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Collision Physics Problem

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Two identical objects with the same initial speed collide and stick together. If the composite object moves with half the initial speed of either object, what was the angle between the initial velocities?


    2. Relevant equations
    Law of momentum conservation: m1v1 + m2v2 = (m1+m2)v_final


    3. The attempt at a solution
    I attempted to look at the momentum conservation equation to see if it's possible to incorporate angle. Looked at the dot product as well (v1 dot v2) but it does not make use of the information that the final speed is half the initial of either object so it must be wrong.

    I am basically blank-minded at this point. Please guide me through. Thanks for your help in advance.
     
  2. jcsd
  3. Feb 8, 2009 #2

    Hootenanny

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    Welcome to Physics Forums.

    You have the right idea in applying conservation of momentum. So we have:

    [tex]m_1\bold{v}_1 + m_2\bold{v}_2 = \left(m_1+m_2\right)\bold{v}_3[/tex]

    Knowing that the two masses are identical and are initially traveling at the same speed (note that vi above are vector velocities) can you relate m1 and 2 as well as the three velocities?
     
    Last edited: Feb 8, 2009
  4. Feb 8, 2009 #3
    Aw ha, since all the masses are equal, the equation of law of conservation of momentum can be reduced to [itex]v_1 + v_2 = v_3[/itex]


    This is something new I can add into my diagram to see if I can figure out more, thanks
     
    Last edited: Feb 8, 2009
  5. Feb 8, 2009 #4

    Hootenanny

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    Not quite! You need to be careful, the expression actually reduces to:

    [tex]\bold{v}_1 + \bold{v}_2 = 2\bold{v}_3[/tex]

    Can you see why?
     
  6. Feb 8, 2009 #5
    Please tell me if my following deduction is correct.
    I first have

    [itex]P_1 + P_2 = P_3[/itex]

    [itex]m_1 v_1+m_2 v_2=(m_1 + m_2)v_3[/itex]

    Since [itex]m_1 =m_2[/itex] and [itex]v_1 =v_2[/itex], this follows that

    [itex]P_1 =P_2[/itex] which gives me:

    [itex]2(P_1)=P_3[/itex].

    where [itex]P_3=2(m_1 )\frac{1}{2}(v_1 )=m_1 v_1 =P_1[/itex]

    I then have [itex]2P_1 =P_1[/itex]


    Now Intuitively, I know that to go from [itex]2(P_1 ) to (P_1 )[/itex] I am only off by a factor of an angle, cosx.
    So the equation then I really have is

    [itex]2(P_1)Cos(x)=P_3[/itex]

    [itex]Cos(x)=\frac{P_1}{2P_1}=\frac{1}{2}[/itex]
     
    Last edited: Feb 8, 2009
  7. Feb 8, 2009 #6

    Hootenanny

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    I'm not quite sure what your doing there, but as I said in my previous post you should be careful and note that vi is the velocity vector. The question states that the initial speed (scalar) of both particles is equal and equal to half of the final speed (scalar). Note, simply because the speed of the two particles are equal doesn't necessarily mean that their velocities are.

    Do you follow?
     
  8. Feb 8, 2009 #7
    ok ><

    I will start over then
     
  9. Feb 8, 2009 #8

    Hootenanny

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    Try using my hint from earlier: Can you find a relationship between the three velocities (or their components)?
     
  10. Feb 8, 2009 #9
    The only relationship I can determine is that their magnitude are equal. So if I take the square root of both the object's vertical component squared + horizontal component squared, I would have their magnitude, or speed. This speed is twice than the resulting speed after the collision.

    The problem is, I am not given any numerical value for each corresponding velocity component to each object.

    I thought I had to express one variable in terms of another.

    But it's gonna be even more complicated once I do these in addition to adding and squaring.

    I arrived at the above before but I can make no further application to solve for anything. Some how I think I must incorporate angle, or the use of cosine from this point onward since it's the angle that I am trying to solve for. I am quite confused here because I have a set of several equations and relationships. And that's about when I came to post here.
     
    Last edited: Feb 8, 2009
  11. Feb 8, 2009 #10
    Oh wait I think I see something.

    The magnitude of [itex]v_1[/itex] minus the magnitude of [itex]v_2[/itex] times CosX equals [itex]v_3[/itex], which is half of [itex]v_1[/itex]. So I am solving for x.

    since the speed of [itex]v_2=v_1[/itex],

    [itex]v_1-v_1 cos(x)=\frac{1}{2}v_1[/itex]

    [itex]cosx=\frac{1}{2}[/itex]

    x = 45 degrees
     
    Last edited: Feb 8, 2009
  12. Feb 8, 2009 #11

    Hootenanny

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    Okay, let me ask you two questions:

    (1) How do you normally find that angle between two vectors?

    (2) Can you write out v3 in terms of the components of vi if v1 = v1xi + v1yj and v2 = v2xi + v2yj?
     
  13. Feb 8, 2009 #12
    ok give me 1 sec to try you questions and I'll be right back! thanks for helping and staying with me!
     
  14. Feb 8, 2009 #13
    ok, I recall that
    (1)
    to find the angle between 2 vectors, I will take the dot product of them.

    v1 v2 = |v1 v2| [itex]cos(x)[/itex]

    so x is just arcos([v1 v2]/[|v2 v2|])


    (2)
    v3 = v1 + v2
    v3 = (v1x + v2x)i + (v1y + v2y)j

    and I think I have the answer to the question soon lol
     
    Last edited: Feb 8, 2009
  15. Feb 8, 2009 #14

    Hootenanny

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    Good :approve:.

    Now, using that last relation can you find |v3|2?
     
  16. Feb 8, 2009 #15
    |v3|[itex]^2[/itex] = (v1x+v2x)[itex]^2[/itex] + (v1y+v2y)[itex]^2[/itex]

    hmm.. are the terms on the right side suppose to be another set of dot products?
     
  17. Feb 8, 2009 #16
    It is useful to learn/understand this formalism. If X is a vector, then denote by X^2 the inner product of X with itself. If X and Y are vectors, then we have:

    (X + Y)^2 = X^2 + Y^2 + 2 X dot Y

    This is, of course, equivalent to the good old cosine rule from elementary geometry.

    So, if you have:

    v1 + v2 = 2 v3,

    then squaring both sides gives:

    v1^2 + v2^2 + 2 v1 dot v2 = 4 v3^2
     
  18. Feb 8, 2009 #17

    Hootenanny

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    Just to expand on the Count's post, one can of course prove this relation by mechanically calculating the square of the norm using the components.
     
  19. Feb 8, 2009 #18
    yeah i see how it's done, but pardon me, for I do not see still how squaring |v3| gives me the solution to the angle
     
  20. Feb 8, 2009 #19

    Hootenanny

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    Now you know that

    [tex]\begin{aligned}\left|\bold{v}_3\right|^2 & = \left|\bold{v}_1\right|^2 + \left|\bold{v}_2\right|^2 + \bold{v}_1\cdot\bold{v}_2 \\
    & = \left|\bold{v}_1\right|^2 + \left|\bold{v}_2\right|^2 + \left|\bold{v}_1\right|\left|\bold{v}_2\right|\cos\theta \end{aligned}[/tex]

    Does that help?
     
  21. Feb 8, 2009 #20
    aw i see.

    well i need to have the values for each term except for theta in order to solve for the angle. my mind is not clear right now to whether I I have those values or not.

    I am tire right now because i've been working on the physics problems from sun set last night to sun rise just now and I need to sleep lol.

    Thanks again Hootenanny, I feel welcomed to the forum and I really appreciate your help. But I gotta sleep so see you again!
     
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