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Two masses attached by inextensible string?

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data
    okay this is probably extremely basic stuff but I'm having a lot of trouble in this class and i hope someone can help! :/

    A 10 kg mass is placed on a fricitonless surface and is attached to a 5 kg mass as shown by a light and inextensible string.
    ( here is a drawing of the model: http://i53.tinypic.com/ekne6e.png )

    a) Find the acceleration with which the system moves.
    b) What is the speed of the masses when the 5 kg mass strikes the floor?
    c) After the 5 kg mass strikes the floor, the 10 kg mass on the table continues to move to the right. How far from the edge of the table does it land?
    d) How would your answer to part c change if there was friction between the mass and the table? Explain.

    2. Relevant equations

    3. The attempt at a solution
    a) I set up the equations:
    T = 10a
    49-T = 5a

    49-10a = 5a
    so i got a = 3.266 m/s^2

    b) I used the formula Vf^2 = Vi^2+2ad but I've got a feeling I'm approaching this problem totally wrong. But anyway:
    Vf^2 = (3.266^2)+(2*9.8*0.6)
    Vf^2 = 22.43
    Vf = 4.74 m/s^2 (i put that as my answer)

    c) I have no clue how to approach this, if anyone can guide me please? All I can think of is using the same equation as the last part but I don't think I'm right?

    d) I put: If the surface had friction, the answer to part (c) would be less than the answer obtained in part (c), because the friction would be an opposing force against where the mass is trying to move.
    Not the best explanation but it answers the question, I think.

    if anyone can help I would appreciate it SO much! I'm currently failing ap physics and I really don't want to be :'(
  2. jcsd
  3. Oct 23, 2011 #2


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    Homework Helper

    (a) looks good
    (b) "Vf^2 = Vi^2+2ad
    Vf^2 = (3.266^2)+(2*9.8*0.6) "
    Since Vi = 0, you should have Vf^2 = 2*3.266*0.6

    (c) The 10 kg mass will no longer accelerate because the 5 is no longer pulling on it. So you know its initial velocity when it flies off the table. I trust it doesn't hit the pulley. This leaves you with a 2D trajectory problem to solve. My approach for them is to write two headings for horizontal and vertical, ask yourself what kind of motion you have in each and write the formulas for that motion in each case. Put in the knowns in all three formulas and hope one of them can be solved for something.

    Good luck!
  4. Oct 23, 2011 #3
    Yay! Okay so

    Vf^2 = 0+(2*3.266*0.6)
    Vf^2 = 3.91
    Vf = 1.97 m/s^2

    And for part (c).... I separated the x component and y compenent of the motion into separate charts, though I completely failed the 2d motion test so if my work can be checked/corrected, I'd appreciate it! heres my work: http://i55.tinypic.com/250k1fk.png
  5. Oct 24, 2011 #4


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    Homework Helper

    I got 1.98 in place of your 1.97, but I use g = 9.81 so we agree.

    Oops, in the horizontal calculation you used an acceleration of 9.81!
    Gravity only acts in the vertical direction. Horizontally, just use x = v*t.
    I agree with your time to 2 digit accuracy. Just multiply it by 1.97.

    Sorry to hear you had trouble with 2D motion. I am a veteran of 29 years teaching high school physics, and my students had really good success with it once I got them into that routine beginning with the 2 headings and asking what type of motion in each direction. Usually it is accelerated motion vertically and constant speed motion horizontally.
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