Relative speed with respect to mirror image in two dimensional motion

In summary: Oh , in that case ,Vimage wrt ground = 25-(-20)=45 .So , relative speed = 45+25=70Is that right ?Oh , in that case ,Vimage wrt ground = 25-(-20)=45 .So , relative speed = 45+25=70Is that right ?Yep, that's correct. You are calculating the magnitude of the relative velocity, not the direction. So the final answer is 70m/s.
  • #1
agoogler
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0

Homework Statement



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.
zExdN.png

Homework Equations



VA with respect to B = VA - VB

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 traveling 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.
 
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  • #2
agoogler said:
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.

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.
 
  • #3
agoogler said:
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 traveling in y direction with the same velocity , and we want to find relative velocity , we need not consider the y direction . ( Is that correct?)
Yep, 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?
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.
 
  • #4
Pranav-Arora said:
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.
Okay.
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 .
 
  • #5
agoogler said:
Okay.
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 ).
Correct! :approve:
So Velocity of particle with respect to image is -25-25=-50 .
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?
 
  • #6
Pranav-Arora said:
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?
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 ?
 
  • #7
agoogler said:
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 ?

No. Don't you know how to calculate velocities in different reference frames? This must be in your text.

http://en.wikipedia.org/wiki/Relative_velocity
Check the section on Classical Mechanics.
 
  • #8
Pranav-Arora said:
No. Don't you know how to calculate velocities in different reference frames? This must be in your text.

http://en.wikipedia.org/wiki/Relative_velocity
Check the section on Classical Mechanics.
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 ?
 
  • #9
agoogler said:
Anyways , I checked the wikipedia page so,
Vimage wrt ground = Vimage - Vground
:confused:

You have ##\vec{v}_{\text{image wrt mirror}}##. This is equal to ##\vec{v}_{\text{image wrt ground}}-\vec{v}_{\text{mirror wrt ground}}##.
 
  • #10
Pranav-Arora said:
:confused:

You have ##\vec{v}_{\text{image wrt mirror}}##. This is equal to ##\vec{v}_{\text{image wrt ground}}-\vec{v}_{\text{mirror wrt ground}}##.
Oh , okay .
SO 25=Vimage wrt ground - (-20)
So , Vimage wrt ground=25-20=5 ??
I got the same answer again .
Please help.
 
  • #11
agoogler said:
Oh , okay .
SO 25=Vimage wrt ground - (-20)
So , Vimage wrt ground=25-20=5 ??
I got the same answer again .
Please help.

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}}##?
 
  • #12
Pranav-Arora said:
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}}##?
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 ?
 
  • #13
agoogler said:
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 ?

Yep, looks good to me.

I hope you can reach the final answer now.
 
  • #14
Pranav-Arora said:
Yep, looks good to me.

I hope you can reach the final answer now.
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 ?
 
  • #15
agoogler said:
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 ?

What is the correct answer?
 
  • #16
Pranav-Arora said:
What is the correct answer?
The correct answer is 25 . Did you also got 50 as answer ?
 
  • #17
agoogler said:
Did you also got 50 as answer ?

Yes. I don't see where we went wrong. We should wait for other members to reply.
 
  • #18
I agree with you guys that the answer is 50 m/s.
 
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  • #19
TSny said:
I agree with you guys that the answer is 50 m/s.

Thanks TSny! :)
 
  • #22
agoogler said:
But what about this post made by phinds -
https://www.physicsforums.com/showpost.php?p=4448647&postcount=2

Is what he wrote true? If so why is our answer more than 30 ?

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.
 
  • #23
Pranav-Arora said:
Remember, you are still working in the reference frame of mirror.
So what? A relative velocity is going to be the same viewed from any inertial frame.
 
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  • #24
haruspex said:
So what? A relative velocity is going to be the same viewed from any inertial frame.

Oops, sorry about that. I replied hastily there. :redface:
 

What is relative speed with respect to mirror image in two dimensional motion?

Relative speed with respect to mirror image in two dimensional motion refers to the speed of an object as it moves in relation to its mirror image. This is typically measured in terms of velocity, which takes into account both the speed and direction of the object's motion.

How is relative speed calculated in two dimensional motion?

To calculate relative speed in two dimensional motion, you must first determine the velocity of the object and its mirror image. Then, you can use vector addition to find the relative velocity of the object in relation to its mirror image. This involves adding the x and y components of both velocities to determine the overall relative speed and direction.

What is the significance of relative speed with respect to mirror image in two dimensional motion?

Relative speed with respect to mirror image is important in understanding how an object appears to move in relation to its mirror image. It can also help in predicting the behavior of objects in certain situations, such as collisions or reflections.

Does the angle of incidence affect relative speed with respect to mirror image in two dimensional motion?

Yes, the angle of incidence can affect the relative speed with respect to mirror image in two dimensional motion. When the angle of incidence is perpendicular to the mirror, the relative speed will be the same as the actual speed of the object. However, as the angle of incidence increases, the relative speed will decrease.

What are some real world applications of understanding relative speed with respect to mirror image in two dimensional motion?

Understanding relative speed with respect to mirror image in two dimensional motion can be useful in fields such as optics, robotics, and computer graphics. It can also be applied in everyday situations, such as when driving a car and using mirrors to gauge the speed and distance of other vehicles.

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