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Direct central impact occurs between a 300N body moving to

  1. Jun 12, 2012 #1
    Hi all

    1 )
    Direct central impact occurs between a 300N body moving to the right with velocity of 6 m/s and 150 N body moving to the left with velocity of 10 m/s . Find the velocity of each body after impact if the coefficient of restitution is 0.8 .

    my answer :

    Given ::
    m1 = 300 /9.81 = 30.58 kg
    u1 = 6 m/s
    m2 = 150/9.81 = 15.2 kg
    u2 = -10 m/s
    e = 0.8
    V 1 ? v2 ?


    m1u1+m2u2 = m1v1+m2v2
    30.58X 6 + 15.2 X (-10) = 30.58v1+ 15.2 v2
    183.48 -152 = 30.58v1+ 15.2 v2
    31.48 = 30.58v1+ 15.2 v2 ===== this is first equation (1)

    e(u1 - u2) = (v2 - v1)
    0.8(6 + 10) = v2 - v1
    4.8 + 8 = v2 - v1
    12.8 = v2 - v1
    v2 = 12.8 + v1 ===== this is first equation (2)

    make v2 in fist equation
    31.48 = 30.58v1+ 15.2 (12.8 + v1)
    31.48 = 30.58v1+ 194.56 + 15.2 v1)
    31.48 - 194.56= 30.58v1 + 15.2 v1
    -163.08 = 45.78
    v1 = -3.6 m/s

    to find v2 = 12.8 + (-3.6) = 9.2 m/s
     
  2. jcsd
  3. Jun 12, 2012 #2
    Hi manal950! :smile:

    Were you unsure you solved it correctly? Your approach looks correct to me.
     
  4. Jun 12, 2012 #3
    thanks so much ..
     
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