Problem 6.66

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

î

\([m_{1}v_{1}\cos \theta + m_{2}v_{2}\cos (90º - \theta )]\)

ĵî

\(+ [m_{1}v_{1}\sin \theta - m_{2}v_{2}\sin (90º - \theta )] = m_{1}u\)

îĵî

\((v_{1}\cos \theta + v_{2}\sin \theta ) + (v_{1}\sin \theta - v_{2}\cos \theta ) = u\)

\(\therefore v_{1}\cos \theta + v_{2}\sin \theta = u\cdots \cdots (1)\)

และ \(v_{1}sinθ - v_{2}cosθ = 0 \cdots \cdots (2)\)

จาก \((2)\) จะได้ \(v_{2} = v_{1}tanθ \cdots \cdots (3)\)

\((1)^{2} + (2)^{2} ; v_{1}^{2} + v_{2}^{2} = u^{2}\)

\(v_{1}^{2} + (v_{1}\tan \theta )^{2} = u^{2}\)

\(v_{1}^{2}(1 + \tan ^{2}\theta ) = u^{2}\)

\(v_{1}^{2}\sec ^{2}\theta = u^{2}\)

\(\therefore v_{1} = ucos\theta \;\blacksquare\)

\(v_{1} \to (3) ; v_{2} = (ucos\theta )\tan \theta = usin\theta \;\blacksquare\)