Answer to Question #83309 in Classical Mechanics for yosef

Question

An H2 molecule can be formed in a collision that involves three hydrogen atoms. Suppose that before such a collision, each of the three atoms has speed 2000 m/s , and they are approaching at 120 ∘ angles so that at any instant, the atoms lie at the corners of an equilateral triangle. Find the speeds of the H2 molecule and of the single hydrogen atom that remains after the collision. The binding energy of H2 is Δ=7.23×10−19J, and the mass of the hydrogen atom is 1.67×10−27kg.
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Answer 1

In the process of collision, the total momentum of the system is conserved. The initial total momentum is equal to zero by symmetry. After the collision, the remaining hydrogen atom and the H2 molecule are receding from the common center of mass in opposite directions with velocities v1 and v2, respectively. Their masses are m1 = 1.67 × 10−27 kg and m2 = 2m1 – Δ/c2, where c is the speed of light, and the last term is the relativistic correction to the H2 mass. Since Δ/c2 = 7.23 × 10−19 J/(299 792 458 m/s)2 = 8 × 10−36 kg is much smaller than the mass of hydrogen atom, this relativistic correction can be neglected; hence, we can take m2 = 2m1 with high precision. The initial speed v0 = 2000 m/s is also much smaller than the speed of light, so we can use non-relativistic expressions for energy and momentum. The conservation of total zero momentum then implies v2 = v1/2. It remains to find v1. The initial kinetic energy of the system is equal to E_0=3m_1 (v_0^2)⁄2; the final kinetic energy after the collision is then, by conservation of energy, E_1=E_0+Δ=3m_1 (v_0^2)⁄2+Δ=2m_1 (v_2^2)⁄2+m_1 (v_1^2)⁄2=3m_1 (v_1^2)⁄4. Hence, v_1^2=2v_0^2+4 Δ⁄((3m_1 ) ), and v_1=√(2v_0^2+4 Δ⁄((3m_1 ) )). Substituting here the numerical values, we obtain v1 = 24192 m/s and v2 = v1/2 = 12096 m/s.