Problem 6.40

กำหนดให้ \(m = 1.00 kg, M = 10.0 kg\) และ \(u = 550 m/s\)

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

\((M + m)v = mu\)

\(v =\frac{mu}{M + m}\)

จาก \(ΔK = K_{f}- K_{i}\)

\(\Delta K=\frac{1}{2}(M+m)v^{2}-\frac{1}{2}mu^{2}\)

\(\Delta K=\frac{1}{2}(M+m)(\frac{mu}{M + m})^{2}-\frac{1}{2}mu^{2}\)

\(\Delta K =\frac{m^{2}u^{2}}{2(M + m)}-\frac{m(M + m)u^{2}}{2(M + m)}\)

\(\Delta K = -\frac{mMu^{2}}{2(M + m)}\)

\(\Delta K = -0.138 MJ\)

∴ ระบบมีพลังงานจลน์ลดลง \(= 0.138 MJ \;\blacksquare\)