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Homework Help: Mechanics question-Momentum&F=ur

  1. Jan 14, 2008 #1
    [SOLVED] Mechanics question-Momentum&F=ur

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

    An ice hockey player is required to strike a puck having a mass of 1.2 kg, so that it is
    accelerated uniformly from rest across a smooth ice rink surface so as to cover a distance of 25 m in 8 s.
    The puck travels another 8 m before coming to a rest. Determine (a) the
    acceleration that the puck has to experience, (b) the minimum force the ice hockey player has to apply to the puck to achieve this acceleration, (c) the momentum of the puck at the end of the 8 s period, (d) the frictional force that brings the puck to rest, and (e) the coefficient of kinetic friction between the ice and the puck.

    2. Relevant equations

    [tex]F=\mu R[/tex]

    3. The attempt at a solution

    (a)use the formula shown above u=0 t=8s s=25m therefore simply substitue into equation a=-0.78125
    (b)use 3rd equation m=1.2kg a=-0.78125 f=0.9375N
    (c)momentum is g=mv . Once again substitute values into equation. m=1.2kg v=6.25m/s found out by using s=1/2 (u+v)t g=7.5kg m/s
    (d)&(e). Both use the 2nd formula but i'm unsure as how to go about it. I know R=mg. but F can't be the same as in f=ma can it?

    Also check over the other stuff too please for any mistakes thanks.
  2. jcsd
  3. Jan 14, 2008 #2


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    For (d) you know the puck comes to rest in 8m from the speed you calculated above. You need to use the kinematic equations to find a and then F. Once you find F you can use the second equation to get [itex]\mu[/itex].
  4. Jan 14, 2008 #3
    oh ok. that means that the new acceleration [tex]a_2=-0.59185606[/tex] found by using the equation [tex]v^2=u^2 + 2as[/tex] (u=6.25m/s s=33 v=0) when inserted into f=ma gives me [tex]f_2=-0.710227272[/tex] then inserting that into [tex]F=\mu R[/tex] gives me [tex]\mu=-0.060331912[/tex].

    Is the working out for the other parts of the question right too?
    Last edited: Jan 14, 2008
  5. Jan 14, 2008 #4


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    Seems to be but I'm not sure why your acceleration in the first is negative. The question is a bit weird looking at it again. It assumes no friction for the first part then suddenly friction comes in to slow it down. Oh well. :smile:
  6. Jan 14, 2008 #5
    i probably messed up somewhere. Thanks for the help much appreciated.
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