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Physics 65 Online
OpenStudy (anonymous):

Someone help me with this physics question please? So lost! A bouncy ball is dropped from a height, h, and rebounds off the floor, returning to essentially the same height. During the actual bounce (while it is in contact) the ball is flattened and deformed by amount, y, before returning to its original shape. a) derive an expression for the acceleration, a, while the ball is in contact with the floor. b) use the result to calculate the acceleration if h = 1.50m and y=0.50cm

OpenStudy (anonymous):

wow what does deforming got to do with this problem?

OpenStudy (anonymous):

Actually: it is only after it has deformed by y that the acceleration is 0. When it is compressing, the acceleration is decreasing, when expanding, the acceleration is increasing again, then decreases as gravity becomes stronger than the expansion force of the ball.

OpenStudy (anonymous):

No idea how such a thing could be modeled, though

OpenStudy (amistre64):

kinetic and potential equations perhaps?

OpenStudy (amistre64):

|dw:1348427412516:dw|

OpenStudy (amistre64):

how fast is it moving once it hits the ground? how fast does it accelerate to go from Vf to 0 in the space of y?

OpenStudy (amistre64):

\[g=\frac{2d}{t^2}\] \[t=\sqrt{\frac{2d}{g}}\] this should give us the time it takes to hit the floor so we can calculate the velocity with\[v=gt\] \[v=g\sqrt{\frac{2d}{g}}\] \[v=\sqrt{\frac{2dg^2}{g}}\] \[v=\sqrt{2dg}\]

OpenStudy (amistre64):

so, v as it hits the ground is: 5.42 m/s and it acclerates to 0 in a distance of y

OpenStudy (amistre64):

\[(v_f)^2-(v_i)^2=2ay\] \[\frac{(v_f)^2-(v_i)^2}{2y}=a\]

OpenStudy (amistre64):

.5 cm = .005 meters right? to keep things kosher

OpenStudy (amistre64):

filling in gives us:\[a=\frac{5.42^2}{.01}\]

OpenStudy (amistre64):

with any luck, someone else can vouch for this :)

OpenStudy (anonymous):

You might be able to do the same thing with kinetic and potential (gravitational and elastic), but it may be harder.

OpenStudy (amistre64):

without knowing the mass of the ball, i wouldnt know how to apply them

OpenStudy (anonymous):

Or the elasticity, but it's a possible way if you just leave it as m and k.

OpenStudy (amistre64):

using\[-4.9t^2+1.5=0;~t=0.5533\] and \[t=\sqrt{\frac{2*1.5}{9.8}}=0.5533\]so im confident that that was right :)

OpenStudy (anonymous):

I'll admit, your way is simpler.

OpenStudy (amistre64):

9.8 times .5533 = 5.422, so i like the results for velocity at time of impact :)

OpenStudy (amistre64):

started phy201 this term. just seeing how much of the stuff i can fill out :)

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