Question

The position vector $\overrightarrow{r}$ of a particle of mass m is given by the following equation
$\overrightarrow{r}\left(t\right)={\alpha t}^3\hat{i}+\beta t^2\hat{j}$
Where $\alpha =10/3{ms}^{-2},\beta =5{ms}^{-2}$ and $m=0.1kg$. At $t=1$ s, which of the following statements is/are true about the particle?

• A. The velocity $\overrightarrow{v}$ is given by $\overrightarrow{v}=(10\hat{i}+10\hat{j}){ms}^{-1}$
• B. The angular momentum $\overrightarrow{L}$ with respect to the origin is given by $\overrightarrow{L}=(-5/3)\hat{k}$ N ms
• C. The force $\overrightarrow{F}$ is given by $\overrightarrow{F}=(\hat{i}+2\hat{j})N$
• D. The torque $\overrightarrow{\tau }$ with respect to the origin is given by $\overrightarrow{\tau }=-\left(20/3\right)\hat{k}$ Nm

Answer 1

Answer: (A), (B), (D)

The velocity $\overrightarrow{v}$ is given by $\overrightarrow{v}=(10\hat{i}+10\hat{j}){ms}^{-1}$
The torque $\overrightarrow{\tau }$ with respect to the origin is given by $\overrightarrow{\tau }=-\left(20/3\right)\hat{k}$ Nm
The angular momentum $\overrightarrow{L}$ with respect to the origin is given by $\overrightarrow{L}=(-5/3)\hat{k}$ N ms

The position vector is $\overrightarrow{r}\left(t\right)=\alpha t^3\hat{i}+\beta t^2\hat{j}$

$\therefore$ Velocity of particle $\overrightarrow{v}\left(t\right)=\frac{d\overrightarrow{r}\left(t\right)}{dt}=3\alpha t^2\hat{i}+2\beta t\hat{j}$

$⟹$ $\overrightarrow{v}(t=1)=3\times \frac{10}{3}\times 1^2\hat{i}+2\left(5\right)\left(1\right)\hat{j}=10\hat{i}+10\hat{j}$

Thus option A is correct.

Position vector at $t=1$ s $\overrightarrow{r}(t=1)=\frac{10}{3}\times 1^3\hat{i}+5\left(1^2\right)\hat{j}=\frac{10}{3}\hat{i}+5\hat{j}$

Angular momentum $\overrightarrow{L}(t=1)=m\left[\overrightarrow{r}\right(t=1)\times \overrightarrow{v}(t=1\left)\right]$

$\therefore$ $\overrightarrow{L}(t=1)=\left(0.1\right)\left[\right(\frac{10}{3}\hat{i}+5\hat{j})\times (10\hat{i}+10\hat{j}\left)\right]=\left(0.1\right)[\frac{100}{3}\hat{k}-50\hat{k}]$

$⟹\overrightarrow{L}(t=1)=-\frac{5}{3}\hat{k}$ N ms

Thus option B is correct.

The acceleration of the particle $\overrightarrow{a}\left(t\right)=\frac{d\overrightarrow{v}\left(t\right)}{dt}=6\alpha t\hat{i}+2\beta \hat{j}$

$\therefore$ $\overrightarrow{a}(t=1)=6\times \frac{10}{3}\times 1\hat{i}+2\left(5\right)\hat{j}=20\hat{i}+10\hat{j}$

Force on the particle $\overrightarrow{F}(t=1)=m\overrightarrow{a}(t=1)=0.1(20\hat{i}+10\hat{j})=2\hat{i}+\hat{j}$ N

Thus option C is incorrect.

Torque w.r.t origin $\overrightarrow{\tau }(t=1)=\overrightarrow{r}\times \overrightarrow{F}$

$\therefore$ $\overrightarrow{\tau }(t=1)=\left[\right(10/3)\hat{i}+5\hat{j}]\times [2\hat{i}+\hat{j}]$

$⟹$ $\overrightarrow{\tau }(t=1)=\frac{10}{3}\hat{k}-10\hat{k}=\frac{-20}{3}\hat{k}$ Nm

Thus option D is correct.