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Math Help - Physics: Force Question...

  1. #1
    Member looi76's Avatar
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    Physics: Force Question...

    Question:


    Use data from Figure to determine,
    (a) the distance travelled by the ball during the first 0.40s,
    (b) the change in momentum of the ball, of mass 45g, during contact of the ball with the surface,
    (c) the average force acting on the ball during contact with the surface.

    Attempt:
    (a) 0.8m
    (b) Change \ in \ momentum = mv - mu
    = (45 \times 10^{-3} \times 4.2) - (45 \times 10^{-3} \times 0)
    = 0.19Ns

    Answer in text book is 0.35Ns, where did I go wrong?
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by looi76 View Post
    Question:


    Use data from Figure to determine,
    (a) the distance travelled by the ball during the first 0.40s,
    (b) the change in momentum of the ball, of mass 45g, during contact of the ball with the surface,
    (c) the average force acting on the ball during contact with the surface.

    Attempt:
    (a) 0.8m
    (b) Change \ in \ momentum = mv - mu
    = (45 \times 10^{-3} \times 4.2) - (45 \times 10^{-3} \times 0)
    = 0.19Ns

    Answer in text book is 0.35Ns, where did I go wrong?
    It's hard to read the scale .....

    Your answer to (a) looks fine.

    (b) I get the time of intial contact to be 0.43 s (when v = 4 m/s) and the time when the ball leaves the surface (speed starts to decrease) to be 0.58 s (when v = -3.6 m/s).

    Then \Delta p = m \Delta v = ....... where \Delta v = v_{final} - v_{initial} = -3.6 - 4 = -7.6 m/s.

    (c) F_{average} = \frac{\Delta p}{\Delta t}. Get \Delta p from (b). \Delta t = 0.58 - 0.43 = 0.15 s.


    Scale reading uncertainty: You should assume all my values for t are \pm 0.01 and my values for v are \pm 0.1 ........
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Then \Delta p = m \Delta v = ....... where \Delta v = v_{final} - v_{initial} = -3.6 - 4 = -7.6 m/s.
    I haven't looked at any of the numbers, but note that v_{initial} is negative and v_{final} is positive. So \Delta v = +7.6~m/s according to your working.

    -Dan
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