 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019/2/solution

Let

w = v_{1} - v_{2}

V = (M_{1} v_{1} + M_{2} v_{2}) / (M_{1} + M_{2})

v_{1} = V + \frac{M _{2}}{M _{1} + M _{2}} w, v_{2} = V - \frac{M _{1}}{M _{1} + M _{2}} w .

Substitute into the sum of the two kinetic energies. The cross terms cancel, while the coefficient of

∣ w ∣^{2} /2

μ = M_{1} M_{2} / (M_{1} + M_{2})

. Hence

T = \frac{1}{2} (M_{1} + M_{2}) ∣ V ∣^{2} + \frac{1}{2} μ ∣ w ∣^{2} .

The first term is translation of the centre of mass; the second is the kinetic energy measured in the centre-of-momentum frame. For an isolated perfectly inelastic collision, the bodies finish with a common velocity, while their total momentum and thus

V

are unchanged. The kinetic energy converted to internal forms is therefore

T_{i} - T_{f} = \frac{1}{2} μ v^{2},

where

v

is their initial relative speed. Total energy remains conserved; it is the translational-plus-relative kinetic energy that decreases.

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