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ランダウ=リフシッツ『力学』 Landau & Lifshitz Mechanics §16 問題 Problem1 解答solution

V + v_{10} cos {\theta}_0 = v_{10} sin {\theta}_0 cot {\theta}_{1} (*)
V - v_{20} cos {\theta}_0 = v_{20} sin {\theta}_0 cot {\theta}_{2} (**)

(*),(**)より 
 cos {\theta}_0 = \frac{V(v_{10} cot {\theta}_{1} - v_{20} cot {\theta}_{2})}{v_{10} v_{20}(cot {\theta}_{1} + cot {\theta}_{2})}
 sin {\theta}_0 = \frac{V(v_{10}+ v_{20} )}{v_{10} v_{20}(cot {\theta}_{1} + cot {\theta}_{2})}
 
 sin ^2 {\theta}_0 + cos ^2 {\theta}_0 = 1に代入して,
( \frac{V(v_{10}+ v_{20} )}{v_{10} v_{20}(cot {\theta}_{1} + cot {\theta}_{2})}) ^2+ (\frac{V(v_{10} cot {\theta}_{1} - v_{20} cot {\theta}_{2})}{v_{10} v_{20}(cot {\theta}_{1} + cot {\theta}_{2})}) ^2   = 1
 (v_{10}+ v_{20} ) ^2 + (v_{10} cot {\theta}_{1} - v_{20} cot {\theta}_{2}) ^2 = \frac{v_{10} ^2 v_{20} ^2(cot {\theta}_{1} + cot {\theta}_{2}) ^2}{V ^2}

両辺に  sin ^2 {\theta}_1 sin ^2 {\theta}_2をかけると,
 (v_{10}+ v_{20} ) ^2 sin ^2 {\theta}_1 sin ^2 {\theta}_2+ (v_{10} cos {\theta}_1 sin {\theta}_2- v_{20} sin {\theta}_1 cos {\theta}_2) ^2 = \frac{v_{10} ^2 v_{20} ^2(cos {\theta}_{1} sin {\theta}_2+ sin {\theta}_1cos {\theta}_{2}) ^2}{V ^2}
 (v_{10}+ v_{20} ) ^2 sin ^2 {\theta}_1 sin ^2 {\theta}_2+ (v_{10} cos {\theta}_1 sin {\theta}_2- v_{20} sin {\theta}_1 cos {\theta}_2) ^2 = \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}

左辺第2項を展開して,
 (v_{10}+ v_{20} ) ^2 sin ^2 {\theta}_1 sin ^2 {\theta}_2+ v_{10} ^2 cos ^2 {\theta}_1 sin ^2 {\theta}_2 + v_{20} ^2 sin ^2 {\theta}_1 cos ^2 {\theta}_2 -2v_{10} v_{20} sin {\theta}_1 sin {\theta}_2 cos {\theta}_1 cos{\theta}_2= \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}
 (v_{10}+ v_{20} ) ^2 sin ^2 {\theta}_1 sin ^2 {\theta}_2+ v_{10} ^2 (1-sin ^2 {\theta}_1)sin ^2 {\theta}_2 + v_{20} ^2 sin ^2 {\theta}_1 (1- sin ^2{\theta}_2) -2v_{10} v_{20} sin {\theta}_1 sin {\theta}_2 cos {\theta}_1 cos{\theta}_2= \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}
 2v_{10}v_{20}  sin ^2 {\theta}_1 sin ^2 {\theta}_2+ v_{10} ^2 sin ^2 {\theta}_2 + v_{20} ^2 sin ^2 {\theta}_1  -2v_{10} v_{20} sin {\theta}_1 sin {\theta}_2 cos {\theta}_1 cos{\theta}_2= \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}
  v_{10} ^2 sin ^2 {\theta}_2 + v_{20} ^2 sin ^2 {\theta}_1  -2v_{10} v_{20} sin {\theta}_1 sin {\theta}_2( cos {\theta}_1 cos{\theta}_2- sin {\theta}_1 sin {\theta}_2)= \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}
  v_{10} ^2 sin ^2 {\theta}_2 + v_{20} ^2 sin ^2 {\theta}_1  -2v_{10} v_{20} sin {\theta}_1 sin {\theta}_2 cos ( {\theta}_1 + {\theta}_2)= \frac{v_{10} ^2 v_{20} ^2 sin ^2 ({\theta}_1 + {\theta}_2)}{V ^2}

両辺を, v_{10}v_{20}で割って, v_{10} = p_0/m_1,v_{20} = p_0/m_2を代入すると,   \frac{m_2}{m_1} sin ^2 {\theta}_2 + \frac{m_1}{m_2} sin ^2 {\theta}_1  -2sin {\theta}_1 sin {\theta}_2 cos ( {\theta}_1 + {\theta}_2)= \frac{p_0 ^2  sin ^2 ({\theta}_1 + {\theta}_2)}{m_1 m_2V ^2}

  \frac{m_2}{m_1} sin ^2 {\theta}_2 + \frac{m_1}{m_2} sin ^2 {\theta}_1  -2sin {\theta}_1 sin {\theta}_2 cos ( {\theta}_1 + {\theta}_2)= p_0 ^2(\frac{1}{m_1} + \frac{1}{m_2})\frac{  sin ^2 ({\theta}_1 + {\theta}_2)}{(m_1+m_2)V ^2}

  \frac{m_2}{m_1} sin ^2 {\theta}_2 + \frac{m_1}{m_2} sin ^2 {\theta}_1  -2sin {\theta}_1 sin {\theta}_2 cos ( {\theta}_1 + {\theta}_2)= \frac{2 \epsilon  }{(m_1+m_2)V ^2}sin ^2 ({\theta}_1 + {\theta}_2)