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    question:
    a smooth sphere S of mass m is moving with speed u on a smooth horizontal plane. the sphere S collides with another smooth sphere T, of equal radius to S but of mass km, moving in the same straight line and in the same direction with speed \lambda u,0<\lambda<\frac{1}{2}. the coefficient of restitutioni between S and T is e. Given that S is brought to rest by the impact and

    e=\frac{1+k\lambda}{k(1-\lambda)}

    deduce that k>1.
    to be honest, i have put zero input into this question since i really dont know how to approach it. i pretty much can deduce that lambda =/=1, but i dont know what to do to the coefficient of restitution. rearranging won't help!
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    Hi KeineHeldenMehr,

    Ok lets try and get you started .

    Now first off we need to make sure were aware of certain observations, that we recognise ALL the inequalities we will need. So we have, from the question

    0 < \lambda < \frac{1}{2}
    also from the question, although not explicitly stated,
    0 < k

    We can infer this from that fact that, as far as M2 is concerned , there is no such thing as negative mass, so if m > 0, k must be > 0.

    Now will we need that? who knows perhaps, its worth noting. And finally
    0 < e < 1

    Now that final one we should just know already, but again it is very easy to just over look such a simple inequality that we can use. Ok so that all in place lets have a crack at the question.

    So in the question we have
    e = \frac{1 + k\lambda}{k(1-\lambda)}

    Now we know from our inequalities that k > 0, and λ > 0. Note just because they give us the fact that λ < 1/2, doesn't mean we actually have to use that fact, at least not here . So we can infer that kλ > 0. This means that,

    1 + k\lambda &gt; 1

    Now we now look to the e inequality, we have observed that e < 1. We have however just shown that the numerator of the fraction equal to e is > 1, so what does this say about the denominator of that fraction, specifically can we using that information derive an inequality for k( 1 - λ ).

    I will leave you there KeineHeldenMehr. This is the way I would have approached the question, some may go other ways. But I think really the thing to take away from it is that when dealing with inequalities questions like this, I would describe it as being very Heuristic. By this I mean we have no set method, its very much more on 'Best Guess' and that Mathematical Intuition, that you will develop over time.

    I hope that helps KeineHeldenMehr, have a really good go at the question now, it might take a good while for anything to reveal itself, but have a crack at it. If you do run in to a complete impass though post back, and well see how we can guide you
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    (Original post by Galadirith)
    Hi Galadirith,

    Now we now look to the e inequality, we have observed that e < 1. We have however just shown that the numerator of the fraction equal to e is > 1, so what does this say about the denominator of that fraction, specifically can we using that information derive an inequality for k( 1 - λ ).
    i dont think this is a bad approach at all, it's actually quite fun.

    e>1, implies that the denominator must be negative, in other words, since ,  k&gt;0, 1-\lambda&lt;0\implies \lambda&gt;1,

    this then follows that if 1+k\lambda&gt;1, that k \not= 1 and hence must be k>1, which i think solves it.

    any flaws in that argument? otherwise, thank you!
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    (Original post by KeineHeldenMehr)
    i dont think this is a bad approach at all, it's actually quite fun.
    Well I am very glad of that

    e>1, implies that the denominator must be negative
    Ok rite, well you started looking in the right place, but came out with the wrong answer . Lets try and work this though. So we know that e > 1, so from this we can produce the inequality

    1 &gt; \frac{1 + k\lambda}{k(1-\lambda)}

    Now you said that the denominator must therefore be negative for this inequality to be true. Now in fairness that is not completely incorrect, if the denominator is negative then it will indeed be less that 1. However, I think I did not included something that has slightly lead you astray . So I'm going to back track slightly to fix this. So we know that 0 < e < 1, so we can then derive the inequality

    0 &lt; \frac{1 + k\lambda}{k(1-\lambda)} &lt; 1

    Now can you see in this inequality, considering that 1 + kλ > 1, if the denominator IS negative, then it violates this inequality, as it would then be < 0. So you weren't wrong to say it was negative in first case, but that actually wasn't the entire inequality, I missed out explaining the 0 < bit, I just assumed it would be obvious, which I realise now it actual isn't .

    So let actually make this simpler for a second, instead of having a complicated fraction, let have a simplified version of it. Because right now we are only looking at the numerator and denominator as a whole, let us write the numerator as A, and the denominator as B, so

    0 &lt; \frac{A}{B} &lt; 1

    So we have already determined that A > 1, so we now need to find an inequality or, inequalities for B. Let us split the inequality into 0 < A/B and A/B < 1. Now with that first on, can you see that we know that A > 1, therefore A is positive, this means that B cannot be negative, otherwise it will violate this inequality. Now that means we have B > 0, and that is all that we can gather from that one. Now I want you to consider the other inequality A/B < 1. What can we gather about B from this inequality?

    in other words, since ,  k&gt;0, 1-\lambda&lt;0\implies \lambda&gt;1,
    Just as one final note, you said here that it implies λ>1, but we were told in the question that 0 < λ < 1/2. So just be aware of what your doing, because in a question, lets say you didn't quite know what to do, in this case, you would clearly be able to see that something you have calculated actual goes against what they have said in the question, so you can see that you must have gone wrong somewhere.

    Try and progress with that, hopefully that should get you to the correct answer
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    \frac{A}{B}&lt;1 \implies B&gt;A if we already know that A>1.

    therefore: k(1-\lambda)&gt;1+k\lambda

    simplify and make k the subject:

    k(1-2\lambda)&gt;1 \implies k&gt;\frac{1}{1-2\lambda},0&lt;\lambda&lt;\frac{1}{2}

    due to that range, k>1?
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    Yep that looks great , you can go deeper if you choose, for example you devided by ( 1 - λ ) without first determining that it is positive, because as you will know multiplying, and hence dividing by negative numbers in inequalities flips the direction of the inequality, but in this case it is relativly simple to see that it isn't negative, so thats fine . But depth is completly dependant on how much the question really wants you to anylise the problem, and what you have done would be more than enough for most.

    (Original post by KeineHeldenMehr)
    \frac{A}{B}&lt;1 \implies B&gt;A if we already know that A>1.
    k(1-2\lambda)&gt;1 \implies k&gt;\frac{1}{1-2\lambda},0&lt;\lambda&lt;\frac{1}{2}
    One final thing, just to draw all to a conclusion, Im sure it was an unintential mistake, but you wrote "k( 1 - 2λ )", when it should be "k( 1 - λ )" . But that 2 of course doesnt change your logic.

    Anyway, great work KeineHeldenMehr, have fun with your Maths
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    (Original post by Galadirith)
    One final thing, just to draw all to a conclusion, Im sure it was an unintential mistake, but you wrote "k( 1 - 2λ )", when it should be "k( 1 - λ )" . But that 2 of course doesnt change your logic.

    Anyway, great work KeineHeldenMehr, have fun with your Maths
    hmm... i think there might be a slight misunderstanding

    from here: B>A, k(1-\lambda)&gt;1+k\lambda i expanded the brackets.

    k-k\lambda&gt;1+k\lambda then pretty much collected all the k's the lhs.

    k-2k\lambda&gt;1 which is where this 2 arises.

    k(1-2\lambda)&gt;1 factorised,

    i think that i've got the wrong end of the stick, considering since we know that A>1, therefore if we just consider:

    k(1-\lambda)&gt;A, then divided through by 1-\lambda we get what you mentioned. i just want to clarify whether that's the point you're trying to make.

    otherwise, i must have made another mistake elsewhere.
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    Ah sorry KeineHeldenMehr,

    You are absolutly correct. This I suppose enphasises the diversity of such a question. Yes I concluded it using a different method, but you way is absolutly correct, that was me being silly
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    (Original post by Galadirith)
    Ah sorry KeineHeldenMehr,

    You are absolutly correct. This I suppose enphasises the diversity of such a question. Yes I concluded it using a different method, but you way is absolutly correct, that was me being silly
    no worries, was very useful! thank you, you're a star!
 
 
 
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