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amishra

  • 3 years ago

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.

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  1. CliffSedge
    • 3 years ago
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    You can use impulse-change-in-momentum relation.

  2. amishra
    • 3 years ago
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    I used it but I didn't get the answer. Can you show me?

  3. CliffSedge
    • 3 years ago
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    *sorry, 'impulse = change-in-momentum relation'

  4. CliffSedge
    • 3 years ago
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    Sure, what did you get for the change in momentum?

  5. Yahoo!
    • 3 years ago
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    F = change in momentum 0.5*12 - (-6*0.5)

  6. amishra
    • 3 years ago
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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. :/

  7. Yahoo!
    • 3 years ago
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    Note:...the Direction is opposite..)

  8. Zekarias
    • 3 years ago
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    Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]

  9. CliffSedge
    • 3 years ago
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    Yes, the change in velocity is 12+6 since it changes direction.

  10. CliffSedge
    • 3 years ago
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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.

  11. Yahoo!
    • 3 years ago
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    That is Impulse @CliffSedge

  12. CliffSedge
    • 3 years ago
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    Impulse divided by time = force.

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

  14. CliffSedge
    • 3 years ago
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    \[\large impulse=F_{avg} \cdot \Delta t\]

  15. CliffSedge
    • 3 years ago
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    \[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]

  16. Yahoo!
    • 3 years ago
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    lol...it is same as Finding change in momentum

  17. Yahoo!
    • 3 years ago
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    \[F=\frac{ dP }{ dT }\]

  18. CliffSedge
    • 3 years ago
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    Correct, but the question asked for average, not instantaneous.

  19. Yahoo!
    • 3 years ago
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    wat is ur answer @CliffSedge

  20. amishra
    • 3 years ago
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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.

  21. CliffSedge
    • 3 years ago
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    I already provided it (in formula form, anyway)

  22. Yahoo!
    • 3 years ago
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    wat is ur answer @amishra

  23. amishra
    • 3 years ago
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    6 Newtons

  24. CliffSedge
    • 3 years ago
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    That's what I got too. :-)

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

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

  27. amishra
    • 3 years ago
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    @CliffSedge Oh, thank you for sharing that :D

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