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Simple momentum question

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is moving along the x-axis
    with speed v when it collides with a particle of
    mass 2m initially at rest. After the collision, the
    first particle has come to rest, and the second
    particle has split into two equal-mass pieces that
    move at equal angles q > 0 with the x-axis.

    What is true about the speed of the new particles?

    2. Relevant equations

    conservation of momentum:
    mv = mv

    3. The attempt at a solution

    According to the answer, each new particle moves with a speed greater than v/2. This seems to violate conservation of momentum to me...but as most people get this question right, I'm probably missing something obvious...
     
  2. jcsd
  3. Jun 22, 2010 #2
    Ok, so conservation of momentum, m1v1 + m2v2 = m1v'1 + m2v'2. Lets call m1 the mass that hits and m2 the mass initially at rest. Since the target is initially at rest and after collision the mass that m1 is at rest, our equation reduces to: m1v1=m2v'2
    v'2=(m1/m2)*v1, and since m2=2*m1. v'2 = (1/2)*v1
     
  4. Jun 22, 2010 #3
    According to the answer, each new particle moves with a speed greater than v/2. This seems to violate conservation of momentum to me...but as most people get this question right, I'm probably missing something obvious...[/QUOTE]


    Before collision total momentum is mv. (It is rigt)

    After collision , 2m split 2 particle into equal mass m but moving at equal angles q > 0 with the x-axis.
    Let say after collison velocity V'. You must divide 2 componets for each particle and consider for x components. Use law of conservation of momentum for x axis , find V' then think!!!!
     
  5. Jun 22, 2010 #4
    got it. i'm dumb haha.
     
  6. Jun 22, 2010 #5
    yeah you have to break in into components because the x components together add to the initial.. the y components cancel each other out. This is why v > 1/2v
     
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