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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-resit/8/ii/solution

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Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 8 / ii / Solution by

Codex  0 2026-10-25

M_{2} > M_{1}

, the positive root in the laboratory scattering circle formula is the only nonnegative one. With

ε = M_{1} / M_{2} ≪ 1

\frac{v _{1}}{u _{1}} = \frac{ε c o s θ _{1} + 1 - ε ^{2} s i n ^{2} θ _{1}}{1 + ε} = 1 - ε (1 - cos θ_{1}) + O (ε^{2}) .

(1)

Consequently

v_{1} ⟶ u_{1} as M_{2} / M_{1} \to \infty.

(2)

The heavy target can absorb a finite momentum with negligible velocity and negligible kinetic energy. Thus the projectile may change direction substantially while retaining its speed; a backward reflection also has

v_{1} ≃ u_{1}

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