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Newton Third Problem Ball on a Scale - part 2

  1. Jun 27, 2008 #1
    1. The problem statement, all variables and given/known data

    You are standing on scales which read weight in Newtons. A 0.50 kg ball is dropped from a height of 1 m into your hands. Your hands drop from chest level to waist level during the catch, a distance of about 25 cm. Your mass is 62 kg. Assuming that you decelerate the ball uniformly during the catch, what would be the maximum reading on the scales? (Hint: The scales read 607.6 N before you caught the ball.)

    Properties of the Ball
    Displacement upon being released: ?
    Displacement while being decelerated: ?
    Mass: 0.50 kg (given)

    Properties of Me
    Weight: 607.6 Newtons (given)
    Mass: 62 kg (given)


    2. Relevant equations

    V (final) ^ 2 = V (initial) ^ 2 + (2 * acceleration * displacement)

    F (weight) = g * mass

    Force = mass * acceleration




    3. The attempt at a solution

    The maximum reading on the scale will occur when I catch the ball, because at that time the scale will record my weight + the ball's weight + the acceleration of the ball whilst being caught.

    F (weight me) + F (weight ball) + mass (ball) * acceleration

    First find the acceleration of the ball after it is dropped:

    V (final) ^ 2 = V (initial) ^ 2 + (2 * acceleration * displacement)
    V (final) ^ 2 = 0^2 + 2 * 9.8 * 1
    V (final) ^ 2 = 19.6
    V (final) = 4.43 m/s

    So therefore, the final velocity is 4.43 m/s

    So, the velocity of the ball upon being caught is 4.43 m/s downward. Now, let's find out the acceleration as it is being decelerated by my hand.


    V (final) ^ 2 = V (initial) ^ 2 + (2 * acceleration * displacement)
    0^2 = 4.43 ^ 2 + (2 * acceleration * 0.25)
    0 = 19.6 + 0.5a
    0.5a = -19.6
    a=39.2 m/s^2

    So, therefore the acceleration of the ball is -39.2 m/s^2

    What force is that? F=ma; F= (0.5 * 39.2); F=19.6 N


    So, let's return to my original statement: The maximum reading on the scale will occur when I catch the ball, because at that time the scale will record my weight + the ball's weight + the acceleration of the ball whilst being caught.

    Therefore:

    F (weight me) + F (weight ball) + mass (ball) * acceleration
    607.6 Newtons + 4.9 Newtons + 19.6 Newtons = 632 Newtons

    Therefore the maximum number on the scale will read 632 Newtons.
     
  2. jcsd
  3. Jun 27, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I didn't double-check your arithmetic, but your solution looks good to me. Your reasoning can be tightened up a bit.

    A better way to think of this is to consider the forces acting on you, under the assumption that you do not accelerate as you catch the ball and thus the net force on you is always zero. As you catch the ball, the forces on you are: (1) Your weight = mg, (2) The force that the ball exerts against your hand, and (3) The upward force that the scale exerts on you (which equals the scale reading). You are trying to find the maximum value of that third force.

    To find the force that the ball exerts on you, analyze the forces on the ball (and apply Newton's 3rd law).

    Of course, you get the same answer as you did, but thinking of it this way might give you some additional insight.
     
  4. Jun 27, 2008 #3
    Thank you very much, DocAl. The reason why I posted this is because my teacher said that I forgot to account for the ball's acceleration and she docked me marks for it. I didn't understand why she said that.
     
  5. Jun 27, 2008 #4

    Doc Al

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    Staff: Mentor

    That makes no sense, since you obviously calculated and made use of the ball's acceleration.
     
  6. Jun 27, 2008 #5
    Yeah, I thought so. I actually sent in that assignment twice. The first time I made a mistake with the acceleration and so i re-did it and sent it in a second time as I posted it on this thread.

    I think I will ask if the correct submission one was evaluated. Thanks again, Doc Al.
     
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