[tex]m_1 = m_2 = m\\v_1, v_2\\u_1, u_2 = ?[/tex]
W przypadku zderzenia sprężystego zachowaniu ulega pęd i energia.
Z zasady zachowania pędu:
[tex]mv_1 + mv_2 = mu_1 + mu_2 \ \ /:m\\\\v_1 + v_2 = u_1 + u_2\\\\(1) \ \underline{v_1-u_1 = u_2 - v_2}}[/tex]
Z zasady zachowania energii:
[tex]\frac{1}{2}mv_1^{2} + \frac{1}{2}mv_2^{2} = \frac{1}{2}mu_1^{2}+\frac{1}{2}mu_2^{2} \ \ /\cdot\frac{2}{m}\\\\v_1^{2}+v_2^{2} = u_1^{2}+u_2^{2}\\\\\underline{v_1^{2}-u_1^{2} = u_2^{2}-v_2^{2}}[/tex]
Rozwiązujemy układ równań:
[tex]v_1^{2}-u_1^{2} = u_2^{2}-v_2^{2}\\v_1 - u_1 = u_2-v_2\\--------- \ \ dzielimy \ stronami\\\\v_1+u_1 = u_2-v_2 \ \ \rightarrow \ \ u_2 = v_1+u_1-v_2 \ - \ podstawiamy \ do \ (1)\\\\v_1-u_1 = v_1+u_1-v_2-v_2\\\\-u_1-u_1 = v_1-v_1-2v_2\\\\-2u_1 = -2v_{2} \ \ /:(-2)\\\\\boxed{u_1 = v_2}[/tex]
[tex]i\\u_2 = v_1+v_2-v_2\\\\\boxed{u_2 = v_1}[/tex]
[tex]Odp. \ u_1 = v_2, \ u_2 = v_1[/tex] [tex]u_1 = v_2, \ u_2 = v_1[/tex]
