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Velocity after Impact

  1. Feb 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Just prior to impact with a golf ball (0.042 kg), a clubhead (0.195 kg) is travelling with a velocity of 33.7 m/s horizontally towards the target and a vertical velocity of 0 m/s. The coefficient of restitution between the ball and club is 0.79. Impact lasts for 0.00043 s. The launch angle of the ball after impact is 13.1°. During impact the ball behaves as a linear spring with a stiffness of, K= 1182019, and is maximally compressed from its resting state by Δx = 0.01 m.

    What is the velocity of the club, towards the target (in the typical horizontal X direction), immediately after impact?

    I'm stuck on this one and I need to know how to do this one in order to do the last four questions on my bonus assignment. Help would be greatly appreciated! Thanks in advance!

    2. Relevant equations
    v'={M[V(1+e)-ev]+mv}/M+m

    3. The attempt at a solution
    v'={0.042[0(1+0.79)-(0.79)(33.7)]+(0.195)(33.7)}/(0.042+0.195)
    v'=23.01 (incorrect)
     
  2. jcsd
  3. Feb 18, 2016 #2

    haruspex

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    What do you know about coefficient of restitution? (The tricky part is stating the relevant velocities accurately.)
     
  4. Feb 18, 2016 #3
    e=v'/v

    OR
    e=(V'-v')/(V-v)

    I know the golf ball is at rest at the beginning, but the club is not so I am inclined to think I use the second version. so:

    e=(V'-v')/(V-v)
    0.79=(V'-v')/(V-v)
     
  5. Feb 18, 2016 #4

    haruspex

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    As I posted, the tricky part is specifying the velocities correctly in that equation. This is not a simple head on collision. The ball moves off in a different direction from the club's prior velocity. Precisely what velocities should you plug into the equation?
     
  6. Feb 18, 2016 #5
    Should I get the x and y components of the club head's velocity?
     
  7. Feb 19, 2016 #6

    haruspex

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    It depends what you mean by x and y. You are told that the club head is moving horizontally prior to impact. What does the angle of the ball's flight from the impact tell you about the face of the club?
     
  8. Feb 19, 2016 #7
    IMG_0131.jpg

    This is what I picture happening when I read the question...
    Perhaps the club head is angled? Or the club begins moving upward at an angle of 13.1 degrees before it hits the ball?
     
  9. Feb 19, 2016 #8

    SammyS

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    It does not move upward. According to the given information it moves horizontally before impact.

    Hint: Use conservation of momentum.
     
  10. Feb 19, 2016 #9
    MV+mv=MV'+mv'
    (MV+mv-mv')/M=V'
    (0.195)(33.7)+(0.042)(0)-(0.042)(48.34)/0.195=V'
    23.29m/s=V'

    This is the horizontal velocity of the golf ball (didn't show the work because it was calculated in a previous question on my assignment).

    I was very confused about how to get v', a classmate helped me out with that bit.

    Thank you both for your help!
     
    Last edited: Feb 19, 2016
  11. Feb 19, 2016 #10
    1. The problem statement, all variables and given/known data
    Just prior to impact with a golf ball (0.042 kg), a clubhead (0.195 kg) is travelling with a velocity of 33.7 m/s horizontally towards the target and a vertical velocity of 0 m/s. The coefficient of restitution between the ball and club is 0.79. Impact lasts for 0.00043 s. The launch angle of the ball after impact is 13.1°. During impact the ball behaves as a linear spring with a stiffness of, K= 1182019, and is maximally compressed from its resting state by Δx = 0.01 m.

    What was the average force, exerted in the target (x, horizontal) direction, on the ball by the clubhead during the time of impact?

    2. Relevant equations
    F=ma
    MV+mv=MV'+mv'
    ΣFΔt=mΔv
    F=kx

    3. The attempt at a solution
    We're currently doing momentum/impacts/collisions/projectiles in lecture, but my professor hasn't mentioned anything about k or Hooke's Law yet. I kiiiind of remember it from high school physics and googled a bit about it and found the equation above.

    First I tried this:
    F=kx
    F=(1182019)(0.01)
    F=118201.9

    Which turned out to be incorrect so I tried this:
    ΣFΔt=mΔv
    ΣF=(mΔv)/Δt
    ΣF=[(0.195)(33.7)]/0.00043
    ΣF=15282.56

    Which ended up also being incorrect so I tried this:
    F=ma
    F=m(Δv/Δt)
    F=(0.195)(33.7/0.00043)
    F=15282.56

    Which is also incorrect...
    I have the following from previous questions:
    • the speed of the ball immediately after impact=49.63
    • velocity of the ball, towards the target (in the typical horizontal X direction), immediately after impact=48.34
    • the vertical velocity of the ball immediately after impact=11.25
    • the velocity of the club, towards the target (in the typical horizontal X direction), immediately after impact=23.29
    • vertical velocity of the clubhead immediately after impact=-2.42
     
  12. Feb 19, 2016 #11
    Never mind I got it!

    ΣFΔt=mΔv
    ΣF=(mΔv)/Δt
    ΣF=[(0.195)(33.7-23.29)]/0.00043
    ΣF=4720.81

    Forgot to use the change in velocities... Silly mistake on my part!

    I am curious however if there is a way to do this using the k value? I didn't have to use it at all for any of the parts of this question... Wondering if maybe my professor just gave it to us to throw us off?
     
  13. Feb 19, 2016 #12

    haruspex

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    Had I known you already had the golf ball velocity I would not have banged on about the coefficient of restitution. However, I disagree with your answer for that. I get 48.34m/s for the speed of the golf ball, not just its horizontal component. The horizontal component would be 47.1.
     
  14. Feb 19, 2016 #13

    haruspex

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    The trouble with the k value is that it is only the same during compression and decompression for an ideal spring. Since the coefficient of restitution of the club+ball is less than 1, we cannot tell if it is ideal.
    Assuming the given k value applies to decompression, the energy equation gives a ball speed of 53.05m/s. At the other extreme, we can suppose k applies to compression and all losses are in the non-ideal nature of the ball, so multiply by 0.79 to get 41.91m/s.
     
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