Ray optics: velocity of image reflected off a moving mirror

[Zayan]

Homework Statement

The mirror velocity is 2 î m/s, object velocity is 4 î m/s. We have to find the velocity of the image.

Relevant Equations

Vm= (Vo + Vi)/2.
Here Vm= Vel. of mirror, Vo= Vel. of object, Vi= Vel. of image. All in ground reference.

Using equation, Vm=(Vo+Vi)/2;
2= (4+Vi)/2
=>Vi=0.
I.e, image velocity is zero.
I can't understand how it is possible that the image is at rest wrt ground. If the object moves with more velocity than the mirror, wouldn't it collide with the mirror at some point? And plus the object is moving 2m/s wrt the mirror so how is the image at rest wouldn't it move with 2m/s in the opposite direction?

 

[kuruman]

You are looking for the instantaneous velocity of the image when the object is at distance $d_o$ from the mirror. What happens in the long run is not an issue.

Secondly, where did you get the equation Vm=(Vo+Vi)/2 ? Why is it relevant to this case?

 

[Zayan]

You are looking for the instantaneous velocity of the image when the object is at distance $d_o$ from the mirror. What happens in the long run is not an issue.

Secondly, where did you get the equation Vm=(Vo+Vi)/2 ? Why is it relevant to this case?

My prof. derived it using relative velocities and it's also in my physics book. It's relevant because two velocities are given so we can easily get the third using this relation.

 

[kuruman]

. . . so how is the image at rest wouldn't it move with 2m/s in the opposite direction?

Having a formula that gives you a number does not necessarily mean that you understand the relative motion of mirrors and images.

Imagine being an observer at rest with respect the mirror, i.e. moving to the right at 2 m/s relative to the ground. You see the object to your left coming towards you at 2 m/s and the image to your right also coming towards you at 2 m/s. You would say that the mirror has velocity $\mathbf{v}_m=0~\mathbf{\hat i}~$m/s, that the object has velocity $\mathbf{v}_o=2~\mathbf{\hat i}~$m/s and that the image has velocity $\mathbf{v}_i=-2~\mathbf{\hat i}~$m/s.

Now imagine that you are at rest relative to the ground. To each of the velocities above you would have to add the mirror's velocity relative to the ground. What do you get?

 

[Zayan]

Having a formula that gives you a number does not necessarily mean that you understand the relative motion of mirrors and images.

Imagine being an observer at rest with respect the mirror, i.e. moving to the right at 2 m/s relative to the ground. You see the object to your left coming towards you at 2 m/s and the image to your right also coming towards you at 2 m/s. You would say that the mirror has velocity $\mathbf{v}_m=0~\mathbf{\hat i}~$m/s, that the object has velocity $\mathbf{v}_o=2~\mathbf{\hat i}~$m/s and that the image has velocity $\mathbf{v}_i=-2~\mathbf{\hat i}~$m/s.

Now imagine that you are at rest relative to the ground. To each of the velocities above you would have to add the mirror's velocity relative to the ground. What do you get?

Thank you. I got a lot confused by using relative velocities. I imagined it to be negative 2m/s as I mixed ground frame with mirror's frame.

 

[Lnewqban]

I got a lot confused by using relative velocities.

In future cases, you could convert the given velocities to distances, in order to make the problem less confusing.
Just multiply each velocity by a common period of time.

The thing here is that a flat mirror acts like a plane of symmetry between object and virtual image.
The virtual distance between you and your mirror image is always double the actual distance between you and the surface of the mirror, regardless any relative movement in the direction perpendicular to the mirror's surface.

That is equivalent to state that if you walk away from a flat mirror (anchored at a fixed location), it will look to you that your image is walking away from you at double your speed rate.