Problem 6.43

จาก \(Σ\vec{p}_{f}= \Sigma \vec{p}_{i}\)

\(m_{1}v_{1} + m_{2}v_{2} = m_{1}u_{1} + m_{2}u_{2}\)

\(m_{1}(v_{1} - u_{1}) = m_{2}(u_{2} - v_{2})\cdots \cdots (1)\)

เนื่องจากเป็นการชนแบบยืดหยุ่น

จะได้ \(ΣK_{f}= \Sigma K_{i}\)

\(m_{1}v_{1}^{2}+\frac{1}{2}\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\)

\(m_{1}(v_{1}^{2} - u_{1}^{2}) = m_{2}(u_{2}^{2} - v_{2}^{2})\)

\(m_{1}(v_{1} + u_{1})(v_{1} - u_{1}) = m_{2}(u_{2} + v_{2})(u_{2} - v_{2})\cdots \cdots (2)\)

\((1)\) แทนใน \((2)\) จะได้ \(v_{1} + u_{1} = v_{2} + u_{2} \cdots \cdots (3)\)

\(v_{2} = v_{1} + u_{1} - u_{2}\cdots \cdots (4)\)

\((4)\) แทนใน \((1)\) จะได้ \(v_{1} = \frac{(m_{1} - m_{2})}{m_{1} + m_{2}}u_{1}+\frac{(2m_{2})}{m_{1} + m_{2}}u_{2}\;\blacksquare\)

v₁ แทนใน \((4)\) จะได้ \(v_{2} = \frac{(2m_{1})}{m_{1} + m_{2}}u_{1}+\frac{(m_{2} - m_{1})}{m_{1} + m_{2}}u_{2}\;\blacksquare\)