Kinematics : Relation between the velocities of 3 particles

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

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Apologies for a bit hazy picture .Let the three particles be in a line after time 't' . If the inclined line (path of v) from the origin is perpendicular to the hypotenuse then we could write vt= v₁cos45°t = v₂cos45°t . But then it also means v₁ = v₂ .

Could someone help me with the problem .

 

[jbriggs444]

So you have three particles released from the origin simultaneously. One with a velocity v₁ along the x axis, a second with velocity v₂ along the y-axis and a third with a velocity v along the 45 degree diagonal in between.

We are not explicitly told that the particles are released from the origin simultaneously. That is an assumption on my part. I make it because the problem is pointless otherwise. All three velocities could be arbitrary.

We are told that the particles are eventually collinear at some time t and asked to find a relation between v₁, v₂ and v. I do not see a statement that the line connecting the three particles need be perpendicular to the path of the diagonally moving particle.

Question for you: If the particles are collinear at some time t (t not equal to 0), are they collinear at all other times t'?

[The point of asking this question is to allow us to dispense with velocities and reason directly about positions instead]

Since this is posed as a multiple choice question, an attractive approach is to start by eliminating the obviously incorrect alternatives.

 

[Jahnavi]

Question for you: If the particles are collinear at some time t (t not equal to 0), are they collinear at all other times t'?

Sorry . I can't think of a condition which either makes the particles collinear or not .

 

[jbriggs444]

Sorry . I can't think of a condition which either makes the particles collinear or not .

OK. Let's ask a simpler set of questions:
1. What are the positions of the three particles at time t?
2. What are the positions of the three particles at time t/2?

 

[Jahnavi]

By position you mean coordinates of the particles ?

 

[jbriggs444]

By position you mean coordinates of the particles ?

Yes.

 

[Jahnavi]

Let us call particle with velocity v₁ , A . Particle with velocity v , B .Particle with velocity v₂ , C .

A at time t = (v₁t,0)

B at time t = (vt/√2,vt/√2)

C at time t = (0,v₂t)

 

[jbriggs444]

Let us call particle with velocity v₁ , A . Particle with velocity v , B .Particle with velocity v₂ , C .

A at time t = (v₁t,0)

B at time t = (vt/√2,vt/√2)

C at time t = (0,v₂t)

A is correct for time t.
B is correct for time t.
C is correct for time t.

You have not answered for time t/2.

Edit: confused myself at first on B. You were correct.

 

[Jahnavi]

You have not answered for time t/2

A at time t/2= (v₁t/2,0)

B at time t = (vt/2√2,vt/2√2)

C at time t = (0,v₂t/2)

 

[jbriggs444]

A at time t/2= (v₁t/2,0)

B at time t = (vt/2√2,vt/2√2)

C at time t = (0,v₂t/2)

Now, a principle of analytic geometry (or linear algebra) is that if you take a geometric figure, multiply all of the coordinates of its vertices by a single fixed multiple and look at the resulting geometric figure, the two figures will be "similar". In particular, if you have three points on a line and multiply their coordinates by the same multiple, you'll have three points on a different, parallel line.

From this you should be able to conclude that if A, B and C are collinear at time t, they are collinear at time t/2 and, in fact, at all times.

 

[Jahnavi]

Thanks . I get the right answer

Since this is posed as a multiple choice question, an attractive approach is to start by eliminating the obviously incorrect alternatives.

Dimensionally all four options look okay . How do we find the obvious incorrect alternatives ?

 

[jbriggs444]

Thanks . I get the right answer
Dimensionally all four options look okay . How do we find the obvious incorrect alternatives ?

Choice 1: v = v1 + v2

If the X axis ball were at (1,0) and the Y axis ball were at (0,1) that would put the diagonal ball at (1,1). That's not a line and never will be a line.

Choice 2: v = $\sqrt{v_1 v_2}$

That means that v is the geometric mean of v1 and v2. So if v1=1 and v2=1 then v=1. But that means that (1,0), (0,1) and ($\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}$) would have to be all on the same line. But no, they're obviously on the same circle instead.

That sort of thing.