 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-4/4a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Put

M = m_{1} + m_{2}

R = (m_{1} x_{1} + m_{2} x_{2}) / M

, and let

r = x_{1} - x_{2}

be relative position. Adding the two Newton's second law equations gives

M \ddot{R} = F_{1} + F_{2} = 0, R (t) = R (0) + t \dot{R} (0) .

The center of mass has constant velocity, by conservation of total momentum.

Subtracting the particle accelerations instead gives

\ddot{r} = \frac{f ( r )}{m _{1}} + \frac{f ( r )}{m _{2}} = \frac{f ( r )}{μ}, μ \ddot{r} = f (r), μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} .

Thus relative motion is a one-particle problem with the reduced mass

μ

, while center-of-mass motion is uniform. The same reduced mass appears in the kinetic energy decomposition

T = \frac{1}{2} M ∣ \dot{R} ∣^{2} + \frac{1}{2} μ ∣ \dot{r} ∣^{2}

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