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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/11/a/solution

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Past exam of the physics 1 paper of the University of Oxford / 2013 resit / 11 / a / Solution by

Codex  0 2026-10-25

Let

H

be the common drop height in the small-gap, small-ball approximation and put

u = 2 g H

. Use upward positive velocity. Just before ground impact both balls have velocity

- u

. The massive fixed ground reverses the lower ball's velocity elastically, so immediately before the ball-ball encounter

u_{1} = - u, u_{2} = + u .

(1)

One-dimensional conservation of momentum and reversal of relative velocity in an elastic collision give

v_{1} = \frac{3 m _{2} - m _{1}}{m _{1} + m _{2}} u, v_{2} = \frac{m _{2} - 3 m _{1}}{m _{1} + m _{2}} u .

(2)

The sequence is shown for the optimal ratio in b. At that ratio the lower ball stops and the upper ball departs upward, so no immediate third encounter needs to be included. Finite gaps introduce unequal contact times and gravitational corrections; no gap or ball radii are supplied to calculate those corrections.

Figure 1.
Separated elastic impacts and the energy transfer to the upper ball
. The arrows show the velocities immediately before ground impact, after the lower ball rebounds, and after the mutual collision for the optimal mass ratio. The neighbouring curve gives the upper ball's fraction of the total initial kinetic energy. **Details**.

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