anonymous
  • anonymous
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.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
You can use impulse-change-in-momentum relation.
anonymous
  • anonymous
I used it but I didn't get the answer. Can you show me?
anonymous
  • anonymous
*sorry, 'impulse = change-in-momentum relation'

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anonymous
  • anonymous
Sure, what did you get for the change in momentum?
anonymous
  • anonymous
F = change in momentum 0.5*12 - (-6*0.5)
anonymous
  • anonymous
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. :/
anonymous
  • anonymous
Note:...the Direction is opposite..)
anonymous
  • anonymous
Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]
anonymous
  • anonymous
Yes, the change in velocity is 12+6 since it changes direction.
anonymous
  • anonymous
That change in momentum divided by the time of contact (1.5s,really? That's a long contact time . . .) = the average force.
anonymous
  • anonymous
That is Impulse @CliffSedge
anonymous
  • anonymous
Impulse divided by time = force.
anonymous
  • anonymous
Oh THANK YOU ALL! I got it now! I didn't see the change in direction! :D
anonymous
  • anonymous
\[\large impulse=F_{avg} \cdot \Delta t\]
anonymous
  • anonymous
\[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]
anonymous
  • anonymous
lol...it is same as Finding change in momentum
anonymous
  • anonymous
\[F=\frac{ dP }{ dT }\]
anonymous
  • anonymous
Correct, but the question asked for average, not instantaneous.
anonymous
  • anonymous
wat is ur answer @CliffSedge
anonymous
  • anonymous
I'm sorry, I used the impulse formula, and I got the correct answer. :) It doesn't really make a difference.
anonymous
  • anonymous
I already provided it (in formula form, anyway)
anonymous
  • anonymous
wat is ur answer @amishra
anonymous
  • anonymous
6 Newtons
anonymous
  • anonymous
That's what I got too. :-)
anonymous
  • anonymous
\[\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}\]
anonymous
  • anonymous
I also recommend getting in the habit of checking the units in your formulas. \[\large N=kg\frac{m}{s^2}\] \[\large kg\frac{m}{s^2} = \frac{(kg)(m/s)}{(s)}\]
anonymous
  • anonymous
@CliffSedge Oh, thank you for sharing that :D

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