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amishra Group Title

A ball of mass 0.5kg is thrown vertically down at 12m/s onto a driveway. If it is in contact with the driveway for 1.5 seconds, and rebounds with a velocity of 6m/s, find the average force of the driveway on the ball.

  • one year ago
  • one year ago

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  1. CliffSedge Group Title
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    You can use impulse-change-in-momentum relation.

    • one year ago
  2. amishra Group Title
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    I used it but I didn't get the answer. Can you show me?

    • one year ago
  3. CliffSedge Group Title
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    *sorry, 'impulse = change-in-momentum relation'

    • one year ago
  4. CliffSedge Group Title
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    Sure, what did you get for the change in momentum?

    • one year ago
  5. Yahoo! Group Title
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    F = change in momentum 0.5*12 - (-6*0.5)

    • one year ago
  6. amishra Group Title
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    I took the change in velocity to be 6 m/s. And then I multiplied mass into this change in velocity. So I got 3 as my momentum. :/

    • one year ago
  7. Yahoo! Group Title
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    Note:...the Direction is opposite..)

    • one year ago
  8. Zekarias Group Title
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    Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]

    • one year ago
  9. CliffSedge Group Title
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    Yes, the change in velocity is 12+6 since it changes direction.

    • one year ago
  10. CliffSedge Group Title
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    That change in momentum divided by the time of contact (1.5s,really? That's a long contact time . . .) = the average force.

    • one year ago
  11. Yahoo! Group Title
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    That is Impulse @CliffSedge

    • one year ago
  12. CliffSedge Group Title
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    Impulse divided by time = force.

    • one year ago
  13. amishra Group Title
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    Oh THANK YOU ALL! I got it now! I didn't see the change in direction! :D

    • one year ago
  14. CliffSedge Group Title
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    \[\large impulse=F_{avg} \cdot \Delta t\]

    • one year ago
  15. CliffSedge Group Title
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    \[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]

    • one year ago
  16. Yahoo! Group Title
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    lol...it is same as Finding change in momentum

    • one year ago
  17. Yahoo! Group Title
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    \[F=\frac{ dP }{ dT }\]

    • one year ago
  18. CliffSedge Group Title
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    Correct, but the question asked for average, not instantaneous.

    • one year ago
  19. Yahoo! Group Title
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    wat is ur answer @CliffSedge

    • one year ago
  20. amishra Group Title
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    I'm sorry, I used the impulse formula, and I got the correct answer. :) It doesn't really make a difference.

    • one year ago
  21. CliffSedge Group Title
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    I already provided it (in formula form, anyway)

    • one year ago
  22. Yahoo! Group Title
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    wat is ur answer @amishra

    • one year ago
  23. amishra Group Title
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    6 Newtons

    • one year ago
  24. CliffSedge Group Title
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    That's what I got too. :-)

    • one year ago
  25. CliffSedge Group Title
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    \[\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}\]

    • one year ago
  26. CliffSedge Group Title
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    I also recommend getting in the habit of checking the units in your formulas. \[\large N=kg\frac{m}{s^2}\] \[\large kg\frac{m}{s^2} = \frac{(kg)(m/s)}{(s)}\]

    • one year ago
  27. amishra Group Title
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    @CliffSedge Oh, thank you for sharing that :D

    • one year ago
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