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Block 1 slides down and collides with block two pushing it onto a friction surface.

  1. Oct 26, 2011 #1
    Block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.1 m and then collides with stationary block 2, which has mass m2 = 5m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction µk is 0.35 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?

    Answers
    (a) 0.984 m
    (b) 0.246 m


    Relevant Equations
    m1u1 + m2u2 = m1v1 + m2v2
    vf^2 = vi^2 + 2as (where s is the Δx)
    _______________________________________________

    This is what I attempted.
    I started the problem by working on part (b) first because it seemed that finding the distance traveled through and inelastic collision would be easier due to the shared final velocity. I know that,
    h = 3.1m
    m2 = 5m1
    μk = 0.35.

    so I use the formula m1u1 + m2u2 = (m1 + m2)v. However, since block two has no initial velocity the equation becomes m1u1 = (m1 + m2)v. Replacing m1u1 with mgh I get,
    v(m1 + m2) = mgh
    v = m1gh/(m1 + m2)
    v = m1(9.8)(3.1)/(5m1+m1)
    v = 30.38m1/6m1
    v = 5.06

    From there I use the kinematic equation
    vf^2 = vi^2 + 2as
    knowing vf = 0 and, by searching the web I found that F = ma = -μkmg making, a = -μkg I plug in my knowns into the equation.
    s = -vi^2/-μkg
    s = -(5.06^2)/-0.35(9.8)
    s = 7.46

    Now thank goodness I took geometry because 7.46 = 0.246 does not pass the reflexive property! A joke of course but this brings me to my question: what did I do wrong? Obviously something horrible which will dishearten physics lovers everywhere but please, show pity. I'm having difficulty understanding this.

    Can somebody explain to me what I did wrong and what I should have done to get the right answer instead and why (especially why! Baby words please!). I will very much appreciate your help. Thank you for taking the time to review my question.
     
  2. jcsd
  3. Oct 26, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Block 1 slides down and collides with block two pushing it onto a friction surfac

    There's your problem. m1u1 is a momentum; mgh is an energy. You cannot set one equal to another.

    Use mgh to find m1's speed at the bottom, then you can calculate m1u1.
     
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