- 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.

- chestercat

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

That is Impulse @CliffSedge

- anonymous

Impulse divided by time = force.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

wat is ur answer @CliffSedge

- anonymous

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

- anonymous

I already provided it (in formula form, anyway)

- anonymous

wat is ur answer @amishra

- anonymous

6 Newtons

- anonymous

That's what I got too. :-)

- anonymous

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

- 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

@CliffSedge Oh, thank you for sharing that :D

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