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Unable to solve Mechanics problems watch

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    (Original post by Zacken)
    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.
    Thanks! I didn't think of taking u and v in the duration of it's fall. It's led me to the right answer (0.3125 m above the window).
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    I'm trying to deal with another problem. The answer I got is close, but I think I made a mistake somewhere.

    'An object is projected vertically upwards with a velicty u m s-1. T seconds later another object is projected vertically upwards from the same point and with the same speed. Find, in terms of u, T and g, the further time which elapses before the objects collide.'

    Using the equation 's = ut + (1/2)at2' I got a displacement for each ball:

    First ball: s = ut + (1/2)gt2
    Second ball: s = u(t-T) + (1/2)g(t-T)2

    Making them equal, I got:

    t = -u/g + T/2

    As mentioned earlier, it's close but not the answer. Have I made a mistake?
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    (Original post by fablereader)

    First ball: s = ut + (1/2)gt2
    Second ball: s = u(t-T) + (1/2)g(t-T)2
    You seem to be taking positive as upwards, so your acceleration due to gravity acts downwards, hence the acceleration is -g, i.e: you should have s = ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}g(t-T)^2.
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    (Original post by Zacken)
    You seem to be taking positive as upwards, so your acceleration due to gravity acts downwards, hence the acceleration is -g, i.e: you should have s = ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}g(t-T)^2.
    I always forget the simple things. Thank you for reminding me! Even after that, I got the wrong answer (u/g + T/2), so I tried changing (t - T) to (t + T), and that got me the right answer.

    Thanks!
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    (Original post by fablereader)
    I always forget the simple things. Thank you for reminding me! Even after that, I got the wrong answer (u/g + T/2), so I tried changing (t - T) to (t + T), and that got me the right answer.

    Thanks!
    Hmm, we have ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}g(t^2 - 2tT + T^2) \Rightarrow 0 = -uT + gtT - \frac{1}{2}T^2 \Rightarrow \frac{u}{g} + \frac{1}{2}T = t

    Why is this wrong? It should be t -T.

    I think your wrong answer is arising from the fact that it is the "further time elapsed" so you need to do \text{further time}= \frac{u}{g} + \frac{1}{2}T - T = \frac{u}{g} - \frac{T}{2}, i.e: subtract T from the "answer".
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    (Original post by Zacken)
    Hmm, we have ut - \frac{1}{2}gt^2 = ut - uT - \frac{1}{2}g(t^2 - 2tT + T^2) \Rightarrow 0 = -uT + gtT - \frac{1}{2}T^2 \Rightarrow \frac{u}{g} + \frac{1}{2}T = t

    Why is this wrong? It should be t -T.

    I think your wrong answer is arising from the fact that it is the "further time elapsed" so you need to do \text{further time}= \frac{u}{g} + \frac{1}{2}T - T = \frac{u}{g} - \frac{T}{2}, i.e: subtract T from the "answer".
    Sorry, I just assumed that producing an equivalent of t would be the answer, but you rightly point out that they called it 'further time'. However, I don't fully understand that, so can you explain why that would require subtracting T from the equivalent of t to get the answer? I'd like to understand it fully.
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    (Original post by fablereader)
    Sorry, I just assumed that producing an equivalent of t would be the answer, but you rightly point out that they called it 'further time'. However, I don't fully understand that, so can you explain why that would require subtracting T from the equivalent of t to get the answer? I'd like to understand it fully.
    So you know that the time from the very start of the entire thing to when they collide is t = \frac{u}{g} + \frac{T}{2}, the second particle, however - is released a a time T. Imagine you have a stopwatch and as soon as the first particle is released you pressed it. This stopwatch is magical, so it displays \frac{u}{g} + \frac{T}{2} when the two particles collide.

    The question asks for further time, i.e: how much time after the second particle has been released do they collide? i.e: how much time after T seconds has already elapsed do they collide? Basically, they want you to find the time from if you had pressed the stopwatch or started the stopwatch from when the second particle is released.

    So to get this, we take the time that our stopwatch displays, i.e: the time from the first release and subtract the time that we had to wait for the lazy second particle to be released. i.e: this is \frac{u}{g} + \frac{T}{2} - T, i.e: our total time and then subtract the T seconds that has already elapsed.

    Does that make sense? If not, I'd be happy to explain more.
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    (Original post by Zacken)
    So you know that the time from the very start of the entire thing to when they collide is t = \frac{u}{g} + \frac{T}{2}, the second particle, however - is released a a time T. Imagine you have a stopwatch and as soon as the first particle is released you pressed it. This stopwatch is magical, so it displays \frac{u}{g} + \frac{T}{2} when the two particles collide.

    The question asks for further time, i.e: how much time after the second particle has been released do they collide? i.e: how much time after T seconds has already elapsed do they collide? Basically, they want you to find the time from if you had pressed the stopwatch or started the stopwatch from when the second particle is released.

    So to get this, we take the time that our stopwatch displays, i.e: the time from the first release and subtract the time that we had to wait for the lazy second particle to be released. i.e: this is \frac{u}{g} + \frac{T}{2} - T, i.e: our total time and then subtract the T seconds that has already elapsed.

    Does that make sense? If not, I'd be happy to explain more.
    Ah, yes, that makes sense. I had to read it over a few times, but I got it. Thank you!
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    (Original post by fablereader)
    Ah, yes, that makes sense. I had to read it over a few times, but I got it. Thank you!
    Awesome! Good going with all your questions, by the way - first class work. :-)
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    I'm unsure how to answer this problem. Could anyone help?

    'A particle A is thrown vertically upwards from the bottom of a tower. At the same instant a second particle, B, is dropped from the top of the tower. Given that when the particles collide they are travelling at the same speed, find the ratio between the distances they have travelled.'
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    (Original post by Zacken)
    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!
    Whoever wrote this question was an effing idiot - its wording is utterly incorrect and misleading.
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    (Original post by fablereader)
    I'm unsure how to answer this problem. Could anyone help?

    'A particle A is thrown vertically upwards from the bottom of a tower. At the same instant a second particle, B, is dropped from the top of the tower. Given that when the particles collide they are travelling at the same speed, find the ratio between the distances they have travelled.'
    You don't know the initial speed u of the lower particle, but you can express it in terms of the time taken t (which is the same for both particles) by equating the speeds of the particles after time t (use SUVAT to do so). Then eliminate u from another appropriate SUVAT eqn expressing s in terms of t.
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    (Original post by atsruser)
    You don't know the initial speed u of the lower particle, but you can express it in terms of the time taken t (which is the same for both particles) by equating the speeds of the particles after time t (use SUVAT to do so). Then eliminate u from another appropriate SUVAT eqn expressing s in terms of t.
    OK, got it. Thanks!
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    I generally have difficulty with bearings. Now, before I start studying vectors, I need to really try to 'understand and use bearings to solve problems involving angles'. Could anyone help me with this problem, so I can better learn how to solve these kinds of problems?

    'A boat sets off from a harbour at point O. It travels 7 km at a bearing of 140 degrees to a point A. The boat then travels to point B which is 3.5 km on a bearing of 260 degrees from point A. Find

    a the bearing the boat should take to travel backt o the harbour in a straight line from B

    b the distance of the homeward journey described in part a.

    Could anyone help me understand bearings, with this problem as an example?
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    (Original post by fablereader)
    I generally have difficulty with bearings. Now, before I start studying vectors, I need to really try to 'understand and use bearings to solve problems involving angles'. Could anyone help me with this problem, so I can better learn how to solve these kinds of problems?

    'A boat sets off from a harbour at point O. It travels 7 km at a bearing of 140 degrees to a point A. The boat then travels to point B which is 3.5 km on a bearing of 260 degrees from point A. Find

    a the bearing the boat should take to travel backt o the harbour in a straight line from B

    b the distance of the homeward journey described in part a.

    Could anyone help me understand bearings, with this problem as an example?
    Draw a standard Cartesian axes, north, west, south, east. Put your rule on the y-axis facing upwards. Whenever you are given a bearing, move the ruler clockwise x^{\degrees} (where x is the given bearing). So 90^{\circ} will make your rule lie on the x-axis pointing to the right. 180^{\circ} will make your bearing lie on the y-axis pointing downwards, etc...

    Now you should automatically convert any question about bearings into a question about triangles and geometry by drawing the required lines and points.

    Draw a line 7 cm long at a bearing of 140, this will be in the fourth quadrant, starting from the origin. Label the end point A.

    Now pretend A is your "north" so put your ruler facing the direction of A and rotate it 260 degrees. Then draw a line from A at an angle of 260 degrees away from A (so it should be pointing back inwards, kinda - you get me?) and it should be 3.5 cm long.
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    (Original post by Zacken)
    Draw a standard Cartesian axes, north, west, south, east. Put your rule on the y-axis facing upwards. Whenever you are given a bearing, move the ruler clockwise x^{\degrees} (where x is the given bearing). So 90^{\circ} will make your rule lie on the x-axis pointing to the right. 180^{\circ} will make your bearing lie on the y-axis pointing downwards, etc...

    Now you should automatically convert any question about bearings into a question about triangles and geometry by drawing the required lines and points.

    Draw a line 7 cm long at a bearing of 140, this will be in the fourth quadrant, starting from the origin. Label the end point A.

    Now pretend A is your "north" so put your ruler facing the direction of A and rotate it 260 degrees. Then draw a line from A at an angle of 260 degrees away from A (so it should be pointing back inwards, kinda - you get me?) and it should be 3.5 cm long.
    OK, I get it. I've drawn the lines and also deduced that angle OAB is 60 degrees (the external angle is 300, so 360 - 300 = 60). However, I'm not sure where to go from here.
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    (Original post by fablereader)
    OK, I get it. I've drawn the lines and also deduced that angle OAB is 60 degrees (the external angle is 300, so 360 - 300 = 60). However, I'm not sure where to go from here.
    So now draw alone from B to the origin and find the angle between AB and OB, then from that, you can deduce the bearing that B needs to be able to go in a straight line from B to O.
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    (Original post by Zacken)
    So now draw alone from B to the origin and find the angle between AB and OB, then from that, you can deduce the bearing that B needs to be able to go in a straight line from B to O.
    Thanks for the help! Got the right answers (a is 350 degrees, and b is 6.06 km).
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    (Original post by fablereader)
    Thanks for the help! Got the right answers (a is 350 degrees, and b is 6.06 km).
    First class work!
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    I'm having trouble wtih a vector problem.

    'The diagram shows a trapezium ABCD, with AB parallel to DC and AB twice as long as DC.
    E is the mid-point of BC. AD-> = p and DC-> = q.
    Find, in terms of p and q

    a AB->
    b AC->
    c CD->
    d DB->
    e AE->
    f ED->'


    -> signifies it is a translation instead of a distance. Unfortunately, I can't put in the diagram. I've solved a-d (2q, p + q, -q, 2q - p respectively), but I've stalled on e. Can anyone help?
 
 
 
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