Relative speed with respect to mirror image in two dimensional motion

[agoogler]

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.

Homework Equations

V_{A with respect to B} = V_A - V_B

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.

 

[phinds]

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.

 

[Saitama]

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.

 

[agoogler]

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 .

 

[Saitama]

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!

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?

 

[agoogler]

Correct!

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 ?

 

[Saitama]

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.

 

[agoogler]

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,
V_{image wrt ground} = V_image - V_ground
= 25 + 20 = 45 . ( since V_{ground wrt mirror} = 20 ).
Is that right ?

 

[Saitama]

Anyways , I checked the wikipedia page so,
V_{image wrt ground} = V_image - V_ground

You have $\vec{v}_{\text{image wrt mirror}}$. This is equal to $\vec{v}_{\text{image wrt ground}}-\vec{v}_{\text{mirror wrt ground}}$.

 

[agoogler]

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=V_{image wrt ground} - (-20)
So , V_{image wrt ground}=25-20=5 ??
I got the same answer again .
Please help.

 

[Saitama]

Oh , okay .
SO 25=V_{image wrt ground} - (-20)
So , V_{image 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 $\vec{v}_{\text{mirror wrt ground}}$?

 

[agoogler]

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 V_{mirror wrt ground} is also towards left.
Taking towards right as positive -
-25 = V_{image wrt ground} - ( -20)
-45 = V_{image wrt ground}
Am I finally right ?

 

[Saitama]

The direction of V_{mirror wrt ground} is also towards left.
Taking towards right as positive -
-25 = V_{image wrt ground} - ( -20)
-45 = V_{image wrt ground}
Am I finally right ?

Yep, looks good to me.

I hope you can reach the final answer now.

 

[agoogler]

Yep, looks good to me.

I hope you can reach the final answer now.

Let me try .
V_{Particle wrt image}=V_particle - V_image
V_{Particle wrt image}= 5-(-45)
So , V_{Particle 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 ?

 

[Saitama]

Let me try .
V_{Particle wrt image}=V_particle - V_image
V_{Particle wrt image}= 5-(-45)
So , V_{Particle 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?

 

[agoogler]

What is the correct answer?

The correct answer is 25 . Did you also got 50 as answer ?

 

[Saitama]

Did you also got 50 as answer ?

Yes. I don't see where we went wrong. We should wait for other members to reply.

 

[TSny]

I agree with you guys that the answer is 50 m/s.

 

[Saitama]

I agree with you guys that the answer is 50 m/s.

Thanks TSny! :)

 

[agoogler]

Thanks TSny! :)

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 ?

 

[phinds]

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 ?

I was basing my numbers on the MIRROR. Maybe the IMAGE needs a different calculation because it will be moving twice as fast as the mirror.

 

[TSny]

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.

 

[haruspex]

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.

 

[Saitama]

So what? A relative velocity is going to be the same viewed from any inertial frame.

Oops, sorry about that. I replied hastily there.