# Energy Conservation: Determining Forces on Particles

• cleggy
In summary, to determine the forces acting on particles in an isolated system, one can use energy conservation and the energy function E= 1/2m1(v1)^2 + 1/2m2(v2)^2 - (k/r^2), where k is a positive constant and r is the magnitude of the separation vector. This can be differentiated to get dE/dt = 0, which leads to F= [-2k/mod(x1-x2)^3](v1-v2) according to the superposition principle and the chain rule. This expression contains both kinetic and potential energy terms.

#### cleggy

1. I have to use energy conservation to determine the forces acting on the particles.

2. An isolated system consists of two particles of masses m1 and m2, whose position vectors in an inertial frame are x1 and x2 and velocity vectors are v1 and v2.

The interaction of the particles can be described by the energy function :

E= 1/2m1(v1)^2 + 1/2m2(v2)^2 - (k/r^2)

k is a positive constant
r = mod(x1-x2) and is the magnitude of the separation vector.

3. Do I have to differentiate E to get 1/2m1(a1)^2 + 1/2m2(a2)^2 - (k/r^2) and equate it to zero?

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The units of the first two expressions involving the mass are not energy but momentum. Check the expression again.

Ooops. Thanks chrisk for pointing that out.

Indeed the velocity vectors should each be squared.

Differentiating with respect to time does lead to

dE/dt = 0

because the total energy of the system is constant. Take into account that r is a function of time.

So dE/dt = m1a1 + m2a2 +2k/mod(v1-v2)^3 ?

Would I be right in saying that due to superposition principle then the forces on the particles is F = -2k/mod(v1-v2)^3 ?

Check how you differentiated. Use the chain rule.

Ah.

So dE/dt = m1a1 +m2a2 + [2k/mod(x1-x2)^3](v1 -v2) = 0

Recall that

F=-dU/dx

when F is conservative. The given expression contains kinetic and potential energy (U) terms.

Then F= [-2k/mod(x1-x2)^3](v1-v2)