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Homework Help: Relative speed with respect to mirror image in two dimensional motion

  1. Jul 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle P is moving in a straight line with velocity u=10 m/s along the ground. Its line of motion makes angle θ=60° with the X axis drawn on the ground. A flat mirror oriented perpendicular to the ground as well as the X axis is moving with constant velocity of v=20 m/s. This velocity is directed parallel to X axis from right to left. Calculate the relative speed (in m/s) between the particle and its image.

    2. Relevant equations

    VA with respect to B = VA - VB

    3. The attempt at a solution

    The velocity of particle is 10m/s , and angle is 60°. So the velocity in terms of unit vectors is 5i + 5√3j. But since the image will be travelling in y direction with the same velocity , and we want to find relative velocity , we need not consider the y direction . ( Is that correct?)
    Now in x direction , I've a few problems. First how to calculate the velocity of image ? My friend told me that it is 2m-o ( where m=velocity of mirror and o = velocity of object ( in this case particle ) ) . Is that right?
    So velocity of image is = -40i-5i=-45i ( since v. of mirror is -20i) .
    Now Velocity of image with respect to particle is v= -45i-5i = -50i .
    So relative speed is 50 m/s.
    But the answer is wrong.
    Please help me.
  2. jcsd
  3. Jul 18, 2013 #2


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    x component of particle motion is certainly LESS than the 10m/s total velocity and mirror motion is 20m/s in x direction, so how could you POSSIBLY get a combined velocity of more than 30m/s?

    What's with the "i" and "j" stuff? Just get the x component of the particle motion.
  4. Jul 18, 2013 #3
    Yep, correct.
    I don't see how your friend arrived at that relation.

    Anyways, to solve the problem, work in the reference frame of mirror first. From there you can find the velocity of image wrt mirror.
  5. Jul 18, 2013 #4
    So velocity of particle with respect to mirror is -20-5 = -25 .
    Also the velocity of image with respect to mirror is then 25 ( I.E. in opposite direction to that of velocity of particle with respect to image ).
    So Velocity of particle with respect to image is -25-25=-50 .
    So again I get the relative speed between particle and image as 50 , Where I went wrong ?
    Please help .
  6. Jul 18, 2013 #5
    Correct! :approve:
    Remember, you are still working in the reference frame of mirror.

    You have the velocity of image in mirror frame. Can you find the velocity of image wrt ground now?
  7. Jul 18, 2013 #6
    Um , I'll try .
    Since velocity of image with respect to mirror is towards right 25 and velocity of mirror with respect to ground is towards left 20 I guess the velocity of image with respect to ground should be 25-20=5 .
    Is that correct ? Also is there a formula for changing reference frames ?
  8. Jul 18, 2013 #7
    No. Don't you know how to calculate velocities in different reference frames? This must be in your text.

    Check the section on Classical Mechanics.
  9. Jul 18, 2013 #8
    Haha , actually this is not a homework question . In my class , we haven't even learned about vectors . Because I love physics I just study it online. I found this question online and just wanted to solve it.

    Anyways , I checked the wikipedia page so,
    Vimage wrt ground = Vimage - Vground
    = 25 + 20 = 45 . ( since Vground wrt mirror = 20 ).
    Is that right ?
  10. Jul 18, 2013 #9

    You have ##\vec{v}_{\text{image wrt mirror}}##. This is equal to ##\vec{v}_{\text{image wrt ground}}-\vec{v}_{\text{mirror wrt ground}}##.
  11. Jul 18, 2013 #10
    Oh , okay .
    SO 25=Vimage wrt ground - (-20)
    So , Vimage wrt ground=25-20=5 ??
    I got the same answer again .
    Please help.
  12. Jul 18, 2013 #11
    What are the direction of velocities? ##\vec{v}_{\text{image wrt mirror}}## is towards left. What is the direction of ##\vec{v}_{\text{mirror wrt ground}}##?
  13. Jul 18, 2013 #12
    The direction of Vmirror wrt ground is also towards left.
    Taking towards right as positive -
    -25 = Vimage wrt ground - ( -20)
    -45 = Vimage wrt ground
    Am I finally right ?
  14. Jul 18, 2013 #13
    Yep, looks good to me.

    I hope you can reach the final answer now.
  15. Jul 18, 2013 #14
    Let me try .
    VParticle wrt image=Vparticle - Vimage
    VParticle wrt image= 5-(-45)
    So , VParticle wrt image =50
    And............ this is a wrong answer. ( And same as the one in my attempted solution in original first post ).
    Where have I went wrong ?
  16. Jul 18, 2013 #15
    What is the correct answer?
  17. Jul 18, 2013 #16
    The correct answer is 25 . Did you also got 50 as answer ?
  18. Jul 18, 2013 #17
    Yes. I don't see where we went wrong. We should wait for other members to reply.
  19. Jul 18, 2013 #18


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    I agree with you guys that the answer is 50 m/s.
  20. Jul 18, 2013 #19
    Thanks TSny! :)
  21. Jul 18, 2013 #20
  22. Jul 18, 2013 #21


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  23. Jul 18, 2013 #22


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    Draw a picture showing the position of the object, mirror, and image at some instant of time. If you want, let the distance from the object to the mirror be, say, 60 m at this instant. Note how far the image is from the object.

    Now redraw the picture for a time of 1 second later showing the new locations of the object, mirror, and image. Determine from the picture the new distance from the object to the image.

    So, you can see how much the object-image distance decreased during 1 second of time.
  24. Jul 18, 2013 #23


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    So what? A relative velocity is going to be the same viewed from any inertial frame.
  25. Jul 19, 2013 #24
    Oops, sorry about that. I replied hastily there. :redface:
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