Time of closest approach between two particles

[Davidllerenav]

Homework Statement

Two objects $1$ and $2$ move at constant speeds $v_1$ and $v_2$ along of two mutually perpendicular lines. At the moment $t = 0$ the particles are located at distances $l_1$ and $l_2$ from the point of intersection of the lines. At what time will the two objects have a minimum distance? And
what is its expression?

Homework Equations

None

The Attempt at a Solution

I tried solving this using Galilean transformation. So I set the coordinate system on de first object, like this.

Then. I ended up with something like this.

After that, I calculated $\sin \theta = \frac{min}{d}$ thus $min=d* \frac{v_2}{\sqrt{v_1^{2}+v_2^{2}}}$. Am I right?

 

[BvU]

Only see [IMG} twice. The other (below) is unexeplained (what is v_2/1 ?).

I find it hard to believe d is not depnding on l₁ or l₂

 

[Davidllerenav]

Only see [IMG} twice.

Yes, sorry. That's why I put two links. This is the firts one.

The other (below) is unexeplained (what is v2/1 ?).

It is the velocity vector of the second object with respect to the first one.

 

[haruspex]

Homework Statement

Two objects $1$ and $2$ move at constant speeds $v_1$ and $v_2$ along of two mutually perpendicular lines. At the moment $t = 0$ the particles are located at distances $l_1$ and $l_2$ from the point of intersection of the lines. At what time will the two objects have a minimum distance? And
what is its expression?

Homework Equations

None

The Attempt at a Solution

I tried solving this using Galilean transformation. So I set the coordinate system on de first object, like this.

Then. I ended up with something like this.

After that, I calculated $\sin \theta = \frac{min}{d}$ thus $min=d* \frac{v_2}{\sqrt{v_1^{2}+v_2^{2}}}$. Am I right?

I am not able to figure out what your second diagram represents or how you arrive at it.
What is d? Only the given variables should appear in the answer, and it should be symmetric between the subscripts 1 and 2.

 

[haruspex]

It is the velocity vector of the second object with respect to the first one.

Then it points in the wrong direction. It looks more like their sum.

 

[Davidllerenav]

Then it points in the wrong direction. It looks more like their sum.

Yes, I know. In class we did something similar. We set the coordinate system on the first particle, so its velocity would be applied on the other particle, but with oposite direction.

 

[haruspex]

Yes, I know.

No matter, it can be fixed by making the v2 vector point down instead. But I still do not understand the diagram. Please reply to post #4.

Instead of a diagram using relative velocities, you could simply write down the equation for the vector of their relative positions and minimise its magnitude, but I guess you are trying to use a method you have been taught.

 

[Davidllerenav]

I am not able to figure out what your second diagram represents or how you arrive at it.
What is d? Only the given variables should appear in the answer, and it should be symmetric between the subscripts 1 and 2.

I arrived at it by setting the coordinate system on the first particle, so its velocity would be applied on the other particle, but with oposite direction. d is the distance between the particles at the beginning.

 

[Davidllerenav]

Instead of a diagram using relative velocities, you could simply write down the equation for the vector of their relative positions and minimise its magnitude, but I guess you are trying to use a method you have been taught.

How can I solve this usign relative velocity? And how can I solve it without using relative velocity?

 

[TSny]

@Davidllerenav , In your diagram in post #1 you have the direction of V₁ parallel to the initial separation d. I don't think that would generally be a valid assumption.

 

[haruspex]

@Davidllerenav , In your diagram in post #1 you have the direction of V₁ parallel to the initial separation d. I don't think that would generally be a valid assumption.

Yes, I think that was why I could not interpret it. There is a confusion between representations of displacements and representations of velocities.

 

[Davidllerenav]

@Davidllerenav , In your diagram in post #1 you have the direction of V₁ parallel to the initial separation d. I don't think that would generally be a valid assumption.

Sorry if I messed it up. How can I solve this correctly?

 

[haruspex]

Sorry if I messed it up. How can I solve this correctly?

You need to find the direction of the relative velocity vector in relation to that of the initial relative position vector.

 

[Davidllerenav]

You need to find the direction of the relative velocity vector in relation to that of the initial relative position vector.

I will end up with a triangle, right?

 

[haruspex]

I will end up with a triangle, right?

Yes, but the angle will depend on both the velocity ratio and the ratio of the initial distances.

 

[Davidllerenav]

Yes, but the angle will depend on both the velocity ratio and the ratio of the initial distances.

What do you mean by velocity ratio and ratio of the inicial distances?

 

[haruspex]

What do you mean by velocity ratio and ratio of the inicial distances?

In your earlier diagram you had tan(θ) as the ratio of the velocities, v1 and v2. If you forget the velocities and just draw a diagram of the initial positions you will see an angle whose tangent is the ratio of l1 to l2.

The diagram you need makes everything relative to one of the objects, object 1, say. There will be the initial vector position of object 2 relative to object 1, and the velocity vector of object 2 relative to object 1. The angle you need is the angle between these two vectors.

 

[Davidllerenav]

In your earlier diagram you had tan(θ) as the ratio of the velocities, v1 and v2. If you forget the velocities and just draw a diagram of the initial positions you will see an angle whose tangent is the ratio of l1 to l2.The diagram you need makes everything relative to one of the objects, object 1, say. There will be the initial vector position of object 2 relative to object 1, and the velocity vector of object 2 relative to object 1. The angle you need is the angle between these two vectors.

I tried doing it like this. I don't know if I'm right. In case I'm wrong, could you please show how to do it correctly?

 

[BvU]

I refuse to look at picures on a site that wants to cookie me and then deletes the picture after a while, so that the thread becomes worthless. PF should ban imgur and the like. If you can upload to them, you can upload directly to PF too.

 

[Davidllerenav]

I refuse to look at picures on a site that wants to cookie me and then deletes the picture after a while, so that the thread becomes worthless. PF should ban imgur and the like. If you can upload to them, you can upload directly to PF too.

Here it is.

 

[BvU]

Better

but mysterious to me.
The arrows are all over the place
What's 1 and 2 if $l_1$ and $l_2$ are at the curly brackets ?
and where are the angles @haruspex mentions ?

 

[Davidllerenav]

The arrows are all over the place
What's 1 and 2 if $l_1$ and $l_2$ are at the curly bracjeets ?
and where are the angles @haruspex mentions ?

1 and 2 are the object 1 and object 2.. $l_1$ and $l_2$ are the distances that the particles are at time $t=0$. I don't know which are the angles. How would the correct diagram be?

 

[TSny]

Your idea of switching to the rest frame of one of the particles is a nice approach.

I have particle 1 as moving up the y-axis, instead of particle 2. But, you can make the necessary adjustments. You should be able to get the distance of closest approach by simple geometric constructions and some trig. Note that the angle θ and the magnitude of v_2/1 are determined by v₁ and v₂.

 

[BvU]

Giving it away, eh ? But: beautiful graphics !

 

[TSny]

Giving it away, eh ? But: beautiful graphics !

Yeah, when the number of posts gets fairly large (over 20 now), then I tend to give in some. Hope I'm not violating PF rules. But thanks for the comments.

 

[BvU]

I agree. @Davidllerenav, can you now see what previous posts from helpers were aiming at ?
And, probably more important to you: has the penny dropped ?

 

[Davidllerenav]

YYou should be able to get the distance of closest approach by simple geometric constructions and some trig. Note that the angle θ and the magnitude of v_2/1 are determined by v₁ and v₂.

Ok, so the distance between the two objects would be the magnitude of the vector from 1 to 2, right?

 

[BvU]

At t=0, yes. What about later times ?

 

[Davidllerenav]

And I find the magnitude like this: $|\vec r| = \sqrt{(\vec r_1)^2 +(\vec r_2)^2}$, and each vector would be equal to: $\vec r_1 = (-l_1 + v_1*t)\vec j$ and $\vec r_2 = (-l_2 + v_2*t)\vec i$, right?

 

[TSny]

Ok, so the distance between the two objects would be the magnitude of the vector from 1 to 2, right?

Yes. But you are only looking for the minimum distance between them. In the figure on the right in post #23, can you draw the line of motion of 2 in this frame of reference? Can you identify the point on the line of motion that corresponds to the minimum distance between them?

 

[Davidllerenav]

Yes. But you are only looking for the minimum distance between them. In the figure on the right in post #23, can you draw the line of motion of 2 in this frame of reference? Can you identify the point on the line of motion that corresponds to the minimum distance between them?

It would be a line that goes throught the velocity vector. I think that the point that would corresponds to the minimum distance would be when the point 1 and the line of motion of 2 make a prepedicular line, right?

 

[TSny]

It would be a line that goes throught the velocity vector. I think that the pint that would correspond to the minimum distnace would be when the point 1 and line of motion of 2 make a prepedicular line, right?

Yes, if I'm interpreting what you said correctly.

 

[Davidllerenav]

Yes, if I'm interpreting what you said correctly.

Ok. so I have to make a triangle, right? But what would be the sides? One would be unknown since it is the minimum distance. What about the other two? I think that one would be the distance at time $t=0$ and the other the line of motion of 2.

 

[TSny]

Ok. so I have to make a triangle, right? But what would be the sides? One would be unknown since it is the minimum distance. What about the other two? I think that one would be the distance at time $t=0$ and the other the line of motion of 2.

I don't believe that the line segment representing the initial distance between the objects will be of use. If you've drawn the trajectory of 2 and the line segment representing d_min, you might see two right triangles that you can work with.

 

[BvU]

when the point 1 and the line of motion of 2 make a prepedicular line

show us in the drawing ...