# Math Help - Potential question

1. ## Potential question

Hello. We have a potential V= (x-3) / (x^2 + 16) And the question is if a particle is released at x=2 with initial velocity u find the values of u such that x(t) -> minus infinity as t-> infinity.

My attempt: I sketched the graph first of all. Then I was thinking that if we have a negative initial velocity then the particle moves to the left and to the right if we have a positive initial velocity. Then I said that if initial velocity is positive then 0<=E<1/16 (where 1/6 is the maximum turning point of V and E is the total energy) so the particle will move to the right reach the left turning point of V=E and then returns back so x(t) -> minus infinity as t->infinity as we wanted. Now if initial velocity is negative then we must have E>=0. Then we solve the inequalities w.r.t the speed v (which is always positive) since at x=2 E=0.5*v^2+V(2) and deduce from there the values of velocity (which can be negative or positive).

Am I thinking too complicated? Is there an easier way? My attempt was based on the sketch of V(x), which I couldn't upload, but the sketch is easy. [maximum at (8, 1/16) minimum at (-2, -1/4), crosses the axes at (3,0) and ( 0, -3/16), and also as x->infinity V-> 0 from above and as x-> minus infinity V->0 from below.]

Any help and guidance will be greatly appreciated. Thank you very much!

2. ## Re: Potential question

What is the mass of the particle?

I think your thoughts are generally on target. But I don't think you can answer the question fully without knowing the mass.

3. ## Re: Potential question

The particle has unit mass. That's why I told before at x=2 E=0.5*v^2+V(2). Is there another way to work this exercise? What would you recommend? Thank you again for your reply!

4. ## Re: Potential question

For the negative direction, it will go to negative infinity, period. For the positive direction, if it can "more than climb the hill", it will go to positive infinity. Otherwise, it will go to negative infinity. So you want to solve

$\frac{1}{2}\,u^{2}=\frac{1}{16}$

for $u.$ What will your final answer be?

5. ## Re: Potential question

So for negative direction (negative initial velocity) don't I have to put some limits for energy? I mean if initial velocity is <0 any total energy >=0 would do the job. At point x=2 the potential equal to -1/20. If total energy was -1/20<E<=0 then it would still bound the particle from going to minus infinity since the particle would move to the negative direction and would "collide" with the right turning point of V=E.

Also consider now going to the positive direction don't we need to solve 0<=0.5u^2 - (1/20) <= 1/16?

Moreover by solving the above inequality shouldn't we cross out negative velocities, since we were talking about moving to the positive direction, i.e positive velocity?

6. ## Re: Potential question

You're right. Only potential differences are physically meaningful. So you should set the energy equal to the potential difference that the particle would have to "climb" in order to go right.

7. ## Re: Potential question

So do I have to solve two inequalities for this exercise as I said? Thanks again! If you can find a solution too so we can cross check our answers I would be very thankful!

8. ## Re: Potential question

Well, the potential difference is going to be

$\Delta V=\frac{1}{16}-\left(-\frac{1}{20}\right)=\frac{1}{16}+\frac{1}{20}=9/80.$

Hence, we want to solve

$\frac{1}{2}\,u_{c}^{2}=\frac{9}{80}.$

Since you've assumed that $u>0,$ you can say that you want to solve the inequality $0

But then, the final solution will be $-\infty since all the negative $u$'s are included in your answer.

9. ## Re: Potential question

But this doesn't work! Let u=0.1 where 0<0.1<uc where uc is 3*sqrt(10) / 20. Then E= -0.045 and then E<V since at x=2 V was 1/16 and we need E>V for the particle to be able to move!

10. ## Re: Potential question

I think you might be confusing kinetic energy and total energy. The total energy can be negative, because of the arbitrariness of defining the zero-point of the potential energy. However, the kinetic energy is never negative, because the expression for kinetic energy is

$T=\frac{1}{2}\,m\,u^{2}=\frac{1}{2}\,u^{2}\ge 0,$ in your case. That is, the kinetic energy doesn't depend on the velocity. It depends on the speed (magnitude of the velocity). So that means you basically have to keep track of which direction u is pointed, relative to the potential curve you've got. Does that make sense?

If your velocity is 0.1, then the kinetic energy is positive 0.005. How can adding that to the negative -1/20 be less than -1/20? The answer, which is -0.045, as you've stated, is greater than -1/20 = -0.05.

11. ## Re: Potential question

I am still confused to tell you the truth, let me write again how I would have written the solution to guide me over my way or if you can explain me again the way on your previous post with uc.

Let u (vector) be the initial velocity and v=magnitude(u) i.e speed. Then total enerfy = 0.5*v^2-1/20
Now if a) u>0, we need 0<=E<1/16, b) if u<0 we need E>=0 and if u=0 then E=-1/20 and our condition is not satisfied.

a) 0<=0.5*v^2-1/20<1/16 leads to the solution sqrt(10)/10 <= v < 3* sqrt(10)/20, since v is speed and is always positive! But since we need positive velocity for this case then sqrt(10)/10 <= u (velocity) < 3* sqrt(10)/20
b) 0.5*v^2-1/20>=0 leads to the solution v>= sqrt(10) / 10 but we need negative velocity so u<= -sqrt(10)/10

Our final solution is u exists in [sqrt(10)/10, 3*sqrt(10)/20) or u<= -sqrt(10)/10

12. ## Re: Potential question

Lose the condition that E must be > 0. That simply doesn't apply. It's like I said before: the potential energy y translation is arbitrary. By defining the potential a certain way, I can easily get negative energies. Naturally, I can't get negative kinetic energy. Only changes in potential energy are physically meaningful. Think of your potential energy curve as hills (this is actually a pretty exact metaphor, since gravitational potential energy is directly proportional to the height). If your car is released from rest at the point of interest, what would it do? If it had some initial velocity to the right, what would it do?

Let u (vector) be the initial velocity and v=magnitude(u) i.e speed. Then
total enerfy = 0.5*v^2-1/20
Fine so far.

Now if a) u>0, we need 0<=E<1/16,
No, we don't. As above, the total energy could be negative a million, and it wouldn't change the physics. What we do need is for the total energy to be greater than the potential energy at any point. This might be where your confusion lies. Given that the kinetic energy is never negative, it will always be the case that the total energy is greater than or equal to the potential energy.

b) if u<0 we need E>=0
Again, we do not need E > 0.

and if u=0 then E=-1/20 and our condition is not satisfied.

a) 0<=0.5*v^2-1/20<1/16 leads to the solution sqrt(10)/10 <= v < 3* sqrt(10)/20, since v is speed and is always positive! But since we need positive velocity for this case then sqrt(10)/10 <= u (velocity) < 3* sqrt(10)/20
b) 0.5*v^2-1/20>=0 leads to the solution v>= sqrt(10) / 10 but we need negative velocity so u<= -sqrt(10)/10

Our final solution is u exists in [sqrt(10)/10, 3*sqrt(10)/20) or u<= -sqrt(10)/10
Does that help?

13. ## Re: Potential question

"No, we don't. As above, the total energy could be negative a million, and it wouldn't change the physics. What we do need is for the total energy to be greater than the potential energy at any point. This might be where your confusion lies. Given that the kinetic energy is never negative, it will always be the case that the total energy is greater than or equal to the potential energy"

Initial potential energy = -1/20. Total energy = 0.5*v^2-1/20. Let the velocity be positive and the total energy be <0 i.e the straight line will cut the potential graph at two points in the third quadrant. The particle would move with the positive initial velocity to the right hitting the right turning point of V=E then go to the left reach the left turning point of V=E and then the movement will be oscillatory between the two turning points of V=E so x(t) would not go to minus infinity as t goes to infinity!

To be exact potential energy could not be <-1/20 so I am talking about the case -1/20<E<0!

14. ## Re: Potential question

Originally Posted by Darkprince
Initial potential energy = -1/20. Total energy = 0.5*v^2-1/20. Let the velocity be positive and the total energy be <0 i.e the straight line will cut the potential graph at two points in the third quadrant.The particle would move with the positive initial velocity to the right hitting the right turning point of V=E then go to the left reach the left turning point of V=E and then the movement will be oscillatory between the two turning points of V=E so x(t) would not go to minus infinity as t goes to infinity!

To be exact potential energy could not be <-1/20 so I am talking about the case -1/20<E<0!
Hmm. I think I mistook the scale on my plot. I was thinking one tick mark = one unit, but it's one tick mark equals two units. Right. So, at x = 0, you could have negative total energy, which would, indeed, mean that the particle was trapped. That's certainly true if the initial kinetic energy is zero. But now, if you have enough initial kinetic energy such that the particle could reach the zero energy point (right at x = 3), then when it reaches that turning point and starts going in the negative direction, it will have just enough energy to escape the potential well and go to negative infinity. So you want to solve

$\frac{1}{2}\,u_{c}^{2}-\frac{1}{20}=0.$

If the initial velocity is to the right, with a speed less than $u_{c},$ then the particle will not go to negative infinity.

So here's the question: how much energy does the particle have to have if it's initially moving in the negative direction? Well, it would actually be the same speed $u_{c},$ because you must solve the same equation. So, if $|\mathbf{v}|=u,$ then the particle is trapped for $0 and it will go to negative infinity if it has an initial total energy equal to zero. However, if it has an initial positive velocity, and enough energy to get over that hump, then it will go to positive infinity. So you must also solve for

$\frac{1}{2}\,u_{d}^{2}-\frac{1}{20}=V(8)=\frac{1}{16}.$

If the initial velocity is to the right, and the speed is greater than $u_{d},$ then the particle will also not go to negative infinity.

I must admit to being lazy, and not checking if this answer is what you have or not.

$\mathbf{v}\in(-\infty,-u_{c})\cup(u_{c},u_{d}).$