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Energy of conservation question

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A package of mass m is released from rest at a warehouse loading dock and slides down a h = 3.7 m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

    (a) Suppose the packages stick together. What is their common speed after the collision?

    (b) Suppose the collision between the packages is elastic. To what height does the package of mass m rebound?

    2. Relevant equations

    energy of conservation eqn

    3. The attempt at a solution

    the velocity acquired that mass m when it reaches the bottom of the path
    v = √2gh
    h = height of fall

    So I got v= 7.668 m/s (I went sqrt(2*9.8*3) )

    the kinetic energy of the mass by the time it collides with the big mass 2m will be
    E = mv^2 / 2
    after collsion the kientic energy of the combined mass will be equal to the initil kinetic energy of the mass m

    ( m +2m) * v'^2 / 2 = mv^2 / 2

    v ' = is the final velocity of the combined mass .

    So I got v= 4.4 m/s

    Does that look right???

    b) velocity of the mass m after collision
    v' = (m2 - m1) v / (m1+m2)
    v = √gh
    m2 = 2m
    m1 = m
    v = √gh .
    the height reached by teh mass after collsion
    h = v'^2 / g


    ARE THE STEPS TAKEN IN QUESTION B CORRECT? or should it be like this:

    where vf= final velocity of package with mass m right after collision
    vi= the velocity from part a (7.668 m/s)

    vf= ((m1 - m2)*vi) / (m1 + m2)

    = ((m - 2m)*vi) / (m + 2m)

    = (-m*vi)/ (3m)

    = -vi/3

    = -7.668/3 = -2.556 m/s


    So for mass m:

    where Kf= final kinetic energy
    Ki= initial kin energy
    Ugf= final gravitational potential energy
    Ugi = initial grav. pot. energy
    yf = final height
    vi = initial velocity (THIS time I used the 2.556 m/s I just found not 7.668 m/s)


    Kf + Ugf = Ki + Ugi

    Kf and Ugi = 0 so,

    solving for yf I get:

    yf = vi^2 / 2g

    = ((-2.556)^2) / (2*9.8)

    = 0.33 m


    IS THIS RIGHT? OR ARE NEITHER CORRECT?


    need some quick help :/

    thanks a lot in advance!
     
  2. jcsd
  3. Feb 27, 2009 #2

    LowlyPion

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    Homework Helper

    The initial speed is wrong, because the height is 3.7, not 3 as you used.
    Secondly, Kinetic energy isn't conserved in an inelastic collision (where they stick together).

    But momentum is conserved. So

    V = mv/3m = 1/3*v

    For part b, you have an elastic collision, and here kinetic energy is conserved.

    2 equations (momentum and kinetic energy) and 2 unknowns the 2 final velocities ... solve.
     
  4. Feb 27, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    In part A your method is correct, however you used the wrong value of height to calculate v.

    I am not following part B. Are you using conservation of momentum to set up an equation? Also I do not see the 2m mass's after-collision velocity anywhere.
     
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