PHY.6:5.1 - Derive coupled motion and identify where mechanical energy goes

Two bodies move along one line. Their masses are m1 and m2. Displacements x1 and x2 are measured from a configuration in which their connecting spring is unstretched, so its extension is x2-x1. Velocities are v1 and v2. A spring and a viscous damper act between them; the connector’s inertia is neglected in the chosen regime.

The momentum balances need the connecting force. Use the ideal response

f = k*(x2-x1) + c*(v2-v1),

where k>0 is stiffness and c>=0 is damping. The force on body 1 is f; that on body 2 is -f.

When a response parameter is missing. Suppose this linear response form is already justified, but its parameters are not yet supplied. A static test with extension 1 m, zero relative velocity and force 3 N determines k=3 N/m; it leaves c undetermined. At extension 1 m and relative velocity -2 m/s, the choices c=0 and c=0.5 N*s/m predict 3 N and 2 N. Both reproduce the static test. A prediction of either body’s acceleration therefore remains conditional on c; a supplied value or useful bound can settle a stronger question. A question about total momentum change can already be answered from the external forces, independently of c. The single test does not establish the assumed linear form.

With the response parameters and external forces u1 and u2 supplied, the connected evolution is

x1' = v1; x2' = v2; m1*v1' = f+u1; m2*v2' = -f+u2.

Adding the momentum balances gives p' = u1+u2 for p=m1*v1+m2*v2. This removes the internal force from the total momentum change while retaining it in the relative motion.

The retained mechanical energy is

H = (m1*v1^2 + m2*v2^2 + k*(x2-x1)^2)/2.

Differentiating and substituting the evolution gives

H' = u1*v1 + u2*v2 - c*(v2-v1)^2.

For the ideal damper whose lost mechanical energy becomes internal energy U, add U' = c*(v2-v1)^2. Then (H+U)' equals the external mechanical power. Omitting U from the motion calculation assumes that its change does not appreciably alter the chosen mechanical response.

Take m1=1 kg, m2=2 kg, k=3 N/m, c=0.5 N*s/m, x1=0 m, x2=1 m, v1=1 m/s, v2=-1 m/s, and no external force. At that instant, f=2 N, the accelerations are 2 m/s^2 and -1 m/s^2, total momentum has zero rate of change, and mechanical energy decreases at 2 W. These are useful initial consequences without computing a full trajectory.

Changed preparation. Body 2 is instead held at x2=1 m from before the initial instant, so v2=0. With the other initial values unchanged, f=2.5 N. The support supplies the reaction u2=f; body 1 accelerates at 2.5 m/s^2. The selected pair now exchanges momentum with the support. Keeping the earlier v2=-1 m/s together with a fixed-position constraint would describe an inconsistent preparation. Suddenly clamping the moving body would be another physical problem, requiring an account of that transition.

The reusable move is to obtain motion by joining balances to a response and preparation, and to revise the relevant exchange when the connection changes.