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!

Quetions of body under collison

  1. Jun 16, 2012 #1
    Hi all

    http://store2.up-00.com/June12/pG127901.jpg [Broken]


    V^2 = u^2 + 2as
    V = 8.85 m/s which will be intial veloctiy befor hits the second body

    Bu using
    e(u1 - u2 )=V2 -V1
    m1u1 +m2u2 =m1v1+m2v2



    for the body which is 20 N
    mass 20.3
    u1 = 8.86 0m/s
    V1 after collision = 0 m/s
    for the second body which is 40 N
    mass = 4.077
    u2 = 0
    v2 we will find final velocity after collion
    m1u1 +m2u2 =m1v1+m2v2
    2.03 X 8.86 + 4.077 X 0 = 2.03X(0) + 4.077V2
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 16, 2012 #2
    v1 after the collision is not 0, this is because the objects have different masses.
    But since u2 is 0, you can solve for v1 using:

    V = (m1-m2)u/(m1 + m2)

    After solving for v, using that for kinetic energy and set that equal to potential energy.
     
  4. Jun 17, 2012 #3
    you mean by ""using that for kinetic energy and set that equal to potential energy""

    this equation
    m1u1 +m2u2 =m1v1+m2v2

    and is the final answer for height
    1.78 and 1.44 m
     
  5. Jun 17, 2012 #4
    Because we have 2 unknown V1 and V2 I will find it by this two equation ...

    m1u1 +m2u2 = m1v1 + m2V2
    2.03X8.8+4.077(0) = 2.03(v1)+4.077V2
    17.6204 = 2.03V1 + 4.077V2 ====== fist equation (1)

    e(u1-u2) = (V2 - V1)
    0.8(8.86 - 0 ) V2 - V 1
    V2 = 7.088 + V 1 this second equation (2)
    so I will make V2 in first equation
    17.6204 = 2.03V1 + 4.077(7.088 + V1 )
    17.6204 = 2.03V1 + 28.89+4.077V1
    V1 = -1.84m/s
    Now I will findV2
    17.06204 = 2.03(-1.84) + 4.077V2
    21.6551 = 4.077V2
    V2 = 5.24 m/s

    so is that correct or not ??
     
    Last edited: Jun 17, 2012
  6. Jun 17, 2012 #5
    I actually gave you the wrong equation, my mistake.

    The correct equation is:

    v2 = (2m1 u1)/(m1+m2)

    Solve for v2, then use that value for kinetic energy and set that equal to potential energy, which is mgh.
     
  7. Jun 17, 2012 #6
    see my post number 4
     
  8. Jun 17, 2012 #7
    Yes, the method is correct, but the velocity is off by a bit.
     
  9. Jun 17, 2012 #8
    m1u1 +m2u2 = m1v1 + m2V2
    2.038735X8.85889+4.07747(0) = 2.038735(v1)+4.07747V2
    18.0609 = 2.038735(v1) + 4.07747V2 ====== first equation (1)

    e(u1-u2) = (V2 - V1)
    0.8(8.85889 - 0 ) V2 - V 1
    V2 = 7.087112 + V 1 this second equation (2)
    so I will make V2 in first equation
    18.0609 = 2.038735(v1) + 4.07747(7.087112 + V 1)
    18.0609 = 2.038735(v1) + 28.89748 + 4.07747V1
    18.0609 - 28.89748 = 2.038735(v1) + 4.07747V1
    -10.83658 = 6.116205 V1
    V1 = -1.77178 m/s

    Now I will findV2
    V2 = 7.087112 + (-1.77178)
    V2 = 5.3153 m/s

    ----------- now for the e = 1 I will get velocity V2 -----------------
    e(u1-u2) = (V2 - V1)
    1(8.85889 - 0 ) =V2 - V1
    V2 = 8.85889 + V 1
    I will make V2 in 1
    18.0609 = 2.038735(v1) + 4.07747(8.85889 + V 1)
    18.0609 - 36.12185 = 6.116205V1
    V1 = -2.9529 m/s
    V2 = 8.85889 + (-2.9529)
    v2 = 5.9059 m/s


    Now finding the height :
    when e = 1
    1/2mv2 = mhg
    0.5(2.03873)(5.9059)^2 = 2.03873 X 9.81 X h
    h = 1.7 m

    when e = 0.8
    1/2mv2 = mgh
    0.5(2.03873)(5.3153)^2 = (2.03873)gh
    m = 1.4 m
     
    Last edited: Jun 18, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook