1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another Collision problem

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A ball of mass = 3-kg moving to the right at 3 m/s collides elastically with a ball of mass m2 = 2 kg moving at 4 m/s to the left. Find (a) the final speed of each object and (b)
    the kinetic energy of each ball before and after the collisions.


    2. Relevant equations
    V1i + V1f = V2f + V2i
    V1i - V2i = - (V1f - V2f)


    3. The attempt at a solution
    I plugged the numbers into the above formulas
    3 + V1f = -4 + V2f
    V2f = 7 + V1f

    3 - (-4) = -(V1f - V2f)
    7 = -V1f + V2f

    substitute V2f
    7 = -V1f + 7 + V1f
    I got 0 = 0 the answer should be V1f = -2.60 ms, V2f = 4.40 m/s
     
  2. jcsd
  3. May 30, 2014 #2

    Nathanael

    User Avatar
    Homework Helper

    Where'd you get those equations from? Conservation of momentum states m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
     
  4. May 30, 2014 #3
    In this problem since it is perfectly elastic kinetic energy is conserved so 1/2m1v1i2 + 1/2m2v2i2 = 1/2m1v1f2 + 1/2m2v2f2
    with some mathematical manipulation you can acquire the above formulas.
     
  5. May 30, 2014 #4

    Nathanael

    User Avatar
    Homework Helper

    Ah, I see now. Still you seem to have made a mistake with your manipulations since there's no possible way to manipulate out of the equation m1 and m2
     
    Last edited: May 30, 2014
  6. May 30, 2014 #5
    I got these equations directly from my physics book.
     
  7. May 30, 2014 #6

    Nathanael

    User Avatar
    Homework Helper

    Well don't trust your physics book. If it comes from conservation of KE then do the manipulations yourself.

    And let me know if you figure it out, because I don't think it's possible to turn
    into:
    The simplest I can make those equations is:
    m[itex]_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})[/itex]
     
    Last edited: May 30, 2014
  8. May 30, 2014 #7

    Nathanael

    User Avatar
    Homework Helper

    But using either
    m[itex]_{1}(V_{1i}+V_1f)(V_{1i}-V_{1f})=m_{2}(V_{2f}+V_2i)(V_{2f}-V_{2i})[/itex]
    or
    m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
    Should lead to the right answer (I think)

    It's just that the equations you used (from your textboook) are wrong for this situation.
     
  9. May 30, 2014 #8

    CAF123

    User Avatar
    Gold Member

    No, those equations are correct for a perfectly elastic collision.

    Brainman, you got 0=0 because your 'two' equations are not independent. One is simply a rearrangement of the other. You can use the result you got in the OP, together with conservation of momentum. This will give the result because the first equation uses the fact that the collision was perfectly elastic, so provides additional information.
     
    Last edited: May 30, 2014
  10. May 30, 2014 #9
    What was my mistake then?
     
  11. May 30, 2014 #10

    CAF123

    User Avatar
    Gold Member

    I just edited my post. See above.
     
  12. Jun 2, 2014 #11
    OK I did that and I am still not getting the correct answer. I am pretty sure I did something wrong when I substituted. Here are my calculations.

    equation 1
    m[itex]_{1}[/itex]V[itex]_{1i}[/itex]+m[itex]_{2}[/itex]V[itex]_{2i}[/itex]=m[itex]_{1}[/itex]V[itex]_{1f}[/itex]+m[itex]_{2}[/itex]V[itex]_{2f}[/itex]
    3(3) + 2(-4) = 3(v1f) + 2(v2f)

    equation 2
    V1i + V1f = V2f + V2i
    3 + v1f = v2f + 4
    v2f = 7 + v1f

    substitute v2f
    1 = 3(v1f) + 2(7 + V1F)
    1= 3V1F + 14 + 2V1f
    1- 3v1f = 14 + 2v1f
    -3v1f = 13 + 2v1f
    -v1f = 13
    v1f = -13

    the answer should be - 2.60 m/s
     
  13. Jun 2, 2014 #12

    CAF123

    User Avatar
    Gold Member

    Small sign error there.
     
  14. Jun 2, 2014 #13
    I am still getting the wrong answer

    1 = 3v1f + 2v1f - 2
    3/5 = v1f

    there must be something else I'm doing wrong.
     
  15. Jun 2, 2014 #14

    CAF123

    User Avatar
    Gold Member

    From conservation of momentum, you correctly obtained the equation 1 = 3v1f + 2v2f.

    From the equation derived by manipulating the equations for conservation of kinetic energy and that of momentum (do you understand the derivation?) you obtain, by fixing the sign error in the previous post, -7 = v1f - v2f

    Solve these two equations for v1f and v2f.
     
  16. Jun 2, 2014 #15
    I am having trouble deriving that formula. What I did was
    V1i + V1f = V2f + V2i
    3 - vif = -4 -v2f // the final velocities are negative because they will both switch directions.
    -1(3 - vif = -4 -v2f)
    -3 + vif = 4 +v2f
    -7 + vif = v2f

    I then tried to substitute this into the other equation.


    1 = 3v1f + 2v2f
    1 = 3v1f + 2v1f - 14
    15 = 5v1f
    v1f = 3
    and I did not get the right answer
     
  17. Jun 2, 2014 #16

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Do what CAF123 says above.

    One way to do this is by elimination: Multiply the 2nd equation by 2, then add the equations to eliminate v2f .
     
  18. Jun 2, 2014 #17
    What did I do wrong in my previous post?
     
  19. Jun 2, 2014 #18

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Don't change signs on these velocities. You didn't on the other equation.

    Algebra will take care of the signs! (That's one of Algebra's most endearing characteristics.)

     
    Last edited: Jun 2, 2014
  20. Jun 3, 2014 #19
    OK I see what I did and now got the right answer. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Another Collision problem
  1. Another problem! (Replies: 1)

  2. Another problem (Replies: 1)

  3. Collision problem (Replies: 5)

Loading...