Momentum; bullet block of wood; distance; friction.

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data
    A 10g bullet is travelling at a speed of 1.5*10^2 m/s parallel to a horizontal surface when it strikes a 5.5kg wooden block. If the bullet becomes lodged in the block, how far will the block move along this surface? The coefficient of friction between the block and the surface is 0.25.

    2. Relevant equations
    p = F(delta)T

    d = 0.5*a*t^2

    p = mv

    3. The attempt at a solution

    p = (0.010kg)*(150m/s) = 1.5 kg*m/s
    Ff = (0.25)(5.51) = 1.3775

    Ff = Fp // when the force created by momentum is equal to the force of friction.

    Ff = p / (delta)t

    (delta)t = p / Ff

    (delta)t = 1.0889s

    a = F/m

    a = p / t*m

    a = 0.25m/s^2

    d = 0.5*0.25*(1.0889)^2

    d =0.15m

    The answer in the book is: 0.015m. Did I make a mistake somwhere or is the answer a misprint?
  2. jcsd
  3. Jul 2, 2009 #2
    You made an error while calculating the frictional force
  4. Jul 2, 2009 #3
    Oh, I see. Thanks.
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