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Speed and coefficeint of friction

  1. Mar 20, 2007 #1
    1. The problem statement, all variables and given/known data

    Three smooth stones, A, B and C are initially at rest, in contact with each other on the smooth surface of a frozen lake. The masses of the stones are A = 400 g, B = 300 g and C = 120 g and the coefficient of sliding friction between the stones and the ice is μslide = 0.0150. An explosion between the stones causes them to fly apart across the surface of the ice. Stone A flies off due North with an initial speed of 4.50 m s−1. Stone B flies off due east with an initial speed of 6.60 m s−1.

    Calculate the speed and direction of motion of stone C immediately after the
    explosion. [10 marks]

    2. Relevant equations

    momentum= mass*velocity
    3. The attempt at a solution

    there isnt a collion taking place...so momentum formula i dont think needs to be used.

    we need to calculate speed of C.

    we know
    mass of C= 0.12kg
    μslide = 0.0150

    F=ma = uR
    0.12a= 0.015R

    dont know R or a.....???

    need help here....
  2. jcsd
  3. Mar 20, 2007 #2
    I think you need to know the amount of time the force of the explosion acted on the stones to solve this problem.
  4. Mar 20, 2007 #3

    Doc Al

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    Staff: Mentor

    It's not a collision; it's an explosion. But in an explosion, just like in a collision, all the forces are internal so momentum is still conserved.
  5. Mar 21, 2007 #4
    well any ideas how to do this????
  6. Mar 21, 2007 #5


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    Follow Doc's advice. Write down total momentum before and total momentum after and set them equal.
  7. Apr 6, 2007 #6
    mometum before collision= momentum after collison
    m1u1+ m2u2 + m3 u3 = m1v1 + m2v2+m3v3

    m1= 0.4
    m2= 0.3

    is this correct so far....?
  8. Apr 6, 2007 #7


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    Before the collision all of the velocities are zero. Your velocities had better be vectors as well - they are heading in different directions.
  9. Apr 6, 2007 #8
    m1= 0.4
    m2= 0.3

    0= 1.8sin 90+ 1.98sin0+ 0.12*v3

    is this correct....and then eqauting to calculate v3.
  10. Apr 6, 2007 #9


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    A vector has two components, x and y. You need to keep track of both. If you call east the positive x axis then the final velocities are (4.5,0) and (0,6.6).
  11. Apr 6, 2007 #10
    taking momentum in x direction-
    o= 0.3*6.6 + 0.12*v3
    v3= -33meters/s in x direction

    taking momentum in y direction
    0= 0.4*4.5 + 0.12*v3
    v3= -15 meters/sec in y direction

    the speed of stone c is west= 33 meter/s
    the speed of stone c is south=15meter/s

    using phythagors theorm to calculate the total velocity of stonce c


    speed of c is aprox 36meters/sec in the south west direction
  12. Apr 6, 2007 #11


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    I think your numbers came out wrong on the x velocity - but otherwise ok.
  13. Apr 7, 2007 #12
    in x direction speed should be -16.5 m/s not -33.

    it this correct, and the rest of the method working out total speed is correct too?
  14. Apr 7, 2007 #13


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    If you are anxious about whether all of your answers are right, it's good to work out ways of checking them yourself. Eg. in this question the sum of the final momenta you get should equal the zero initial total momenta. A good quick check would just be to add them up. That would have caught the 33m/s problem. But yes, it's ok. Of course the pythagorean theorem is also ok.
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