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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
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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CliffSedgeBest ResponseYou've already chosen the best response.2
You can use impulsechangeinmomentum relation.
 one year ago

amishraBest ResponseYou've already chosen the best response.0
I used it but I didn't get the answer. Can you show me?
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
*sorry, 'impulse = changeinmomentum relation'
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Sure, what did you get for the change in momentum?
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
F = change in momentum 0.5*12  (6*0.5)
 one year ago

amishraBest ResponseYou've already chosen the best response.0
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

Yahoo!Best ResponseYou've already chosen the best response.0
Note:...the Direction is opposite..)
 one year ago

ZekariasBest ResponseYou've already chosen the best response.0
Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Yes, the change in velocity is 12+6 since it changes direction.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
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

Yahoo!Best ResponseYou've already chosen the best response.0
That is Impulse @CliffSedge
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Impulse divided by time = force.
 one year ago

amishraBest ResponseYou've already chosen the best response.0
Oh THANK YOU ALL! I got it now! I didn't see the change in direction! :D
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
\[\large impulse=F_{avg} \cdot \Delta t\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
\[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
lol...it is same as Finding change in momentum
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
\[F=\frac{ dP }{ dT }\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
Correct, but the question asked for average, not instantaneous.
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
wat is ur answer @CliffSedge
 one year ago

amishraBest ResponseYou've already chosen the best response.0
I'm sorry, I used the impulse formula, and I got the correct answer. :) It doesn't really make a difference.
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
I already provided it (in formula form, anyway)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
wat is ur answer @amishra
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
That's what I got too. :)
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
\[\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}\]
 one year ago

CliffSedgeBest ResponseYou've already chosen the best response.2
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

amishraBest ResponseYou've already chosen the best response.0
@CliffSedge Oh, thank you for sharing that :D
 one year ago
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