I'm doing a lab experiment and my question is: How does the height that a play-doh ball is dropped affect the diameter of the ball? My prediction is that the higher the play-doh ball is dropped, the bigger the diameter of the ball. But I need to support my reasoning with facts/laws. Which laws/equations/reasons support my hypothesis?
Stacey Warren - Expert brainly.com
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Hm, I think you need a new experiment if I have correctly understood your description of it.
The diameter of the ball is not affected by the height you drop it from. You can take a ball of any diameter and drop it from any height you feel like, can't you? You can't test independent things - things which are not related - in an experiment.
ummm it does affect it...it is a playdoh so when it's dropped, it deforms...
Yeah, it does, but extremely slightly. And only during the collision with the ground.
and when it deforms, it is no longer a perfect sphere. So you shouldn't talk about the diameter, but about the volume.
The larger the height it is dropped form, the larger force it suffers due to collision with the ground. So the volume decreases more.
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You're talking of a ping-pong ball, right? The one which bounces to a great height when thrown?
BTW play-doh means soft-clay that we used to play with in our child hood to create "ingenious" stuff.
hey check this video out to know what happens during a collision that is very to close to completely elastic one:
Its 70K fps!!!
Collision is a very very complicated process... involving many transformations.
now, ofcourse the dropping a clay ball is different, much closer to a inelastic (sticking) collision. here you can notice that when you drop it, it gets flatter from a greater height.
In the case when you drop it from a greater height, it has a higher kinetic energy (due to a higher terminal velocity) just before the collision. and this K.E. is totally used up in the collision. It's almost like the golf ball above in the video stopped mid way after getting flat!