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Behaviour of a ping pong ball dropped in a can full of water.

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data

    I performed the following experiment in my backyard this afternoon.

    A 500ml can is filled with water and a ping pong ball floats on its surface. The can is dropped from a height of 1m onto pavement. The experiment is then repeated with the exception that the second time round, the ping pong ball is covered in vaseline.

    my observations:

    THE NORMAL BALL:
    as the can was falling the ball stayed on the surface of the water. When the can hit the ground the ball flew upwards. Water came out of the can.

    THE VASELINE BALL:
    as the can was falling the ball stayed on the surface of the water. When the can hit the ground the ball sort of 'fell' out of the can. From what i saw, i assume that if the system was perfect it would of stayed in the can.

    I guess what i'm asking is: why did the normal ball bounce out of the can while the vaseline ball remained floating.

    2. Relevant equations



    3. The attempt at a solution

    All variables were constant. The independant variable was the vaseline ball vs. the normal ball.

    THE NORMAL BALL:
    The ball falls with the same velocity as the can. There is no air resistance between the ball and water so resistance is the same as on the can. The ball adheres to the surface of the water (i ensured the ball was covered in water before performing the experiment). As the can falls the ball remains on the surface of the water due to bouyancy. When the can hits the ground it imposes a downwards force on pavement and the pavement causes an upwards force. The upwards force causes the water to move upwards taking the ball with it due to surface tension.

    THE VASELINE BALL:
    The ball falls with the same velocity as the can. There is no air resistance between the ball and water so resistance is the same as on the can. The ball does not adhere to the surface of the water due to its waxy organic nature. As the can falls the ball remains on the surface of the water due to bouyancy. When the can hits the ground it imposes a downwards force on pavement and the pavement causes an upwards force. The upwards force causes the water to move upwards, but the ball stays put because of surface tension so it resists the upwards force.

    LASTLY:
    It was really hard to keep an eye on the ball as it was falling. I kind of feel like the normal ball would of displaced more water and fallen deeper into the can on impact, then somehow a combination of water pressure (i'm not sure how to consider pressure in this case. I know that pressure would be greater at the bottom of the can), the upwards force and bouyancy caused it shoot upwards. However i'm not sure because i didnt actually see it happen.
     
  2. jcsd
  3. Apr 21, 2012 #2
    You are showing a very good attitude at making experiments, drawing hypothesis, making variation, observing, and so on.
    Said that, I think that the falling of an object is a quick movement and any naked eye observation produces rough and unmeasurable data. You'd need an high speed camera, something like 1000 frames/s, but those are expensive objects, out of reach for a domestic use.
     
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