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Homework Help: Elastic collision, one dimension

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data

    bloks b and c have m asses 2m and m respecti vely, and are at rest on a firctionless surface

    black a also of mass m.. is heading at speed v toward block b as show... determine te final velocity of each block after alll subsequent collisions are over, assum all collision are elastic

    aaaaaaa.jpg

    2. Relevant equations


    v_1f = (m_1-m_2/(m_1+m_2)) * v_1i + (2m_2/(m_1+m_2)) * v_2i

    and

    v_2f = (2m_1/(m_1+m_2)) * v_1i + (m_2-m_1/(m_1+m_2)) * v_2i

    3. The attempt at a solution

    i compared block A when it hits block B and then compared block C when block B hits it..

    i got...

    (m-2m / 2m+ m) * v

    block a) - 1/3 v

    found block b's v_i before it hits block c to be 2/3 v

    (2m - m / 3m) *2/3v

    block b) 2/9 v

    (2*2m / 3m) *2/3v

    block c) 8/9v






    i feel confident these answers are right but just trying to see what others think
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2
    Before a and b collide their total momentum is m_a * v

    After the collision of a and b if v_a = -1/3 and v_b = 1/3

    m_a*v_a + m_b*v_b = 1/3 m_a * v so momentum wasn't conserved
     
  4. Mar 2, 2008 #3
    mv + mv = mv + mv
    ai -- bi -- af --bf
    1 + 0 = 1/3 + 2/3

    1 = 1 = momentum conservation...

    negative for direction
     
    Last edited: Mar 2, 2008
  5. Mar 2, 2008 #4
    anyone verify my answer?
     
  6. Mar 2, 2008 #5
    you had -1/3v for the speed of a. -1/3 + 2/3 isn't equal to 1.
    If you substitute the right values in the formula for v2_f that you gave, you should get
    the right value for the speed of b after the first collision.
     
  7. Mar 2, 2008 #6
    v_2f = (2m_1/(m_1+m_2)) * v_1i + (m_2-m_1/(m_1+m_2)) * v_2i

    v_2f = (2*m / 3m) * v + 0

    v_2f = 2/3v
     
  8. Mar 2, 2008 #7
    i think my answers now are correct, and the negitive on the 1/3 is for direction, dosnt mean that momentum is not being conserved
     
    Last edited: Mar 2, 2008
  9. Mar 2, 2008 #8
    True.
     
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