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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 mas

Vedclass · Accepted Answer

$-80\hat{k}$

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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 mass $m = 2\, kg$ and position vector $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$.The velocity vector $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5\hat{i} - 2t^2\hat{j}) = -4t\hat{j}$.At $t = 2\, s$:Position $\vec{r}(2) = 5\hat{i} - 2(2)^2\hat{j} = 5\hat{i} - 8\hat{j}$.Velocity $\vec{v}(2) = -4(2)\hat{j} = -8\hat{j}$.Angular momentum $\vec{L} = m(\vec{r} 	imes \vec{v}) = 2 	imes [(5\hat{i} - 8\hat{j}) 	imes (-8\hat{j})]$.Since $\hat{i} 	imes \hat{j} = \hat{k}$ and $\hat{j} 	imes \hat{j} = 0$:$\vec{L} = 2 	imes [5\hat{i} 	imes (-8\hat{j}) - 8\hat{j} 	imes (-8\hat{j})] = 2 	imes [-40\hat{k} - 0] = -80\hat{k}\, kg\, m^2\, s^{-1}$.

$A$ particle of mass $2\, kg$ is moving such that at time $t$,its position,in meters,is given by $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$. The angular momentum of the particle at $t = 2\, s$ about the origin in $kg\, m^2\, s^{-1}$ is:

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