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Throwing speed question (ball and a box)

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    A ball was thrown into a box. Determine Ek, P, and throwing speed when the:
    mass of softball = 172.7g
    mass of packed box = 2,043.5g
    and the box moved from 0 cm to 2cm.
    and the spring scale that remained parallel to the floor was Ff=8N.

    2. Relevant equations
    Ff=[itex]\mu[/itex]Fn
    Velocity: V=d/t
    Momentum: P=mv
    Work: W=F(cosθ)d
    Kinetic Energy: Ek=1/2mv2
    Collision: m1v1 + m2V2 = (m1 + m2)v'

    3. The attempt at a solution
    I started with Work.

    W=Fd
    =8(0.02)
    =0.16J

    Then tried solving for V
    1/2m1v2=1/2m2v2

    0.5(172.7)v12=0.5(2043)v22
    86.35v12=1021.5v22
    V1 = 11.83V2

    then plug it in to m1v1 + m2V2 = (m1 + m2)v' ?

    Thank you :)
     
  2. jcsd
  3. Nov 8, 2011 #2

    Simon Bridge

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    Good start. Unfortunately you don't know the initial speed of the box or the time it took to come to rest.

    Lynchpin: The kinetic energy of the ball provided energy to move the box.

    So what is the relationship between the work and the kinetic energy?

    What is the relationship between momentum and kinetic energy?
    (gives you p)

    You can find v from p right?
     
  4. Nov 8, 2011 #3
    So, we still need the Work?
    and work done = kinetic energy
    and relationship of momentum and kinetic energy?



    if we are trying to find v, and p=mv is the equation. do we rearrange?

    so, v = p/m? and then do we sub it in here?
    Ek = 1/2mv2

    then, Ek = 1/2m(p/m)2
    then, Ek = 1/2[STRIKE]m[/STRIKE](p[STRIKE]/m[/STRIKE])2

    but we still dont know P or Ek?
     
  5. Nov 8, 2011 #4

    Simon Bridge

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    These two statements are contradictory.
    Is Ek = W ?

    ... that is an m-squared in the denominator, so it does not cancel like that; make p the subject and you get:

    p=√(2mE)
     
  6. Nov 8, 2011 #5
    From what I learned (teaching us), Work done = the change in Ek (my bad on the first post.)

    So, this is what I am thinking now:
    W = Ek2 - Ek1
    W = 1/2m(v2'2 - v1'2)

    and do we add the mass of the ball and the box? or just use the ball?

    and thanks for correcting my math, lol XD
     
  7. Nov 8, 2011 #6

    Simon Bridge

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    All the kinetic energy in the ball does work on the box - we know this because the ball ends up stationary.
    So you can write Ek(ball)=W(box)=Fd=0.16J

    I know you have been taught to solve physics problems by an algorithm like this:
    1. list everything you know
    2. identify the type of problem
    3. list the equations that go with that type
    4. find the equation that has the numbers you know and the unknown you want
    5. plug the numbers into the equation - there is your answer

    Bad news: IRL it almost never works.
    You almost have to unlearn this approach.

    By the time you figure this out, you are supposed to have learned enough physics to cope and this is why you keep getting stuck.
     
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