1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Conservation of momentum and mechanical energy

  1. Feb 4, 2010 #1
    (wasn't sure if I should post this in the general physics or in the "homework" forum.... this question is loosely based on a problem found in a printed book, but I have modified it myself.... regardless, here is the situation)

    1. The problem statement, all variables and given/known data

    A person is sliding down a frictionless incline.

    The person started at rest at an initial height of i meters above the ground, and has slid down some distance (so their vertical elevation is now only 3/5*i, or some such).

    At this point the person throws a block that s/he is carrying in exactly the opposite direction to their velocity.

    2. Relevant equations

    conservation of momentum, conservation of mechanical energy

    3. The attempt at a solution

    Momentum is conserved, and assuming that there are no losses of energy, energy should be conserved too.

    But, for some reason when I try to solve the equations, I only receive complex solutions. I am stumped.

    I am pasting the equations that I used below:

    (values/variables:
    3kg is the mass of the block thrown away
    20kg is the mass of the block and person
    15m is the initial height
    12m is the height above ground at which the block is thrown away
    c is the magnitude of the velocity of the block
    f is the magnitude of the velocity of the person after the block is thrown away
    n is the total mechanical energy of the system


    Total mechanical energy after the block is thrown away

    1/2*3*c^2+20*9.8*12+1/2*(20-3)*f^2=n

    Total mechanical energy at the top of the incline

    20*9.8*15=n

    Conservation of momentum

    20*v=(20-3)*f-c*3

    The magnitude of the velocity of the block-and-human immediately before the block is thrown away

    v^2=2*9.8*15
     
  2. jcsd
  3. Feb 4, 2010 #2
    I think that I know why the equations don't have non-complex solutions.

    The conservation of mechanical energy can't be applied, because the force created by the human is, in fact, an external force, even though it is coming from within the system.

    This force is a result of chemical changes of food into mechanical energy by the person's muscles/body.

    Is this correct?
     
  4. Feb 4, 2010 #3
    An external force does not come from the system.
    However, your velocity of system does not seem correct.
    Change in G.P.E=20*3*g=Increase in K.E
    So, v*v=6g?
     
  5. Feb 4, 2010 #4
    You are right - thanks for catching that (I did have the wrong height). :biggrin:

    I also agree that an external force does not come from the system.

    But, I don't think that mechanical energy is conserved, because the energy required for throwing the block comes from chemical energy (the food that the person ate is converted).

    So,

    ΔEsystem=0=ΔEmechanical + ΔEchemical

    where ΔEmechanical = - ΔEchemical

    That can be reduced to a simpler example.
    An astronaut floating at one point in space with a velocity of 0 takes off his shirt and throws it away. Clearly, there was a change in mechanical energy (v1=0, and v1>0).

    P.S. I think for the above reason I only get solutions that show that no change in velocity occurred after the block was thrown away (i.e., the system keeps going in the same direction).

    e.g.:
    E=2940
    c=-7.66
    f=7.66
    v=7.66
     
    Last edited: Feb 4, 2010
  6. Feb 5, 2010 #5
    The mechanical energy of the block is not conserved because there is an external force on it besides the conservative gravity.Similar arguments hold for the body throwing the block.
    Try conserving the momentum instead.

    But I think you were right. Energy indeed is not conserved.But don't you think a better approach will be to limit the system to the block so that we apply a now external force. In that case W(ext)=Change in mechanical energy?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook