1. Not finding help here? Sign up for a free 30min 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!

Need help using algebra to solve physics equations

  1. Nov 7, 2006 #1
    My question is not so much what to do it is just i have problems solving for a certain equation.

    A 5.00kg ball, moving to the right at a velocity of 2m/s on a frictionless table, collides head-on with a stationary 7.50kg ball. Find the final velocities of the balls if the collision is (a) elastic and (b) completely inelastic

    for part A

    I have m1v1f + m2v2f= m1vi1 +0
    to solve for final velocity 1(the ball for 5kg) you use

    vf1=(m1-m2/m1+m2)vi1

    I really have no i do how to algebraically solve for vf1, can someone explain that to me because when i use the equation i get the right answer which is -.400m/s

    To find vf2 we use the fact that it is an ellastic collision and the kinetic energy before and after is the same

    1/2m1(vf1^2)+1/2m2(vf2^2)=1/2m1(vi1^2)+0
    Using this i have to solve for vf2 but again i have a problem algebraically solving for vf2. But there is a similar example in the textbook so i used what they had to solve for vf1 and vf2 but they do not show the work

    vf2=(2m1/m1+m2)vi1
    answer should be 1.60m/s

    My problem is not understand ellastic equations, it is how to solve the first equation for vf1 and the second for vf2. Can someone show me step by step how to use algebra to solve for vf2 with what i typed above,
     
  2. jcsd
  3. Nov 7, 2006 #2
    so i just did this in lab last week so I am able to help you.....

    1/2 m1(vi1^2) + 0 = 1/2m1(vf1^2) + 1/2 m2(vf2^2)

    all the 1/2s cancel

    m1(vi1^2) = m1(vf1^2) + m2(vf2^2)

    momentum before = momentum after
    m1(vi1) = m1(vf1) + m2(vf2) ---> vf1 = [m2(vf2) - m1(vi1)] / m1

    plug vf1 solved above into equation

    m1(vi1^2) = m1 ([m2vf2 - m1vi1] / m1)^2 + m2(vf2^2)

    m1(vi1^2) = m1 [(m2vf2 - m1vi1)/m1][(m2vf2 - m1vi1)/m1] + m2(vf2^2)

    m1(vi1^2)=m1/m2[m2^2(vf2^2) - 2m2vf2m1vi1 + m1^2(vi1^2)]+m2(vf2^2)

    (m1/m2 = 1/m1 so multiply both sides by m1 to give the left side m1^2 and the right side m1m2(vf2^2) and everything else stays the same)

    m1^2(vi1^2) =m2^2(vf2^2)-2m2vf2m1vi1 + m1^2(vi1^2) + m1m2(vf2^2)

    the m1^2(vi1^2) on both sides cancel

    0 = m2vf2(m2v2f - 2m1vi1 + m1vf2)

    set m2v2f - 2m1vi1 + m1vf2 = 0
    (please dont ask me why bc i didnt really understand bc my teacher just said well one of them must be equal to zero because the equation is equal to 0)

    m2vf2 + m1vf2 = 2m1vi1

    from here its easy to solve for vf2 then plug the solution into the momentum before = momentum after equation and solve for vf1

    I had problems with the algebra for this too....i hate using variables and not numbers. I hope this helps and you can understand each step, its hard to type it all out. Good luck!
     
    Last edited: Nov 7, 2006
  4. Nov 8, 2006 #3
    since yu have the masses sub for them in the 2 equations. you obtain
    5 v1f + 7.5 v2f = 10 ..........(1) and
    (1/2) 5 v1f^2 + (1/2) 7.5 v2f^2 = (1/2) 20 ..........(2)
    now divide eq # (2) by (1/2) also
    divide both eq by 5 to siplify - you get
    v1f + 1.5 v2f = 2 ..........(3) and
    v1f^2 + 1.5 v2f^2 = 4 ..........(4)
    see how simple they became
    solve # (3) for v1f and sub in # (4)
    solve the quadratic you obtained from # (4)
    you will get v2f
    sub this value in # (3)
    and here you go you obtain v1f
    The answers I obtained for
    v1f is - 0.4 m/s and for
    v2f is 1.6 m/s
    Best wishes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?