# Three mass on the vertices of an equilateral triangle, one given an initial velocity.

1. Oct 16, 2009

### Nanyang

Hi everyone, this is a problem I posted here a month ago but it wasn't given any attention from the helpers here. I still cannot solve it. So I will greatly appreciate any help.

1. The problem statement, all variables and given/known data
Three small balls of masses m, 2m and 3m are placed on a smooth horizontal surface so that they lie on the vertices of an equilateral triangle. The masses m and 2m, as well as 2m and 3m, are connected by light inextensible strings. Initially the strings are taut and the mass m is given a velocity v on the surface in a direction parallel to the string connecting 3m and 2m. With what velocity does mass 3m eventually starts moving? (Neglect friction and the mass of the strings.)

Refer to http://www.feynmanlectures.info/exercises/three_balls.html for the diagram.

2. Relevant equations
I guess it would be the conservation of momentum.

3. The attempt at a solution
I wrote down the equation for momentum conservation

mv = mv1 + 2mv2 + 3mv3

Anyway I don't even understand the problem in the first place.

So I am here to seek help, again.

Last edited: Oct 16, 2009
2. Oct 16, 2009

### tiny-tim

Hi Nanyang!

The mass m will move in the same direction until the string is again at 60º.

Take moments about that point, before and after (so that will eliminate mass m completely). Also use conservation of energy.

3. Oct 18, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi tiny-tim

Here's my attempt:

The tension in string connecting m and 2m both will have no contribution to the moment as the perpendicular distance is zero. The tension in that connecting 2m and 3m both will also give zero net moment because the forces are equal and opposite and the perpendicular distances from 3m and 2m to the 'pivot' is the same. So before and after the moments are zero...and that would imply the conservation of angular momentum about that point? Linear momentum must be conserved and also the total kinetic energy of the system. So somehow I will have 3 equations with 3 unknowns for the velocities of each mass right after the mass m reaches 60 deg to the horizontal. But the problem is: how do I write out the equation for the conservation of angular momentum? Maybe I'm already wrong from the start.

4. Oct 18, 2009

### tiny-tim

Hi Nanyang!
I don't understand what you're finding difficult.

Write out what you think the equations for momentum and angular momentum are.

5. Oct 18, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi tiny -tim.

I'm not sure where the velocities of 3m and 2m might be pointing at, so I can't figure out which distance to multiply to the momentum to give the angular momentum.

My guess is that the initial velocity for 3m will be horizontal. But for 2m, my intuition tells me it's diagonally up to the left, at some angle I'm not sure of.

Last edited: Oct 18, 2009
6. Oct 18, 2009

### tiny-tim

Yes, good ol' Newton's second law (F = ma) tells us that a for the 3m will be along the string, since the only force on it is the tension along the string.

For the 2m, of course, there are two tensions, in different directions, so that doesn't help much. Instead, call the velocity vx "horizontal", and vy "vertical".

What equations do you get now?

7. Oct 18, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hmm...

If the velocity of 3m is v3, is it (3v3/2)$$\sqrt{3}$$ + vx$$\sqrt{3}$$ = vy ??

And for linear momentum it will be v = v1 +2vx +2vy +3v3

For kinetic energy it would be v² = v1² +2vx² +2vy² +3v3²

And I think v1² = (v -3v3 -2vx)² + (2vy

8. Oct 18, 2009

### tiny-tim

No, that should be two equations, one for each direction, shouldn't it?

Also, remember that v - v1 will be along the string joining m to 2m (for the same reason that v3 is "horizontal").

9. Oct 19, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi tiny-tim :),

I noticed that since v -v1 is along the string connecting m and 2m, then 2vx +2vy +3v3 is also along the string. So The vertical component divided by the horizontal component will give tan(60 deg)= sqrt(3).

2vy/(2vx +3v3) = sqrt(3) => vy = 3sqrt(3)/2 v3 + sqrt(3) vx which is the equation I got by analyzing the angular momentum.

For the momentum in the x direction it should be

v = 2vx +3v3 +ux -(1)

where ux is the component of v1 in the x direction and similarly for uy. For that in the y direction, I wrote

2vy + uy = 0 -(2)

I substitute in the equation I got from angular momentum to become

3v3 + 2vx + uy/sqrt(3)

I minus uy from both sides so

v-v1 = -(1+ 1/sqrt(3)) uy

But the LHS is pointing along the string and the RHS is definitely vertical, which means that uy = 0?? Then vy = 0 because of (2).

ux = v? (because I substitue vy = 0 into the angular momentum equation and then rearrange to substitue vx into (1))

Then eventually using the equation for kinetic energy, I get 2v2² + 3v3² = 0. Then sustituting the result from the equation for the angular momentum which is 3v3 +2v2 = 0, I get v3 = v2 = 0??

Which makes sense because the initial velocity has to be zero because 3m and 2m was at rest? I just can't get v3 = 2v/15.

10. Oct 19, 2009

### tiny-tim

Hi Nanyang!

mmm … this is what happens if you don't distinguish between ordinary equations and vector equations

your first equation was v = v1 + (2vx+3v3 , 2vy)

your second equation, 3v3 + 2vx + uy/√3 = 0, is not a vector equation …

if you want to substitute it into the first equation, you'll get v = v1 + (-uy/√3 , 2vy)
Yes, you're right … I hadn't noticed that … there's only a limited amount of information to be got from the various methods, and so some methods give the same equation!

11. Oct 19, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Ah I see...

So the equations would be:

(a) (3√3)/2 v3 + √3 vx = vy

(b) v= ux +2vx +3v3 $$\Rightarrow$$ ux= v -2vx -3v3

(c) 0 = uy +2vy $$\Rightarrow$$ uy² = 4vy²

(d) v² = ux² +uy² +2vx² +2vy² + 3v3²

I start by putting (c) into (d) to become (e). Then (a) into (e) to become (f) which is

v² = ux² +2vx² +6[(3√3)/2 v3 + √3 vx]² +3v3²

For the ux in (f) I substitute in (b) to become (g). Eventually after much hard work (forgive me for not typing out exactly my steps as it is rather tedious!) I got (g) to be:

105/2 v3² -6vv3 +66v3vx -4vvx +24vx² = 0

Now the obstacle seems to be finding a relationship between v3 and vx or vx with v. :zzz:

I hope I'm on the right path...

Last edited: Oct 19, 2009
12. Oct 19, 2009

### tiny-tim

Looks good so far!

ok, we've both missed something fairly obvious

the distances between 3m and 2m, and between 2m and m, remain constant, so v3 - v2 must be perpendicular to x3 - x2, and v2 - v1 must be perpendicular to x2 - x1

(and I'm wondering whether conservation of energy is right in this case )

13. Oct 19, 2009

### Nanyang

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

I don't quite understand why they must be perpendicular. Even if they are perpendicular, in what way should I use the information?

Anyway, the energy can't go into the strings, can it? Furthermore it was assumed that the masses have no internal structure, so the energy can't be loss to generating heat. I think.

14. Oct 19, 2009

### tiny-tim

The distance is the square-root of (x2 - x1)2,

which is (x2 - x1).(x2 - x1)

Since it is constant, its derivative must be zero …

2(x2 - x1).(v2 - v1) = 0

in other words, (x2 - x1) is perpendicular to (v2 - v1)

You use it, for example, by saying that vx = v3.
Energy usually isn't conserved in collisions …

generally you shouldn't assume it is unless something in the question says so.

Imagine just two balls on a string, the string starts loose and one ball moves away until the string is tight, when the other also starts moving …

this is exactly the same as two balls colliding without a string, and sticking together

as you know, if they stick together (sometimes called a perfectly inelastic collision), energy is not conserved, and instead we have the equation v1 = v2.

I was confused at first, because I hadn't seen a question like this before , but it's now clear that the two equations for conservation of length (!) replace conservation of energy.

15. May 21, 2010

### SiliconGalley

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi there!

I've hit upon the same physics problem from http://www.feynmanlectures.info/exercises.html and I'm on the case at the moment. Just wanted to ask whether Nanyang and tiny-tim have finally been able to get the v3=2v/15 with their methods and/or if someone could summarize the solution process.

Greetings

Galley

16. May 21, 2010

### tiny-tim

Welcome to PF!

Hi SiliconGalley! Welcome to PF!

On this forum, you have have to show us what you've done, and then we'll comment.

17. May 21, 2010

### SiliconGalley

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi tiny-tim!

That sounds quite reasonable. :-) In this case, I'll check the former discussion in detail and report my insights at a later time.

If you could rate the difficulty of this problem on a 0-10 scale, what would it be (say, for a physics undergraduate at his very beginning)? It seems to be quite puzzling and underestimated - but rather basic on the other hand.

Greetings

Galley

18. May 21, 2010

### tiny-tim

maybe 5

19. May 22, 2010

### SiliconGalley

Re: Three mass on the vertices of an equilateral triangle, one given an initial veloc

Hi there,

m moves with v till the angle is 60° again - but on the other side (in other words the former angle is 120°). The other two, 2m and 3m are at rest before that. Then, the string is taut again and m moves still in direction of the initial v, 2m and 3m begin to move slightly.
To describe this mathematically, I put the situation before and after reaching the 60° angle (the second time, now on the other side) into the equation of conservation of momentum:

in x direction after getting rid of m:
(1) v = v1(x) + 2v2(x) + 3v3 (v3 only horizontal)

in y direction after getting rid of m:
(2) 0 = v1(y) + 2v2(y)

Due to the equation for kinetic energy after getting rid of the masses:
(3) v^2 = v1(x)^2 + v1(y)^2 + 2v2(x)^2 + 2v2(y)^2 + 3v3^2

What I also know: At the moment when 2m and 3m start to move, the angle is 60° again, so:
v2(y) / v2(x) = tan(60°) = sqrt(3)

And, somehow, you made use of the angular momentum (I'm quite ignorant of that, all I know is the definition L = r x mv I could look up and I've a rough layman's idea of what it is. How can it be used in this context, where is an angular momentum in this situation? I can't see it.) Is it necessary to use it?

Greetings

Galley