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Momentum, Impulse and consevation of momentum?

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations
    [tex]\vec{p}[/tex] = [tex]m\vec{v}[/tex]

    [tex]F_{net}[/tex] [tex]\times[/tex] ([tex]t_{2}[/tex] - [tex]t_{1}[/tex] )= [tex]m_{2}[/tex] [tex]v_{2}[/tex] - [tex]m_{1}[/tex] [tex]v_{1}[/tex]


    3. The attempt at a solution
    So part A was easy, All i did was find the Velocity just before it hits using constant acceleration formula and times that velocity by mass.
    So, my attempt went something like this, since [tex]\vec{p}[/tex] = [tex]m\vec{v}[/tex] , I have mass I just need to find velocity just after it hits the table.

    I tried using law of conservation of energy to find velocity, but I can only take the points when it hits the wall and not JUST AFTER because I don't know the height. So this didn't work out too well

    I know I can find the force the ball exerts on the wall (and vice ver sa due to Newton's 3rd law), so because the ball was just released / drop, it's acceleration is just 9.8, that times the mass will give me the force when the ball hits the wall. Now CAN I assume that the force of the ball when it's going back up is the same magnitude of the force just as the ball hits the ground on it's way down (because the wall and the ball remained in contact for .015 seconds)?


    Now, if I was allowed to assume that, the force would be 4.9N (9.8 * .05).
    Momentum Impulse theory states that net force times the difference in time = the difference in momentum,
    [tex]F_{net}[/tex] [tex]\times[/tex] ([tex]t_{2}[/tex] - [tex]t_{1}[/tex] )= [tex]m_{2}[/tex] [tex]v_{2}[/tex] - [tex]m_{1}[/tex] [tex]v_{1}[/tex]


    So I have Fnet, delta t, the masses, and [tex]v_{1}[/tex].
    My second question would be, how do I select the 2 points so I can assign which one is t1 and which one is t2

    again IF my understanding is right (please correct me), I am going to say t1 is when the ball JUST hits the wall and t2 its the point when the ball is about to leave contact with the wall, so t1 will be 0 and t2 will be .015 seconds.

    So plugging all that back into my impulse momentum theory
    I have

    4.9 N (0.015s-0s) = [tex].05kg v_{2}[/tex] - .05kg * 0m/s (because it has JUST hit the wall and remained contact so no velocity)

    Solve for [tex]v_{2}[/tex] I get 1.47m/s
    multiply that number by its mass (.05kg)
    and I get .0735 momentum, which they told me is wrong


    So where did I go wrong / not understanding?
     
  2. jcsd
  3. Apr 14, 2008 #2

    dynamicsolo

    User Avatar
    Homework Helper

    The approach you used to find the ball's velocity just before impact is equivalent to using conservation of mechanical energy (since we are treating gravity as the only force acting on the ball while it falls). You use constant acceleration to find

    (v_f)^2 = (v_i)^2 + 2·g·(h_i) , giving you

    (v_f)^2 = 0 + 2 · 9.81 · 1.5 ,

    from which you found the correct velocity and momentum just before impact. But if you multiply this equation by (1/2)m , you get

    (1/2)·m·(v_f)^2 = (1/2)·m·(v_i)^2 + m·g·(h_i) , or

    KE_f (+PE_f) = KE_i + PE_i ,

    where the final (gravitational) potential energy is in parentheses because we take it as zero at the level of the tabletop.

    You can use this on the trip up because you do know that the final height is 1.0 m , where the ball momentarily comes to rest. So you know the new total of kinetic energy plus potential energy for the ball's climb, and thus can work backwards to find the ball's kinetic energy after the impact. You will find that some kinetic energy has been lost.

    But you just want the velocity right after impact out of this, so that you can find the change in linear momentum of the ball during the impact. This is also called the "impulse" acting on the ball. Since you are given the time duration of the collision with the tabletop, how do you find the net force during the collision?

    (You could also use the constant acceleration kinematics, as you did before -- since this is an equivalent argument, knowing that the ball starts with some unknown velocity and comes to rest at a height of 1.0 m, under a constant deceleration of g. Hint: it's the same speed as if the ball had been dropped from 1.0 m.)
     
    Last edited: Apr 14, 2008
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