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Relative Acceleration

  1. Oct 31, 2003 #1
    Hey, got an interesting little question here that i need some help (and would really appreciate any you can give) on:

    1)The initial acceleration
    2)The velocity of A just before it picks up the mass at C
    3)Using the principle of conservation of momentum, or otherwise, the velocity of A after picking up the mass at C
    4)The distance of A from C at the first position of instantaneous rest.[/QUOTE]

    Ok, so the following picture has my workings so far.
    I took "T" as being the tention in the piece of string. What im getting stuck on is what happens when the particle A picks up the mass at C - it starts to fall back down, right, since its .1kg heavier than the counter mass, or does it have enough acceleration to keep going for awhile, then drop down? Im not sure how i should go about investigating this.

    I think ive gotten parts one and two right (image), part 3, im not so sure how to do, same goes for part 4. I was thinking that part 5 would just be that the particle has fallen back to the starting position, and that that would be the first period of instantaneous rest, or maybe its on its upward journey when it stops for an instant before falling back down? Im just confused

    Please help! Its very much appreciated
    Last edited: Oct 31, 2003
  2. jcsd
  3. Oct 31, 2003 #2


    The way I interpret this question is as follows:

    By the structure of the subquestions 3 and 4, we are being lead to believe that A continues upward to some point at which it is stationary.

    Conservation of momentum (specifically, collision with coupling) can be used here to estimate the velocity of the combined mass the instant after coupling occurs, but before gravity and tension have had any appreciable effect.

    Thereafter, the net force resulting from upward tension and downward gravity will decelerate the combined mass. This net force can be calculated then divided by the combined mass to give the new (downward) acceleration.

    From the initial (upward) velocity of the combined mass, the new acceleration, and a final velocity of zero, you can determine the distance the combined mass rises above point C.
  4. Oct 31, 2003 #3


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    There is a gravitational force of 0.4g pulling down on the 0.4kg mass and a gravitational force of 0.5g pulling down on the 0.5kg mass (in other words, their weights). Since they are connected by a pulley, each is pulling up on the other with its weight. That means that there is a net downward force on the 0.5kg mass of 0.1g and a net upward force of the same magnitude on the 0.4 kg mass. Since they are attached by an "inextensible string", they move as a unit with acceleration 0.1g/(0.4+0.5)= (1/9)g m/s2. That's the answer to part a). Since A rises at that acceleration for one second, its speed at that time is ((1/9)g m/s2)(1 s)= (1/9)g m/s.

    Conservation of momentum does NOT apply since there is an external force (gravity). You can recalculate the acceleration as above by changing the mass of A from 0.4 to 0.6kg.
    It's not a matter of having enough "acceleration"- it has momentum.
    Remember that acceleration is CHANGE in speed. Since in part b, you found A's speed at the instant it aquires the new mass, you can now apply that new acceleration to THAT speed. velocity at time t (in seconds after the motion began) will be v(t)= v(1)- a(t-1) (a is the new acceleration. It is times "t-1" because it didn't apply for the first second.) The "position of instantaneous rest" occurs when v(t) is 0.
  5. Oct 31, 2003 #4
    The gravitational forces are actually 0.4kg X 9.8 N/ kg = 3.92 N and 0.5 kg x 9.8 N/kg = 4.9 N .

    The net force , F1, acting on the 0.5 kg mass is
    F1 = 4.9 - T where T is the tension on the string.

    The net force, F2, acting on the .4 kg mass is
    F2 = T - 3.92 N

    The acceleration of the .5 kg mass, a1, is
    a1 = (F1)/(m) = (4.9 -T)/(0.5 kg)
    a2 = (F2)/m = (T-3.92)/(0.4 kg).

    Since as you pointed out a1 = a2,
    (4.9 -T)/(0.5 kg) = (T-3.92)/(0.4 kg)

    This allows us to calculate T which can then be substitued back into (T-3.92)/(0.4 kg) to find a1 or a2.

    Assuming the masses are starting from rest you can then determine the speed after 1 second.

    This assumes that either the original mass magically changes by .2 kg while maintaining its motion, or that the original mass collides with a 0.2 kg object travelling at a constant speed coinciding with the speed of the 0.4 kg mass at the moment of impact.

    Since the problem is silent on this I used the simpler assumption that the mass that was "picked up" was stationary. Of course there is no reason to agree with my interpretation unless you prefer simpler assumptions. The problem is ambiguous and begs clarification.

    Momentum is conserved when no work is done on the system. Over a period of time work is done and over a period of time momentum is not conserved. However, at the "instant" of impact, gravity has not had the opportunity to work on the system. If the period of impact is short enough to make work done by gravity negligible, then conservation of momentum gives a reasonable estimate for the new velocity of the combined mass.

    I look forward further discussion.
  6. Oct 31, 2003 #5


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    I agree completely. I would just point out that I had said "0.4g" using "g" instead of 9.8.
  7. Nov 1, 2003 #6
    Thanks for all your help phyics247 amd hallsofivy, i think i understand the question now. In our state exam this summer, were expected to be able to do all that in about 15-20 mins. Im not exactly optimistic about it.
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