$_{{{K}_{A}}=\frac{1}{2}{{K}_{B}}\Rightarrow \frac{1}{2}{{m}_{A}}V_{A}^{2}=\frac{1}{2}\times \frac{1}{2}{{m}_{B}}V_{B}^{2}\Rightarrow \frac{1}{2}{{m}_{A}}V_{A}^{2}=\frac{1}{4}\times (\frac{1}{2}{{m}_{A}})\times V_{B}^{2}\Rightarrow 4V_{A}^{2}=V_{B}^{2}\Rightarrow {{V}_{B}}=2{{V}_{A}}}^{{{m}_{B}}=\frac{1}{2}{{m}_{A}}}$ 

با افزايش تندی جسمِ A خواهیم داشت:

$\frac{1}{2}{{m}_{A}}{{({{V}_{A}}+1)}^{2}}=\frac{1}{2}{{m}_{B}}V_{B}^{2}\Rightarrow \frac{1}{2}{{m}_{A}}{{({{V}_{A}}+1)}^{2}}=\frac{1}{2}\times \frac{1}{2}{{m}_{A}}{{(2{{V}_{A}})}^{2}}\Rightarrow {{({{V}_{A}}+1)}^{2}}=2V_{A}^{2}\Rightarrow {{V}_{A}}+1=\sqrt{2}{{V}_{A}}\Rightarrow (\sqrt{2}-1){{V}_{A}}=1\Rightarrow {{V}_{A}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$ $\frac{1}{2}{{m}_{A}}{{({{V}_{A}}+1)}^{2}}=\frac{1}{2}{{m}_{B}}V_{B}^{2}\Rightarrow \frac{1}{2}{{m}_{A}}{{({{V}_{A}}+1)}^{2}}=\frac{1}{2}\times \frac{1}{2}{{m}_{A}}{{(2{{V}_{A}})}^{2}}\Rightarrow {{({{V}_{A}}+1)}^{2}}=2V_{A}^{2}\Rightarrow {{V}_{A}}+1=\sqrt{2}{{V}_{A}}\Rightarrow (\sqrt{2}-1){{V}_{A}}=1\Rightarrow {{V}_{A}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$