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Homework Help: 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)


    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:

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


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    Staff Emeritus
    Science Advisor
    Gold Member

    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?


    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


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    Homework Helper

    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.



    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.

  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


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    Homework Helper

    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 (***):


    Substitute for v in (*)


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


    Isolate v:


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