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    (Original post by EnglishMuon)
    Is that the newer one on the outside of cambridge?
    Yeah, next to the CMS.


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    (Original post by physicsmaths)
    Im doing III 2010 stuff today. If yur interested.
    Im stuck on Q11 III lolz


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    I just started Q11 (had to eat breakfast first) and I am stuck on trying to find deceleration of the block. Isn't it just a_1 = \frac{(M+m) \mu g}{M}? It seems like to get the require value for the relative deceleration you must account for the deceleration caused by R on the block which doesn't really make sense. What am I doing wrong?


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    (Original post by Insight314)
    I just started Q11 (had to eat breakfast first) and I am stuck on trying to find deceleration of the block. Isn't it just a_1 = \frac{(M+m) \mu g}{M}? It seems like to get the require value for the relative deceleration you must account for the deceleration caused by R on the block which doesn't really make sense. What am I doing wrong?


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    Yeh we need to include R, but i don't understand y either, i thought due to 3rd law it cancles with each other. **** knows mate. I ain't good enough to handle questions like these haha 😂


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    (Original post by physicsmaths)
    Yeh we need to include R, but i don't understand y either, i thought due to 3rd law it cancles with each other. **** knows mate. I ain't good enough to handle questions like these haha 😂


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    Oh I figured it out! You are right, it is due to Newton's Third Law. As the bullet travels through the block, a resistive force of magnitude R is exerted on it and simultaneously it exerts a force on the block with the same magnitude but opposite direction. Think about it, when a mass travels through a material surely it exerts a force on the material as it travels while a force is exerted on the mass by the material.

    For second part, by common speed they mean the speed of the block added to speed of the bullet when the speed of the bullet is zero right? You also have to use conservation of linear momentum to find an expression for the initial velocity of both block and bullet as billet collides with the block.


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    (Original post by Insight314)
    Oh I figured it out! You are right, it is due to Newton's Third Law. As the bullet travels through the block, a resistive force of magnitude R is exerted on it and simultaneously it exerts a force on the block with the same magnitude but opposite direction. Think about it, when a mass travels through a material surely it exerts a force on the material as it travels while a force is exerted on the mass by the material.

    For second part, by common speed they mean the speed of the block added to speed of the bullet when the speed of the bullet is zero right? You also have to use conservation of linear momentum to find an expression for the initial velocity of both block and bullet as billet collides with the block.


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    Duno mate i gave up after the first part haha.


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    (Original post by physicsmaths)
    Duno mate i gave up after the first part haha.


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    Have you tried 2014 III Q9. It's a pretty nice one. A bit scary for me since it is mech + vectors which is the worst combo lol but very good for my STEP prep. It is pretty easy tbh, literally only dot products and not that much algebra either. Made a lot of silly mistakes though, even now trying to find out where my sign mistakes came from in showing the expression for \tan \beta.

    I am gonna work through mech section in 2014 and do all the vectors/complex numbers if I find any then move on to 2013 III.


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    a nice question in preparation for STEP:
    Prove that for any integer n, there exist infinitely many primes congruent to 1 mod n.
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    (Original post by gasfxekl)
    a nice question in preparation for STEP:
    Prove that for any integer n, there exist infinitely many primes congruent to 1 mod n.
    Counterexample : for n=1 that's not true .
    Ok yes ignore me
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    (Original post by Vesniep)
    Counterexample : for n=1 that's not true .
    Ok yes ignore me
    Lol no
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    (Original post by gasfxekl)
    Lol no
    What do you mean with no ?
    Every integer prime or not is congruent to 0 mod 1 .
    I'll try to prove it for n>=2 which makes sense
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    (Original post by Vesniep)
    What do you mean with no ?
    Every integer prime or not is congruent to 0 mod 1 .
    I'll try to prove it for n>=2 which makes sense
    Every prime is congruent to any number mod 1. Lol
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    (Original post by gasfxekl)
    Every prime is congruent to any number mod 1. Lol
    Just read what congruent means .
    Sometimes I don't know the exact translation in english (sorry )
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    (Original post by Vesniep)
    Just read what congruent means .
    Sometimes I don't know the exact translation in english (sorry )
    mainland europe pride my brother
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    (Original post by Number Nine)
    mainland europe pride my brother
    not all of greece is in mainland europe bro . how do you know on which side I live ... you know too much about me
    It was not pride it was ignorance
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    STEP II 1992 Q1

    (iii) Don't get this one. Solutions say it's 0, which yeah ok at first glance you'd think that, but I thought \dfrac{\sin x}{x} becomes a \dfrac{0}{0} indeterminate form every 2\pi and so it appears that we do not have:

    'for every \epsilon >0\ \exists\ N such that |f(x)|<\epsilon\ \forall x>N.

    Now I don't know whether there is a subtle extension to this limit to infinity definition that takes into account the domain of the function but I would think that this shows that the limit doesn't exist, which is what I put.

    Also I have to say the last limit (vi) was really tough especially after than many easy limits.
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    (Original post by Insight314)
    Have you tried 2014 III Q9. It's a pretty nice one. A bit scary for me since it is mech + vectors which is the worst combo lol but very good for my STEP prep. It is pretty easy tbh, literally only dot products and not that much algebra either. Made a lot of silly mistakes though, even now trying to find out where my sign mistakes came from in showing the expression for \tan \beta.

    I am gonna work through mech section in 2014 and do all the vectors/complex numbers if I find any then move on to 2013 III.


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    Na im gna do III 14 as a mock again so wnt look it till next week or somethin.


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    (Original post by gasfxekl)
    a nice question in preparation for STEP:
    Prove that for any integer n, there exist infinitely many primes congruent to 1 mod n.
    Should I consider n*p1p2...pn+1 but how am I gonna prove it's prime ?
    tbh it's not an easy question I don't know proofs about primes just the euclidean proof , I'll try it later .
    I would not consider it step preparation though , perhaps the final result of a question with many hints .
    I mean I once saw a proof about infinite primes of the form 4λ+3 it was an interview question and one of the difficult ones (if you haven't read it before of course) and there was a note about 4λ+1 that fails the same contradiction so under no circumstances the general result would be easier to prove .
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    (Original post by IrrationalRoot)
    STEP II 1992 Q1

    (iii) Don't get this one. Solutions say it's 0, which yeah ok at first glance you'd think that, but I thought \dfrac{\sin x}{x} becomes a \dfrac{0}{0} indeterminate form every 2\pi and so it appears that we do not have:

    'for every \epsilon >0\ \exists\ N such that |f(x)|<\epsilon\ \forall x>N.

    Now I don't know whether there is a subtle extension to this limit to infinity definition that takes into account the domain of the function but I would think that this shows that the limit doesn't exist, which is what I put.

    Also I have to say the last limit (vi) was really tough especially after than many easy limits.
    n tends to infinity , so it is k/infinity for -1<k<1 so yes it is 0.
    You can prove it with the squeeze theorem it's very easy
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    (Original post by IrrationalRoot)
    STEP II 1992 Q1

    (iii) Don't get this one. Solutions say it's 0, which yeah ok at first glance you'd think that, but I thought \dfrac{\sin x}{x} becomes a \dfrac{0}{0} indeterminate form every 2\pi and so it appears that we do not have:

    'for every \epsilon &gt;0\ \exists\ N such that |f(x)|&lt;\epsilon\ \forall x&gt;N.

    Now I don't know whether there is a subtle extension to this limit to infinity definition that takes into account the domain of the function but I would think that this shows that the limit doesn't exist, which is what I put.

    Also I have to say the last limit (vi) was really tough especially after than many easy limits.
    I don't think the last limit was tough, it was fairly obvious that you needed to divide top and bottom by \sqrt{n} to get it in an expandable form and then use the binomial expansion to construct a standard Taylor series approach.

    With regards to your first question, you have:

    -\frac{1}{n} &lt; \frac{\sin n }{n} &lt; \frac{1}{n} so application of the squeeze theorem handles it well.

    I'm not sure what you mean by an indeterminate form? First off, n takes on integer values (so not 2pi) (I know it's not specified, but that's what the general usage of n v/s x is).

    Second of all, what's wrong with x=2\pi? We'll get \frac{0}{2\pi}, not an indeterminate form.
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    (Original post by Zacken)
    I don't think the last limit was tough, it was fairly obvious that you needed to divide top and bottom by \sqrt{n} to get it in an expandable form and then use the binomial expansion to construct a standard Taylor series approach.

    With regards to your first question, you have:

    -\frac{1}{n} &lt; \frac{\sin n }{n} &lt; \frac{1}{n} so application of the squeeze theorem handles it well.

    I'm not sure what you mean by an indeterminate form? First off, n takes on integer values (so not 2pi) (I know it's not specified, but that's what the general usage of n v/s x is).

    Second of all, what's wrong with x=2\pi? We'll get \frac{0}{2\pi}, not an indeterminate form.
    There's nothing wrong with it. I spent half an hour confused for no reason, I'm such an idiot...
 
 
 
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