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    OK, after asking for help for some Mechanics problems and creating a thread for each one I'm creating this thread for any time I am unable to solve a problem in the book I am using (Mechancs M1, by Brian Jefferson and published by Oxford). Here is my current problem:

    'P is a point 10 m from the bottom of a slope. A ball is rolled up the slope from P with an initial velocity of 8 m s-1. It undergoes a constant acceleration of -4 m s-2.

    a)
    i) How long does it take to reach the bottom of the slope?
    ii) How far up the slope does it travel?
    b) One second after the first ball is rolled, a second ball is rolled up from the bottom of the slope. It has an initial velocity of 14 m s-1. It also has an acceleration of -4 m s-2. Find when and where the two balls meet.'

    The answer for a)i) is 5 seconds, and the answer for a)ii) is 8 m above P (these are confirmed in the answer sheet). However, I cannot figure out how to get the answer for this second half of the problem. I know the answer (and I can post it if someone wants), but I cannot figure out what process gets to it.

    This question has been asked before on another thread (I'm posting the link below), but they did not explain how they got to the right answer. However, there are a few failed approaches there which I myself also took.

    http://www.thestudentroom.co.uk/showthread.php?t=651221

    Please help!
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    (Original post by fablereader)
    x
    in the other thread they have stated time = 1.6, does this mean 1.6 secs after the second ball starts moving or 1.6 secs from the very start? the former and i can help you, the latter and i cannot
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    (Original post by DylanJ42)
    in the other thread they have stated time = 1.6, does this mean 1.6 secs after the second ball starts moving or 1.6 secs from the very start? the former and i can help you, the latter and i cannot
    1.6 seconds after the second ball starts moving.
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    Is it 23/7 sec from the very start?
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    (Original post by fablereader)
    1.6 seconds after the second ball starts moving.
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    (Original post by DylanJ42)
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    Thank you very much! This has helped clear things up tremendously in my head. I think one of the main problems was that I got t = 2.6, and since it wasn't the number of the answer I ignored the remaining context (after the release of the second ball), but reading this helped me keep that in mind. That got me t = 1.6 s and s = 17.28 m (from the bottom).
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    (Original post by fablereader)
    Thank you very much! This has helped clear things up tremendously in my head. I think one of the main problems was that I got t = 2.6, and since it wasn't the number of the answer I ignored the remaining context (after the release of the second ball), but reading this helped me keep that in mind. That got me t = 1.6 s and s = 17.28 m (from the bottom).
    thats exactly what i got too nice work
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    Funny enough, I thought that t1=t and t2=t-2. I don't know where I found that . Thanks.

    (Original post by fablereader)
    Thank you very much! This has helped clear things up tremendously in my head. I think one of the main problems was that I got t = 2.6, and since it wasn't the number of the answer I ignored the remaining context (after the release of the second ball), but reading this helped me keep that in mind. That got me t = 1.6 s and s = 17.28 m (from the bottom).
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    OK, I'm having trouble with another problem. This one just seems confusing to me, and I'm not entirely sure what they want. Could anyone help with this?

    'A body falls from rest from the top of a tower. During the last second of its motion it falls 7/16 of the whole distance.

    a) Show that the time of descent is independent of the value of g.

    b) Find the height of the tower in terms of g.'

    Note that g is gravity, an acceleration that is approximately 9.8 m s-2.
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    (Original post by fablereader)
    OK, I'm having trouble with another problem. This one just seems confusing to me, and I'm not entirely sure what they want. Could anyone help with this?

    'A body falls from rest from the top of a tower. During the last second of its motion it falls 7/16 of the whole distance.

    a) Show that the time of descent is independent of the value of g.
    You know it is dropped from rest: h = \frac{1}{2}gt^2 where t is the total time and h is the height.

    You know that in t-1 seconds, it falls a distance of \frac{9h}{16}, so we also have our second equation: \frac{9h}{10} = \frac{1}{2}g(t-1)^2

    Now divide the second expression by the first:

    \displaystyle 

\begin{equation*}\frac{\frac{9h}  {16}}{h} = \frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{(t-1)^2}{t^2} \iff \left(\frac{3}{4}\right)^2 = \left(\frac{t-1}{t}\right)^2 \end{equation*}
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    (Original post by fablereader)
    OK, I'm having trouble with another problem. This one just seems confusing to me, and I'm not entirely sure what they want. Could anyone help with this?

    'A body falls from rest from the top of a tower. During the last second of its motion it falls 7/16 of the whole distance.

    a) Show that the time of descent is independent of the value of g.
    This is probably a typo - it's obviously false as written (consider the motion with g = 0 m/s^2 for example) - I guess they meant independent of m, the mass, which is the import of Galileo's famous experiment.
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    (Original post by atsruser)
    This is probably a typo - it's obviously false as written (consider the motion with g = 0 m/s^2 for example) - I guess they meant independent of m, the mass, which is the import of Galileo's famous experiment.
    I think the case for g=0 is the only exception since you get an indeterminate form of \frac{0}{0} in my answer, but otherwise, I think my approach was correct(?).
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    (Original post by Zacken)
    I think the case for g=0 is the only exception since you get an indeterminate form of \frac{0}{0} in my answer, but otherwise, I think my approach was correct(?).
    I'm not sure what you're claiming to show - that the time of descent of an object falling a distance h in a gravitational field g is independent of g? Or something else?

    If it's the former, then it's false - we know that h = 0.5 gt^2 \Rightarrow t = \sqrt{2h/g} which depends on g. Am I confused?
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    OK, to be quite frank I think your discussion was a little bit above my level. I didn't (and still don't) understand it all (I'm no Maths genius). However, doing this division thing got me to:

    9/16 = (t2 - 2t + 1)/t2

    Multiplying it out and putting everything on one side I got:

    0 = 7t2 - 32t + 16.

    Using the quadratic formula I got:

    t = 4 or t = 0.5714285714
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    (Original post by Zacken)
    I think the case for g=0 is the only exception since you get an indeterminate form of \frac{0}{0} in my answer, but otherwise, I think my approach was correct(?).
    OK, I think I understand. The question is very poorly written. What it should say is:

    "Given the information provided, show that we can calculate the time of descent without knowing the value of g".

    It is true that we can make that calculation, but essentially the "value of g" has been encoded in the ratio of the distances fallen.
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    (Original post by fablereader)
    OK, to be quite frank I think your discussion was a little bit above my level. I didn't (and still don't) understand it all (I'm no Maths genius). However, doing this division thing got me to:

    9/16 = (t2 - 2t + 1)/t2
    Just square root both sides first - that's much easier.
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    (Original post by atsruser)
    Just square root both sides first - that's much easier.
    OK, that get's me:

    3/4 = (t-1)/t

    Multiplying it out and putting it all on one side I get:

    t = 4

    Looking at the answers, that's basically what they wanted, I think (their answer: Time is 4 s).

    Thanks for helping!
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    (Original post by atsruser)
    OK, I think I understand. The question is very poorly written. What it should say is:

    "Given the information provided, show that we can calculate the time of descent without knowing the value of g".

    It is true that we can make that calculation, but essentially the "value of g" has been encoded in the ratio of the distances fallen.
    That makes much more sense, the way I originally thought of it was "Here's a situation that happens in a really weird field where the gravitation strength is irrelevant, but we're not going to tell you that, we'll only tell you that the distance fallen in the last second is this much" but your explanation makes much more sense, I originally did a double take when I first read the question; that'll teach me to trust my instincts more. Thanks!
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    OK, having another difficulty. Again, g is gravity (i.e. acceleration).

    'A stone falls past a window of height 2.5 m in 0.5 s. Taking g = 10 m s-2, find the height from which the stone fell.'

    Please help!
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    (Original post by fablereader)
    OK, having another difficulty. Again, g is gravity (i.e. acceleration).

    'A stone falls past a window of height 2.5 m in 0.5 s. Taking g = 10 m s-2, find the height from which the stone fell.'

    Please help!
    So we have: s = -2.5, u = u, v = v, a = -10, t = 0.5 so that you can then find u which will then be your v for the motion starting from release.
 
 
 
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