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Impulsive Tension Question

  1. Jul 18, 2012 #1
    1. The problem statement, all variables and given/known data
    An object of mass 3m is connected by means of a light, inextensible string to a scale pan of mass m. The object rests on the ground. The string passes over a smooth pulley and the scale pan hangs suspended. An object of mass m, falling from rest a distance h above the scale pan, lands on it and does not bounce. Show that the mass of 3m will rise to a height h/5.

    Sorry but there's no diagram given! The answer I got was 4h/5 instead! :(

    2. Relevant equations



    3. The attempt at a solution
    To find the speed of the scale pan before the string becomes taut:
    mgh=0.5u^2
    u=√(2gh)

    To find the speed of the scale pan/speed at which the 3m object jerks in motion:
    Take upwards direction as negative:
    -I = 2mv-2m√(2gh) (this is the change in momentum of the scale pan)
    -I = -3mv

    Equate both to get:
    -3mv=2mv-2m√(2gh)
    -5mv=-2m√(2gh)
    v=2√(2gh)/5

    This speed, v is the initial speed of the upward motion of the 3m mass.
    u=2√(2gh)/5
    v=0 (mass of 3m rise to a height of h/5 and stop)
    To find acceleration, a:

    For the 3m mass (it's going upwards):
    T-3mg=3ma
    For the scale pan+object of mass m (going downwards):
    2mg-T=2ma

    Equate both equations:
    T=3ma+3mg
    T=2mg-2ma
    3ma+3mg=2mg-2ma
    5ma=-mg
    5a=-g
    a=-g/5

    Using an equation of linear motion:
    v^2=u^2+2as
    0=(2√(2gh)/5)^2 + 2(-g/5)s
    s=4h/5

    Did I do something wrong in the middle? Was I on the right track in the first place? Thanks!! :D :smile:
     
  2. jcsd
  3. Jul 18, 2012 #2

    tiny-tim

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    hi jiayingsim123! :smile:

    (try using the X2 button just above the Reply box :wink:)
    no, the 2m√(2gh) is wrong …

    only the mass m was moving, the pan was stationary! :redface:

    (btw, i find this sort of problem easier if i pretend everything is in a straight line …

    in this case, a mass m collides with a stationary mass 4m :wink:)​
     
  4. Jul 18, 2012 #3
    Hi tiny-tim, thanks for helping out again!
    But I thought the mass m drops onto the scale pan, meaning both of it will move downwards as a single entity and therefore the total mass is 2m? So isn't it a mass of 2m colliding with a stationary mass of 3m?
     
  5. Jul 18, 2012 #4

    tiny-tim

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    no!!

    there's 4m of stationary mass, isn't there?? :smile:

    (you're being confused by the fact that the 4m is in two parts)

    only a mass of m was originally moving, and 4m wasn't :wink:
     
  6. Jul 19, 2012 #5
    Okay thanks for clearing that up, tiny-tim!! You're really awesome! :)
     
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