easy enough to solve it
     
  15. May 28, 2012 #14

    ehild

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    Problems of elastic collision can be solved in an easy way without quadratics.

    There are two equations, one for momentum, the other for kinetic energy.

    m1v0+m2u0=m1v+m2u

    m1v02+m2u02=m1v2+m2u2

    Arrange both equation so as the terms belonging to the same object are on the same side.

    m1(v0-v)=m2(u-u0) *

    m1(v02-v2) = m2(u2-u02)-->

    m1(v0-v)(v0+v)=m2(u-u0)(u+u0) **

    Divide eq.(**) by eq.(*). You can do it as the velocities do change during the collision, so you do not divide by zero. You get a third equation:

    v0+v=u+u0 ***

    The system of linear equations (*) and (***) is easy to solve.

    ehild
     
  16. May 28, 2012 #15
    This is a how I got my direct equation for [itex]v_{f1}[/itex] in one of my previous posts. I seem to have forgotten to post how I got it :redface:
     
  17. May 28, 2012 #16

    m1(v0-(u+u0)-v0) = m2(((v0+v)/u0)-u0)

    Or am I completely wrong? - I seem to lack some mathematical skills
     
  18. May 28, 2012 #17

    ehild

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    The unknowns are v (velocity of ball 1 after collision) and u (velocity of ball 2 after collision).

    m1(v0-v)=m2(u-u0) *
    v0+v=u+u0 ***

    Isolate u from (***):

    u=v0+v-u0.

    Substitute for v in (*)

    m1(v0-v)=m2(v0+v-u0-u0)

    Collect the terms with v on one side of the equation:

    v(m1+m2)=(m1-m2)v0+2m2u0

    Isolate v:

    [tex]v=\frac{(m_1-m_2)v_0+2m_1u_0}{m_1+m_2}[/tex]


    ehild
     
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