The position vector of a particle of mass m m | vedclass

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(A) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Given $\v

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(A) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} 	imes \vec{p} = \vec{r} 	imes (m\vec{v})$.Given $\vec{r} = x(t) \hat{i} + b \hat{j}$ and $\vec{v} = v \hat{i}$.$\vec{L} = (x(t) \hat{i} + b \hat{j}) 	imes (m v \hat{i}) = x(t) m v (\hat{i} 	imes \hat{i}) + b m v (\hat{j} 	imes \hat{i})$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$, we get $\vec{L} = -b m v \hat{k}$.The magnitude of the angular momentum is $|\vec{L}| = | -b m v | = b m v$.Since $b$, $m$, and $v$ are all constants, the magnitude of the angular momentum $|\vec{L}|$ is constant and independent of the angle $	heta$.Therefore, the graph of $|\vec{L}|$ versus $	heta$ is a horizontal straight line.

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