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Homework Help: Collision. A moving object hits a stationary object and moves at an angle

  1. Jan 24, 2012 #1
    This is all I have. Some one please please pleeeeeaaaase check this over and see if I'm in any way on the right track. I haven't had much sleep the past few days and am really stressing over my courses. This is a take home part of our unit test. I hope I did this right because it could really save me from bombing the test.

    This is my attempt at a solution. All the formulas are there.


    I really need someone to tell me where I went wrong(if I did) and what to do to correct it. This is due at 10:50 tomorrow morning. I have 10 hours(some of that sleeping time).
    Thank you so very much to anyone who replies. I've never been this desperate.
  2. jcsd
  3. Jan 24, 2012 #2

    Simon Bridge

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    You look like you are saying that the two final momentum vectors sum to the initial one, which is correct, but you seem to have drawn a 30-120-30 triangle - this would only be the case is the final momentum of the heavy ball was the same as the final momentum of the light one (though complimentary directions).

    Your strategy was to find the heavy-ball momentum from the cosine rule, then use the sine rule to find the angle. But your substitution is looks odd: you have, for example, [itex]c^2 \leftarrow 0.5\text{kg}(v^2)[/itex] ... but surely this is supposed to be a [itex]0.25\text{kg}^2(v^2)[/itex]? Arn't you summing momentum vectors?
  4. Jan 24, 2012 #3


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    You are not correctly using conservation of momentum. Momentum is conserved as a vector.

    Write separate momentum conservation equations for x and y components or equivalently use vectors.

    I'd calculate the change in (vector) momentum of the .4kg mass since you know initial and final velocities. That is then the negative change in vector momentum of the other mass.
  5. Jan 24, 2012 #4
    Yes I used the squared 0.500kg. Should I be using the 30, 60, 90 triangle made by the x and y components of the momentum of the .400kg ball?
    I'm not sure how to use ball 1 to figure out ball 2's(the heavier one) direction.
  6. Jan 24, 2012 #5
    Ok so I found the difference in momentum for ball 1(.400kg) is 0.280kg m/s. Then I used that to find that ball 2's velocity is that divided by the mass (.500kg) to get 0.560m/s. Now I just need the direction. I know in a different question we assumed the y components were the same, so can I just do that to get the angle?
  7. Jan 24, 2012 #6
    did you do it in vector components like he said?


    m1v1x + m2v2x = m1v1'x + m2v2'x

    m1v1y + m2v2x = m1v1'y + m2v2'y

    ? cause what do you have when you have a veritcal and horizontal component of something? what are you able to get?
  8. Jan 24, 2012 #7

    Simon Bridge

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    It's harder to stuff up in vector components.

    This is hard to talk about - I'll make up some terms:
    If the initial momentum has magnitude [itex]p=m_1v [/itex] and the 0.4kg and 0.5kg balls final momenta have magnitudes [itex]p_1=m_1v_1[/itex] and [itex]p_2=m_2v_2[/itex] respectively.

    Then ... using your strategy, [itex]\vec{p}=\vec{p_1}+\vec{p_2}[/itex] will give you a triangle [itex]a:b:c = p_1:p_2:p[/itex] where the angle between sides [itex]p[/itex] and [itex]p_1[/itex] is known ... call it [itex]\theta[/itex] ... the other angles are unknown so leave them be. Only, the angle between [itex]p_2[/itex] and [itex]p[/itex] is one you want to know, so call it [itex]\phi[/itex]. That's how you needed to draw that vector diagram.

    As you did, the length of the third side can be determined from the cosine rule - which you did.[tex]p_2^2 = p^2+p_1^2 - 2p_1p\cos\theta[/tex]... then you can use the sine rule to find the angle [itex]\phi[/itex]:[tex]\frac{p_2}{\sin\theta}=\frac{p_1}{\sin\phi}[/tex]

    I don't see anything wrong with that approach.
    The only problem was that what you wrote down, and drew, made it look like you made some mistakes in your calculations.

    However it is harder to stuff up when you use components.

    put the y axis in the north direction, then the x axis points east.
    Using the above notation: [itex]\vec{p}=m_1v\hat{y}[/itex] see?

    Similarly for final:
    [itex]\vec{p}_1 = [p_1]_x\hat{x}+[p_1]_y\hat{y}[/itex]
    [itex]\vec{p}_2 = [p_2]_x\hat{x}+[p_2]_y\hat{y}[/itex]

    (here: [itex][p_1]_x[/itex] is the x-component of [itex]\vec{p}_1[/itex].)

    Use the 30:60:90 triangle to get the components ... that is 1:2:root-3

    The final x components have to add to zero, because the x component of p is zero.
    The final y components have to add to p.
    This will give you two simultaneous equations and two unknowns.
    Get the angle and magnitude you want from the components.

    That looks like more steps, but it is usually easier to think about.
    Which is why the others keep wanting you to do it that way.

    Note: I could go into a lot more detail for the cosine-rule part than I could for the components part because you had done all that work already... it's fine just check that what you wrote down accurately reflects what you actually computed.

    Must be past time to hand it in by now :) good luck.
  9. Jan 25, 2012 #8
    Thanks for the help. Not sure I followed it all exactly, and it was long past when it had to be handed in. I kind of gave him two papers and told him, "Take this one, no this one, wait, maybe this one" and he just took both so we'll see what happens. I don't know why my grasp of vectors is always so fleeting. I just finished Physics 20 in December, so it hasn't been that long, and we did some review. It's a mystery why I can't wrap my head around it in some cases.
  10. Jan 25, 2012 #9

    Simon Bridge

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    I remember it took me a while too - it wasn't until I realised that math was a language that I figured it out.

    The important thing to grok about vectors is components. That's pythagoras and trig.
    Other than that, just practice different methods and one will click.

    From the above, it looks like your main problem is just putting the numbers on the correct side of the brackets. The trick to that is to keep things abstract as long as possible, substitute at the end. Have a rest.
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