A particle of mass m is moving in time t on a | vedclass

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(B) The position vector is $\overrightarrow{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}$.The velocity vector is $\overrightarrow{v} = \frac{d

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$10$

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(B) The position vector is $\overrightarrow{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}$.The velocity vector is $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 20 \alpha t \hat{i} + 5 \beta \hat{j}$.The angular momentum is given by $\overrightarrow{L} = m (\overrightarrow{r} 	imes \overrightarrow{v})$.$\overrightarrow{L} = m [10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}] 	imes [20 \alpha t \hat{i} + 5 \beta \hat{j}]$.Calculating the cross product: $\overrightarrow{L} = m [ (10 \alpha t^2)(5 \beta) \hat{k} - (5 \beta (t - 5))(20 \alpha t) \hat{k} ]$.$\overrightarrow{L} = m [ 50 \alpha \beta t^2 - 100 \alpha \beta t (t - 5) ] \hat{k}$.At $t = 0$,$\overrightarrow{L} = m [ 0 - 0 ] \hat{k} = 0$.We want $\overrightarrow{L} = 0$ at time $t > 0$:$50 \alpha \beta t^2 - 100 \alpha \beta t (t - 5) = 0$.Dividing by $50 \alpha \beta t$ (since $t 
eq 0$):$t - 2(t - 5) = 0$.$t - 2t + 10 = 0 \implies -t = -10 \implies t = 10 	ext{ seconds}$.

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