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Homework Help: Question on velocity changes in free fall

  1. Jun 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume a ping pong of mass 3g is falling, and it has fallen 1m and is now travelling at 5 m/s. If the air resistance force is half its weight, how far will it travel in the next 0.1 seconds. [Hint: you may want to sketch a velocity-time graph)

    2. Relevant equations
    Understanding that area under v-t graph is distance travelled.

    3. The attempt at a solution
    So I thought:
    If mass=3g, then weight is 0.03N
    And as air resistance is half that, then the resultant force on the ping pong ball is 0.03-(0.03/2)=0.015N
    Therefore, as F=ma,
    0.015=m * a
    But we know m=0.003 kg, so,
    0.015=0.003 * a
    a=5 metres per second per second.

    Hence, in 0.1 seconds,

    5 * 0.1 = 0.5 m/s

    So the velocity of the ping pong ball increases by 0.5 m/s in the next 0.1 seconds.
    And we know initial velocity was 5 m/s, so it increased to 5.5 m/s in 0.1 seconds.

    So on a velocity-time graph, this forms a trapezium shape, and you find the area of that trapezium to find the distance the ping pong ball has travelled in 0.1 seconds:

    [ (5+5.5)/2 ] * 0.1 = 0.525 m.

    So the ping pong ball travels 525 centimetres (or 0.525 metres) in the next 0.1 seconds.

    Is that right?

    Thanks :)
  2. jcsd
  3. Jun 10, 2015 #2


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    Hello Physiophysio. Welcome to PF !

    That looks like it's as good an approximation as you can get with the given information.

    Very good analysis.
  4. Jun 10, 2015 #3


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    Yes. (You don't actually need the mass. You know the acceleration will be g/2.)
  5. Jun 10, 2015 #4
    Thanks :)
  6. Jun 10, 2015 #5
    Could you please explain why that is, because, yes, I see how a=5 which is g/2, but how does that come about? Is it because, in free fall with no air resistance and only the weight, a=g, but now air resistance is half of weight, so a would be a/2 which is equal to g/2?
    Or is that logic wrong? Thanks for replying!
  7. Jun 10, 2015 #6


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    If a is defined as the acceleration it undergoes, a is still a, but is now g/2, yes.
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