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Mathematics 12 Online
OpenStudy (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.

OpenStudy (anonymous):

You can use impulse-change-in-momentum relation.

OpenStudy (anonymous):

I used it but I didn't get the answer. Can you show me?

OpenStudy (anonymous):

*sorry, 'impulse = change-in-momentum relation'

OpenStudy (anonymous):

Sure, what did you get for the change in momentum?

OpenStudy (anonymous):

F = change in momentum 0.5*12 - (-6*0.5)

OpenStudy (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. :/

OpenStudy (anonymous):

Note:...the Direction is opposite..)

OpenStudy (anonymous):

Use this formula \[F_{av}{\Delta}t=m{\Delta}v\]

OpenStudy (anonymous):

Yes, the change in velocity is 12+6 since it changes direction.

OpenStudy (anonymous):

That change in momentum divided by the time of contact (1.5s,really? That's a long contact time . . .) = the average force.

OpenStudy (anonymous):

That is Impulse @CliffSedge

OpenStudy (anonymous):

Impulse divided by time = force.

OpenStudy (anonymous):

Oh THANK YOU ALL! I got it now! I didn't see the change in direction! :D

OpenStudy (anonymous):

\[\large impulse=F_{avg} \cdot \Delta t\]

OpenStudy (anonymous):

\[\large impulse = \Delta p \rightarrow F_{avg}=\frac {\Delta p}{\Delta t}\]

OpenStudy (anonymous):

lol...it is same as Finding change in momentum

OpenStudy (anonymous):

\[F=\frac{ dP }{ dT }\]

OpenStudy (anonymous):

Correct, but the question asked for average, not instantaneous.

OpenStudy (anonymous):

wat is ur answer @CliffSedge

OpenStudy (anonymous):

I'm sorry, I used the impulse formula, and I got the correct answer. :) It doesn't really make a difference.

OpenStudy (anonymous):

I already provided it (in formula form, anyway)

OpenStudy (anonymous):

wat is ur answer @amishra

OpenStudy (anonymous):

6 Newtons

OpenStudy (anonymous):

That's what I got too. :-)

OpenStudy (anonymous):

\[\large F_{avg}=\frac{(0.5 kg)(18 m/s)}{1.5 s}\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@CliffSedge Oh, thank you for sharing that :D

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