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Force and Velocity to find Mass (via soccer)

  1. Jul 8, 2015 #1
    Here is the problem, and I'm banging my head against a wall trying to set it up:
    An athlete catches a ball traveling at 20 m/s with her hands. Between the time she makes contact with the ball and the time the ball stops moving, the ball travels a distance of 20 cm. The goaltender feels a force of 500 N against her hands when she stops the ball. What is the mass in kg of the ball?

    I know I need to find the acceleration here, and I started with a = (v-v0)/t. v0 =20 m/s, however I'm stuck on how to get v. I understand there is a huge deceleration going on from when the ball hits her hands to when it stops. I tried doing v=distance/time, so v=0.20m/t, then plugging this in for a, but this seemed to get way too complicated, where a=(.2m/t - 20)*1/t...I think I'm not on the right track here.

    Other equations needing to be used of course are ΣF=ma, but there again I circle back to the acceleration. Also, do I need to take gravity into consideration here, or do I need to break this into scalar x/y components? But if so I don't have an angle to work with, only the 20 m/s.


    I'm just looking for guidance in how to set up this problem. Any help is appreciated. Thank you!
     
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  3. Jul 8, 2015 #2

    SammyS

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    What is 20m/s ?
     
  4. Jul 9, 2015 #3
    Initial velocity (v0) is 20 m/s, right? So I'm trying to find final velocity (v), when the ball is in her hands slowing down and coming to a stop.
     
  5. Jul 9, 2015 #4

    Nathanael

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    The ball comes to a stop. What does it mean to "come to a stop"? :wink:

    The real problem with your method is that you don't know "t" the time that the collision lasts.

    It's still possible with your method, but if you know about energy (specifically the work-energy theorem) I suggest using that for a simpler solution.


    About gravity, that is a good point. I am pretty sure the problem expects you to ignore gravity.
     
  6. Jul 9, 2015 #5
    Thanks Nathanael, but I'm still left a little confused. So you're saying that final velocity is 0 (coming to a stop - that makes sense), which gives me that a=(v-v0)/t = -20/t. If I plug this into ΣF=ma, which would be 500=m(-20/t), I'm still left with 2 unknown variables, nor have I used the 20 centimeters in my equation. Where does that come into play?
     
  7. Jul 9, 2015 #6

    SteamKing

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    Hmm.... initial velocity, final velocity, distance traveled, time: if only there were formulas which used all these quantities together ...
    I dunno, maybe that Newton guy has some ideas ... falling apples and whatnot.
     
  8. Jul 9, 2015 #7

    Nathanael

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    Right, you have to somehow use the 20 cm to replace the time t
    (Meaning you should write another equation involving the distance travelled, 20cm, and the time of the collision, t, then you can eliminate the unknown t)

    Do you know about work and energy? The work done is the change in kinetic energy. That will give you the answer at once.
     
  9. Jul 9, 2015 #8
    Haven't gotten to the chapter on work and energy yet. But just rearranged the equation to get t=.01, which led me to the mass of the ball at .25 kg, hooray! Thank you for not being snarky in regards to me slowly grasping these physics concepts. I appreciate your help!
     
  10. Jul 9, 2015 #9

    Nathanael

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    You're welcome :smile: but I don't think we are finished quite yet. How did you rearrange the equation to solve for t if there are two unknowns (t and m)?

    If you know about the "kinematics equations" (commonly called "suvat equations") then you can solve the problem at once by finding one of those equations that does not involve time (and then plug in the variables and solve for m).

    What I suggested ("write another equation for the distance travelled and eliminate t") was merely a way of deriving that "suvat equation" which doesn't involve time.
    (I am no good at memorizing things so that is the way I work: I re-derive whatever I need.)
     
  11. Jul 9, 2015 #10
    Hi again,
    I understand the idea of blending the two equations to only get one unknown variable.
    I think what I initally did doesn't work, which is to set up the equation v=delta distance/delta time, then switching this to t=delta distance/delta v (which is 0-20). I plugged in t=.2 m/-20m/s, but that doesn't work in that it gives a negative number for t of -.01, right? Because that's what I used then to give me the acceleration, then the mass. But that doesn't seem right.
    So I will look the at substituting the one equation into the other to get one variable. Thank you again!
     
  12. Jul 9, 2015 #11

    Nathanael

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    In this equation (v=Δd/Δt) v is the average speed. So you don't want to use 20 m/s in that equation (it is not traveling 20 m/s the whole time).

    When acceleration is constant (like in this problem) the average speed is given by (vi+vf)/2
     
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