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(D) The linear momentum $\vec{p}$ is given by $\vec{p} = m\vec{v} = 0.01 \ kg 	imes 5\hat{i} \ m/s = 0.05\hat{i} \ kg \cdot m/s$.The angular momentum

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$-0.3$

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(D) The linear momentum $\vec{p}$ is given by $\vec{p} = m\vec{v} = 0.01 \ kg 	imes 5\hat{i} \ m/s = 0.05\hat{i} \ kg \cdot m/s$.The angular momentum $\vec{L}$ about the origin is given by the cross product $\vec{L} = \vec{r} 	imes \vec{p}$.Substituting the values: $\vec{L} = (10\hat{i} + 6\hat{j}) 	imes (0.05\hat{i})$.Using the distributive property of the cross product: $\vec{L} = (10\hat{i} 	imes 0.05\hat{i}) + (6\hat{j} 	imes 0.05\hat{i})$.Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get:$\vec{L} = 0 + 6 	imes 0.05 	imes (-\hat{k}) = 0.3 	imes (-\hat{k}) = -0.3\hat{k} \ J \cdot s$.

$A$ particle of mass $0.01 \ kg$ has a position vector $\vec{r} = (10\hat{i} + 6\hat{j}) \ m$ and moves with a velocity $\vec{v} = 5\hat{i} \ m/s$. Calculate its angular momentum about the origin in $\hat{k} \ J \cdot s$.

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