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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014/8/d/solution

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Past exam of the physics 1 paper of the University of Oxford / 2014 / 8 / d / Solution by

Codex  0 2026-10-25

Put

m_{A} = M = m

and

m_{B} = 2 m

. At reversal the transferred mass is

q_{r} = 0.45 m

, leaving A with

1.55 m

and, after capture of that exhaust, B with

2.45 m

. The first burn gives

v_{A r} = u_{0} ln \frac{2}{1.55} = u_{0} ln \frac{40}{31}, v_{B r} = - \frac{1.55}{2.45} v_{A r} = - \frac{31}{49} u_{0} ln \frac{40}{31} .

(1)

After reversal A emits its remaining exhaust away from B, so B receives no further fuel and eventually coasts at this velocity. Applying the rocket equation with reversed thrust to A gives

v_{A f} = v_{A r} - u_{0} ln 1.55 = u_{0} ln \frac{800}{961} ≃ - 0.18336 u_{0}, v_{B f} ≃ - 0.16126 u_{0} .

(2)

Both now move toward negative x, but A moves faster in that direction. The asymptotic closing speed is

v_{B f} - v_{A f} ≃ 0.0221 u_{0} > 0 : A can catch B.

(3)

This conclusion does not require a burn rate or an initial separation: after a finite burn, the positive eventual closing speed closes any finite remaining gap. Those additional inputs would be required to calculate the meeting time. Any pre-reversal exhaust still in transit affects the timing, not B's eventual momentum after collecting that fixed amount.

Figure 1.
Signed velocities during exhaust collection and thrust reversal
. Original Newtonian calculations versus the fraction of original fuel consumed. Left: the equal-speed event in the ideal negligible-flight-time collection model; the later dashed extension is conditional on collection and can violate ballistic interception. Right: reversing thrust at 45 percent produces a small positive eventual closing speed. **Details**.

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