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