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Question + Problem on Collision/momentum

  1. Apr 8, 2009 #1
    1. A block of mass 1.3kg is resting at the base of a frictionless 40-degree ramp. A bullet of mass 50g is traveling parallel to the ramp's surface at 250 m/s. It collides with the block, enters it, and exits the other side at 100 m/s. How far up the ramp will the block travel?


    http://img26.imageshack.us/img26/7883/60347116.png [Broken]


    2. M1V1+M2V2=M1V1'+M2V2' to solve for velocity. sumFx=M*A to find acceleration, and V'^2=V^2 +2adX, V = initial velocity, V' = final velocity, dX = delta or change in distance, and a = acceleration



    3. So first we need to find the velocity of the block, so we use the first equation (M1V1+M2V2=M1V1'+M2V2' ) to get: (0.05)(250)+(1.3)(0) = (0.05)(100)+(1.3)(V2'), solving it gives us V2'=5.7m/s

    We then use this velocity as the initial because the block starts moving up the ramp. It will stop on the ramp, so final velocity V' = 0 m/s. We have to find the acceleration, and to make this easier we can make the problem one demensional. We set the ramp's surface as the X-axis and then find find Wx (force due to gravity pushing the block down the ramp). The weight = (1.3+0.05)(9.8)= 13.23. I set the problem so moving up the ramp is in the positive direction. so the summation problem looks like -Wx=M*A. I can now solve for acceleration as (-13.23(sin(40)))/1.35 = a = -6.29. We have a negative acceleration because gravity is pushing the opposite direction. So now that i have V, V', and a, i can use the equation V'^2=V^2+2adX to get 0=5.7^2+2(-6.29)dX dX=2.58m


    The answer key shows the answer to be 2.65, so is this due to rounding or a problem during my steps?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 8, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    It's roundoff error--your method is fine.
    Recalculate that speed. (Don't round off until the very last step in your calculations.)
     
  4. Apr 8, 2009 #3
    Thanks Doc
     
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