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Inelastic Collision with friction

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball of clay collides with a block that sits on a frictional surface. If the collision proceeds in 1 dimension and the clay sticks to the block, how far does the block/clay system slide before coming to a stop? The clay has mass m = 50 kg, the block as mass M = 100 kg, the initial velocity of the clay is 20 m/s, and the coefficient of kinetic friction between the block and the surface is 0.2.


    2. Relevant equations

    m1v1+m2v2=(m1+m2)vf
    v_ave=(vi+vf)/2

    3. The attempt at a solution

    50kg(20 m/s) + 100kg(0 m/s) = (50kg+100kg)vf
    = 1000 kgm/s=150kgvf
    = 6.66m/s = vf (speed of both objects after collision)

    Rate of deceleration :
    .2 x 9.8m/s^2 = 1.96 m/s^2

    The velocity is 6.66 m / s, so the time it takes to stop is:
    6.66 / 1.96 ~= 3.40 s

    Average speed with linear deceleration is 1/2 initial speed or:

    6.66 * 1/2 = 3.33 m/s

    So the distance traveled is average speed times the time it took to stop:

    3.33 m/s * 3.40 s ~ 11.33 m.

    Did I do this right or wrong?

    Thanks in advance!
     
  2. jcsd
  3. Mar 2, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Hi scruffles,

    Your solution looks right.

    ehild
     
  4. Mar 2, 2012 #3
    Thanks!
     
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