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Triangle problem

  1. Dec 2, 2006 #1
    1. The problem statement, all variables and given/known data
    A bowling ball mass 3.2 kg rolls west at 4.2 m/s striking a stationary ball mass 3.2 kg. After the collision the second ball travels in a direction 24 degrees south of west at 3.6 m/s. Whats the first balls speed after the collision.


    2. Relevant equations
    mv+mv=mv+mv


    3. The attempt at a solution
    p1=13.44 kgm/s W
    p2=0
    p1'=?
    p2'=11.52 kgm/s 24 degrees south of west

    13.44 W - 11.52 24 S of W = p1'
    I then drew the resulting vectors and subtracted them getting a triangle with the lengths 13.44, 11.52, and 5.52 with an angle of 24 degrees between 13.44 and 11.52. My question is that when i use sine law to get the two remaining angles, all the angles only add up to 160 degrees. What's the deal here? (I got the length of the last angle via: x^2=13.44^2+11.52^2-2(13.44)(11.52)cos24)
     
  2. jcsd
  3. Dec 2, 2006 #2

    radou

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    You probably meant, you got the length of the third side of the triangle, i.e. x should represent the magnitude of the momentum vector p1', which, divided by the mass m1 = 3.2, gives the magnitude of the velocity.
     
  4. Dec 3, 2006 #3
    k thats what i meant the length of the last side, but when I use sine law to find out the 2 remaining angles, they all do not add up to 180
     
  5. Dec 3, 2006 #4

    radou

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    Why are you trying to find the angle? Don't you have to find the speed only? If so, as said, you only have to divide that 'third' triangle side by the mass, since it represents momentum.
     
  6. Dec 3, 2006 #5
    Yes, but I am curious as to why the angles to not make sense. I donno about u but just knowing that the triangle i drew duznt make any sense wuld make me wonder if i even did it right? Ne ways i got 1.7 m/s
     
  7. Dec 3, 2006 #6

    radou

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    Your answer is correct. I got the angles right, but I wouldn't worry about that if I were you.
     
  8. Dec 3, 2006 #7
    k but if it asked for the angle at which it went, how wuld u do that?
     
  9. Dec 3, 2006 #8
    nm i figured it out but instead using a dfferent method
     
  10. Dec 3, 2006 #9

    OlderDan

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    There is nothing wrong with graphical solutions, but they are not the most accurate way to solve such a problem. The most accurate approach is to resolve the known vectors into components using trigonometric ratios, and solve for the components of the unknown vectors.

    Set up a coordinate system with x and y axes overlaying the compass directions, with +y going North and +x going East. Write your known vectors in terms of x and y components and use conservation of momentum to find the components of the third vector. Once you have the components, find the magnitude of the vector and its direction. I know you don't need the direction for this problem, but you said you wanted to get it right.
     
  11. Dec 4, 2006 #10
    yeah i just figured that out right after i posted this, i was just wonder why the triangle didnt make any sense it all. I see how it culdnt make sense, but I guess trig duznt always make sense?
     
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