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anonymous
 4 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.
anonymous
 4 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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can use impulsechangeinmomentum relation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I used it but I didn't get the answer. Can you show me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*sorry, 'impulse = changeinmomentum relation'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sure, what did you get for the change in momentum?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0F = change in momentum 0.5*12  (6*0.5)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0Note:...the Direction is opposite..)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, the change in velocity is 12+6 since it changes direction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That change in momentum divided by the time of contact (1.5s,really? That's a long contact time . . .) = the average force.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is Impulse @CliffSedge

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Impulse divided by time = force.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large impulse=F_{avg} \cdot \Delta t\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol...it is same as Finding change in momentum

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[F=\frac{ dP }{ dT }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Correct, but the question asked for average, not instantaneous.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wat is ur answer @CliffSedge

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I already provided it (in formula form, anyway)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wat is ur answer @amishra

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's what I got too. :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.0@CliffSedge Oh, thank you for sharing that :D
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