A) \[1.73kg{{m}^{2}}/s\]
B) \[3.0kg{{m}^{2}}/s\]
C) \[3.46kg{{m}^{2}}/s\]
D) \[6.0kg{{m}^{2}}/s\]
Correct Answer: C
Solution :
Given : m = 0.160 kg \[\theta ={{60}^{o}}\] v = 10 m/s Angular momentum \[L=\overset{\to }{\mathop{r}}\,\times m\overset{\to }{\mathop{v}}\,\] \[=Hmv\cos \theta \] \[=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g}\cos \theta \] \[\left[ H=\frac{{{v}^{2}}{{\sin }^{2}}\theta }{2g} \right]\] \[=\frac{{{10}^{2}}\times {{\sin }^{2}}{{60}^{o}}\times \cos {{60}^{o}}}{2\times 10}\] \[=3.46kg\,{{m}^{2}}/s\]
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