# A particle of mass

Question:

A particle of mass ' $m$ ' is moving in time ' $t$ ' on a trajectory given by $\overrightarrow{\mathrm{r}}=10 \alpha \mathrm{t}^{2} \hat{\mathrm{i}}+5 \beta(\mathrm{t}-5) \hat{\mathrm{j}}$

Where $\alpha$ and $\beta$ are dimensional constants.

The angular momentum of the particle becomes the same as it was for $\mathrm{t}=0$ at time $t=\ldots$ seconds.

Solution:

$\overrightarrow{\mathrm{r}}=10 \alpha \mathrm{t}^{2} \hat{\mathrm{i}}+5 \beta(\mathrm{t}-5) \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{v}}=20 \alpha \mathrm{t} \hat{\mathrm{i}}+5 \beta \hat{\mathrm{j}}$

$\overrightarrow{\mathrm{L}}=\mathrm{m}(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{v}})$

$=m\left[10 \alpha t^{2} \hat{i}+5 \beta(t-5) \hat{j}\right] \times[20 \alpha t \hat{i}+5 \beta \hat{j}]$

$\overrightarrow{\mathrm{L}}=\mathrm{m}\left[50 \alpha \beta \mathrm{t}^{2} \hat{\mathrm{k}}-100 \alpha \beta\left(\mathrm{t}^{2}-5 \mathrm{t}\right) \hat{\mathrm{k}}\right]$

At $\mathrm{t}=0, \quad \overrightarrow{\mathrm{L}}=\overrightarrow{0}$

$50 \alpha \beta t^{2}-100 \alpha \beta\left(t^{2}-5 t\right)=0$

$t-2(t-5)=0$

$t=10 \sec$