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Falling Mirror

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A plane mirror of length 20 cm is put up in a wall.The upper part of the mirror is 130 cm above the ground.A person is standing 1 m in front of the mirror.The person's eye is 130 cm from the ground.Suddenly,the mirror falls.How long is the person able to see the tip of his feet ?
    Assume that g=10 m/s2


    2. Relevant equations
    S=Vot +1/2 at2
    Laws of Reflection
    V=Vo+at

    3. The attempt at a solution
    I think that the first time the person sees his feet would be when the lower side of the mirror is at height 65 cm from the ground.Therefore,I used S=Vot+1/2 gt2 and found that it takes 3 s to be at 65 cm from the ground.From then on ,the person would be able to see his feet,so the time can be calculated from S = Vot+1/2gt2 with s being the length of mirror ,the speed is Vo=g.3 and using quadratics the solution is 0.06 s .Please help me out plase check my work,I also do not know if the distance of the person from the mirror matters.
     
    Last edited: Mar 26, 2016
  2. jcsd
  3. Mar 26, 2016 #2

    Simon Bridge

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    Stand different distances from the bathroom mirror and see if it changes what you can see.
    Draw a sketch of the situation - if the mirror sits on the ground, how will the person see the reflection of his feet?
     
  4. Mar 26, 2016 #3
    Well ,I can see more objects ,but
    Well ,my field of view is bigger the closer i am .If the mirror is on the ground then the light reflected from my feet cannot reach my eyes,so i cannot see my feet.
     
  5. Mar 26, 2016 #4

    Simon Bridge

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    ... but you said before that the person will see his feet in the mirror at all times after the bottom has reached a height of 65cm ... and yet here you observe that when the bottom is at 0cm the feet are not visible. What does this suggest to you about the time that the feet remain visible?
     
  6. Mar 26, 2016 #5
    Oops,
    I put the wrong sentence then.I guess it should be that the person will only be able to see his feet when the mirror pass 65 cm above the ground,but was my calculations correct ,the distance i put to get the time was the length of the mirror ?
     
  7. Mar 26, 2016 #6

    Simon Bridge

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    ... you mean, while the mirror is passing over the 65cm mark above the ground.

    I can help you check your reasoning but I won't check your maths for you.
    The way to check your reasoning is to draw a diagram: it's just geometry - you should feel confident about drawing triangles.
    You should already have had lots of experience of the maths of triangles so you will be fine there.
    If you are unsure about making a sketch, then make it a scale drawing.
     
  8. Mar 26, 2016 #7
    That is what I mean,so is my reasoning correct that the time the person sees his feeet is equal to the time required for the whole length of the mirror to pass the 65 cm mark ?
     
  9. Mar 26, 2016 #8

    Simon Bridge

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    This is me helping you to check your reasoning:

    The way to check your reasoning is to draw a diagram: it's just geometry - you should feel confident about drawing triangles.
    You should already have had lots of experience of the maths of triangles so you will be fine there.
    If you are unsure about making a sketch, then make it a scale drawing.
     
  10. Mar 26, 2016 #9
     

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  11. Mar 26, 2016 #10
    So is the time when the feet is visible the time needed for mirror in position 1 to reach position 2 ?
     
  12. Mar 26, 2016 #11

    Simon Bridge

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    Did you find anything in your diagrams to contradict what you found mathematically?

    Your problem is that you cannot tell when you get something right.
    This is a serious problem and one of the hardest to overcome... but it is very important because you are training to solve problems where nobody knows the right answers ... so there is nobody to ask. Some people work on problems where nobody even knows the right reasoning to work on it. But it is possible to tell when you get it right if you think about it. One approach is, as here, to try working the problem using an approach you are more familiar with. This is like how you become good at multiplication by, first, checking that repeated addition works out the same way. After a while you get used to just multiplying and move on to the next harder concept.
     
  13. Mar 26, 2016 #12
    I guess it makes perfect sense then.Thanks for the advice.
     
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