How to Find Particle Acceleration Using Conservation of Energy?

[Ron_Gis]

Help

An isolated system of two identical particles, each of mass m, is described by an energy function
E = ½ m |v1|^2 + ½ m |v2|^2 + ½ k |x1 – x2|^2
How can I use conservation of energy to find the acceleration of each particle in terms of the position of the particle? I would be thankful if someone could give me a hint ?

Ron

 

[krab]

Well, the derivative of E with respect to time is zero, so...

 

[Ron_Gis]

Well, the derivative of E with respect to time is zero, so...

Hello krab,

thank you for your help. Your approch to the problem is correct. If the total energy is conserved, the first derivative with respect to time of the given equation should be zero. I have some problems with the maths. Nevertheless, I think that the first derivative is given by

m a1 dot v1 + m a2 dot v2 + 0 where a1 = dv1/dt and a2 = dv2/dt

but I am not sure about this. Furthermore, it is not possible to divide a scalar through a vector, even if the solution is correct. I am too stupid to find the correct expression for the accelerations a1 and a2. I would be thankful if you could help me again.

Ron

 

[krab]

, I think that the first derivative is given by

m a1 dot v1 + m a2 dot v2 + 0 where a1 = dv1/dt and a2 = dv2/dt

but I am not sure about this.

The first part is correct, but x1 and x2 change with time, so why did you get that the time derivative of ½ k |x1 ? x2|^2 is zero?

 

[Ron_Gis]

The first part is correct, but x1 and x2 change with time, so why did you get that the time derivative of ½ k |x1 ? x2|^2 is zero?

Hello krab,

you are right again. The first derivative is given by

m a1 dot v1 + m a2 dot v2 + k (x1 – x2) dot (v1 – v2)

and this expression is equal to zero.

Nevertheless, the dot product is a scalar and I have to find an expression for the accelerations, which are vectors. I think that this is not possible, simply by rearranging this equation. Would you be kind enough to give me another hint.

Regards Ron

 

[krab]

The first derivative is given by

m a1 dot v1 + m a2 dot v2 + k (x1 ? x2) dot (v1 ? v2)

and this expression is equal to zero.

Nevertheless, the dot product is a scalar and I have to find an expression for the accelerations, which are vectors. I think that this is not possible, simply by rearranging this equation. Would you be kind enough to give me another hint.

Regards Ron

You are right. You cannot get it from just this equation. Are you allowed to use the fact that in this isolated system, total momentum is conserved? If so then v1+v2=constant, so a1=-a2 and the result follows.

Alternatively, go back to the beginning and use a "virtual displacement". Pretend that you move say particle 1 by a vector dx1. Use the energy equation to calculate the difference in energy this displaced system would have. From the definition of work, you know that this energy difference is F1 dot dx1.

 

[Ron_Gis]

Hello krab,

I think that I have found the correct solution. Thank you for your help.

Regards Ron