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(B) 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 the

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$-24 \hat{k} \, kg \cdot m^2/s$

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(B) 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 the initial position $\vec{r}_0 = 8 \hat{j} \, m$ and velocity $\vec{v} = 3 \hat{i} \, m/s$.The position of the particle at time $t = 5 \, s$ is $\vec{r} = \vec{r}_0 + \vec{v}t = 8 \hat{j} + (3 \hat{i})(5) = 15 \hat{i} + 8 \hat{j} \, m$.The momentum of the particle is $\vec{p} = m\vec{v} = (1)(3 \hat{i}) = 3 \hat{i} \, kg \cdot m/s$.Now,$\vec{L} = (15 \hat{i} + 8 \hat{j}) 	imes (3 \hat{i}) = 15 	imes 3 (\hat{i} 	imes \hat{i}) + 8 	imes 3 (\hat{j} 	imes \hat{i})$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get $\vec{L} = 0 + 24(-\hat{k}) = -24 \hat{k} \, kg \cdot m^2/s$.

$A$ particle starts from the point $(0, 8) \, m$ and moves with a uniform velocity of $\vec{v} = 3 \hat{i} \, m/s$. What is the angular momentum of the particle about the origin after $5 \, s$ (mass of the particle is $1 \, kg$)?

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