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Two particles in an elastic collision given angle of first find other.

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Particle A has mass m and initial velocity v0. A collides with particle B, which has mass M and is initially at rest. After the collision, the particles follow the paths shown in the figure. Calculate φ in degrees, for an ideal elastic collision.
    Data: m = 0.7 kg; M = 4.7 kg.

    http://imgur.com/UcZciBB

    θ=45

    2. Relevant equations
    K=[itex]\frac{1}{2}[/itex]m1v12


    3. The attempt at a solution
    I'm not sure of the most effective route here. I have three equations but I have four unknowns (v1, v1', v2', and [itex]\phi[/itex])

    First I have momentum in the x direction:
    m1v1=[itex]\frac{1}{2}[/itex]m1v1'cos[itex]\phi[/itex]+m2v2'cosθ

    Second I have momentum in the y direction:
    0=m1v1'sin[itex]\phi[/itex]-m2v2'sin[itex]\theta[/itex]

    or

    m1v1'sin[itex]\phi[/itex]=m2v2'sin[itex]\theta[/itex]

    Lastly I have the conservation of kinetic energy since this is an elastic collision.

    [itex]\frac{1}{2}[/itex]m1v12=[itex]\frac{1}{2}[/itex]m1v1'2+[itex]\frac{1}{2}[/itex]m2v2'2

    I'm not sure of another equation that would help me find reduce this and solve for [itex]\phi[/itex]. Does anyone have any suggestions?

    Thank you!
     
  2. jcsd
  3. Oct 10, 2013 #2

    haruspex

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    You don't care about the exact velocities, only about the ratios between them. So it will turn out that there are really only three unknowns. Just go ahead and solve the equations.
     
  4. Oct 10, 2013 #3

    ehild

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    Remove 1/2 from the first momentum equation.

    Substitute θ=45°, isolate v1' or v2' from the equation for the y components of the momenta, substitute into the other momentum equation, you get both v1' and v2' in terms of Φ and v1. Use these expressions in the energy equation, v1 will cancel.

    ehild
     
  5. Oct 10, 2013 #4
    Not sure where that 1/2 came from to begin with. I definitely did not have that written down! Thank you :)

    I have tried what you suggested a number of times now both by hand and using wolfram alpha and keep getting [itex]\phi[/itex]=[itex]\pi[/itex]n

    I made some substitutions to make entering it into wolframalpha easier.

    v1=v
    v1'=z
    v2'=x

    mv=mzcos([itex]\phi[/itex])+[itex]\frac{Mx}{sqrt(2)}[/itex]

    and from the y-component

    [itex]\frac{Mx}{sqrt(2)}[/itex]=mzsin(phi)

    Take x from the y-component equation and plug it into the x-component equation to solve for z:

    z=[itex]\frac{Mx}{sqrt(2)msin(phi)}[/itex]

    Do the same for but taking z from the y-component equation and plugging that into the x-component equation to solve for x:

    x=[itex]\frac{sqrt(2)mvsin(phi)}{Mcos(phi)}[/itex]

    Plugging these x and z values into the energy equation mv2=mz2+Mx2:

    mv2=[itex]\frac{mv^2}{(cos(phi)+sin(phi))^2}[/itex]+[itex]\frac{2sin^2(phi)}{Mcos^2(phi)}[/itex]

    Plugging in the masses and solving for phi I get p=pi*n

    What am I messing up?
     
  6. Oct 10, 2013 #5

    ehild

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    that is one solution, which means there is no collision, particle A goes on with its initial velocity and B stays in rest. But there is an other solution.

    check z and x , I got different formulas. [tex]x=\frac{mz\sin(\Phi) \sqrt2}{M}[/tex], substitute into the eq. for the x components.



    ehild
     
    Last edited: Oct 10, 2013
  7. Oct 10, 2013 #6
    Thank you for your help.

    I tried your x originally as well. There are a couple ways to simplify when getting rid of z.

    I went through it again and came to the same conclusion on a different path. pi*n

    Plugging that x into the x-components equation:

    mv=mzcos(phi)+mzsin(phi)

    v=z(cos(phi)+sin(phi))

    I can use that in the energy equation. I can also plug in my x to the energy equation and it comes out to pi*n again. Fortunately, I plugged the final equation into my calculator instead of wolfram alpha and came up with the correct answer of about 82 degrees.

    sin(2*phi)=[itex]\frac{1.4}{4.7}[/itex]*sin^2(phi)

    Not sure why alpha doesn't provide the more useful answer.

    Thank you very much for your help and patience!
     
  8. Oct 10, 2013 #7
    Never mind.
     
    Last edited: Oct 10, 2013
  9. Oct 11, 2013 #8

    ehild

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    It is correct now.
    [tex]z=\frac{v}{\cos(\Phi)+\sin(\Phi)}[/tex]
    Sub it back into the equation for x.
    [tex]x=\frac{mz\sin(\Phi) \sqrt2}{M}=\frac{mv\sin(\Phi) \sqrt2}{M(\cos(\Phi)+\sin(\Phi))}[/tex] You can use x and z in the energy equation. m and v2 cancel.

    [tex](\cos(\Phi)+\sin(\Phi))^2=1+\frac{m}{M} 2 (\sin(\Phi))^2[/tex]

    But (cosΦ +sinΦ)2 = 1+2 sinΦ cosΦ.

    1 cancels, remains 2 sinΦ cosΦ=2 m/M sin2Φ--> sinΦ(cosΦ-m/MsinΦ)=0
    Either sinΦ=0 or m/M sinΦ=cosΦ.

    Wolframalpha works for "short" problems. And anyway, it is good for you if you solve the problem yourself.

    ehild
     
  10. Nov 5, 2013 #9

    bobie

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    You can simplify you work if you set m=1 (7/7) and V0 = (M+m) 54/7

    θ = 45° makes thing even easier, as pxm becomes 1 (7/7), px M = py M = pym = 47/7 and vM = √2.

    You can find easily: sin[itex]\Phi[/itex] = pym/ vm; easier is: cos[itex]\Phi[/itex] =vm-1, (√(v02- M*vm2)-1 = 0.1473... [tex]\frac{1}{\sqrt{\frac{2916}{49}- \frac{47*2*7}{7*7}}}= \sqrt\frac{49}{2258}[/tex]
    but you needn't even calculate vm, extremely simple is:

    [tex]tan\Phi = \frac{p_y}{p_x} = \frac{47}{7} = 6.7143, → \Phi = 81°. 52885537[/tex]
     
    Last edited: Nov 5, 2013
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