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Energy and Momentum Physics Questions

  1. Nov 1, 2011 #1
    Question 1 (All basic physics).
    A crane drops a 0.3 kg steel ball onto a steel plate. The ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s, respectively.If the ball is in contact with the plate for 0.03 s, what is the magnitude of the average force that the ball exerts on the plate during impact.

    Equations I thought would be useful (Basic Physics):
    F=ma
    a=v/t

    My ATTEMPT at the question:
    Well first i took my givens which were:
    Mass of Ball: 0.3 kg
    Velocity Before impact: 4.5 m/s
    Velocity After impact: 4.2 m/s
    Accelration of gravity constant: 9.8 m/s^2 (^2=squared[Exponential value as it is written on a calculator)

    So my thinking is that we should find the easiest force equation there, being the first one (before impact):
    F1=ma
    F1=(0.3)x(9.8)
    F1=2.94N (Newtons)
    Then the second one (after impact):
    F2=ma
    F2=(0.3)a
    (Finding acceleration)a=v/t
    a=(4.2-4.5)/(0.03)
    a=-3m/s^2
    (Back to Force of second equation)
    F2=(0.3)x(-3)
    F2=-3N(Newtons)
    So now i assume that force is a scalar quantity and thus negatives don't really apply, only the value so i use:
    ƩF=Average Force=(F1+F2)/2
    ƩF=(3+2.94N)/2
    ƩF=(5.94N)/2
    ƩF=2.97N and the multiple choice answers are:
    (a) 3.0N (MY ANSWER SINCE I ROUNDED 2.97N TO 3.0N)
    (b) 87N
    (c) 3.5N
    (d) 30N
    (e) 133N

    So I ask is how my answer (which was wrong) is not correct, where i messed up, what formulas i should have used, and what the correct answer is. (Note: This isn't homework it is a quiz question that i took and i never understood how it was wrong.)
     
  2. jcsd
  3. Nov 1, 2011 #2
    Before I confuse you with a possibly incorrect explanation, the correct answer was A right?
     
  4. Nov 1, 2011 #3

    SammyS

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    The ball's velocity changes from -4.5 m/s to 4.2 m/s in a time of 0.03 s . What is the average acceleration of the ball during this time?
     
  5. Nov 2, 2011 #4
    @ Vorde No i do not know the correct answer

    @SammyS I also don't know but i would assume that it is around 9.8m/s^2

    I already attempted this question and i don't mind getting confused a bit since i have more help than just the website (actually people help).
     
  6. Nov 2, 2011 #5
    Well, the way I did this was using the equation:

    FΔT = ΔMV
    (Which can be aquired by multiplying F=MA by ΔT)

    Then plug in (.03s) for ΔT, and for ΔMV I put in the difference between the before-collision momentum, and the after-collision momentum. Dividing both sides by ΔT yields a force of exactly 3 newtons, no rounding required.
     
  7. Nov 2, 2011 #6
    I showed that answer to my physics teacher and he says the answer is BOTH (a) and (b) and it just depends on what you perceive what happends after. If it bounces back the answer is 87N, however, if you thought like the way I did where it breaks through (since it is not specified in the question whether it bounces or not) then the answer is 3N. I like that equation set up vorde and i'll use that on any other situations like that. So he was also wrong but he was right too.
     
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