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Car of Mass Momentum Question

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data

    A car of mass 1.2 tonnes collides with a stationary van of mass 2.4 tonnes. After the collision the two vehicles become entangled and skid 15 m before stopping. Police accident investigators estimate that the magnitude of the friction force during the skid was 2880 N. Assume the road is horizontal and that all the motion takes place in a straight line.

    a) Find the speed of the vehicles just after the collision.

    b) Find the speed of the car before the collision.

    2. Relevant equations

    [tex]m_1u_1 + m_2u_2 = (m_1+m_2)v[/tex]

    [tex]p = mv[/tex]

    [tex]v^2 = u^2 + 2as[/tex]

    3. The attempt at a solution

    a) 1200u1 + 0 = 3600v

    to find out the velocity after the collision, use v as the initial velocity to travel 15m with a frictional force of 2880 N.

    v2 = u2 + 2as

    [tex]0 = u^2 + 2as[/tex]

    [tex]0 = u^2 + 30a[/tex]

    [tex]f = ma[/tex]

    [tex]f/m = a[/tex]

    [tex](3600v - 2880)/3600 = a[/tex]

    [tex] v - 288/360 = a[/tex]

    [tex]u^2 + 30(v-288/360) = 0[/tex]
     
    Last edited: Dec 31, 2008
  2. jcsd
  3. Dec 31, 2008 #2

    rock.freak667

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    you can use conservation of energy for the first part:
    The work done by friction to stop the vehicles= loss in kinetic energy of the car-van system
     
  4. Dec 31, 2008 #3

    rl.bhat

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    LaTeX Code: (3600v - 2880)/3600 = a

    LaTeX Code: v - 288/360 = a

    This step is wrong. Check it.
     
  5. Dec 31, 2008 #4
    [tex]-2880/3600 = a[/tex]

    [tex]0 = u^2 + 30(-2880/3600)[/tex]
    [tex]24 = u^2[/tex]
    [tex]u = 4.90[/tex]

    b) [tex] m_1u_1 + m_2u_2 = (m_1 + m_2)v[/tex]
    [tex]1200u_1 = 3600(4.898)[/tex]
    [tex]u_1 = 14.7ms^-^1[/tex]
     
  6. Dec 31, 2008 #5

    rl.bhat

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    That is right.
     
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