Question

Figure shows two identical particles 1 and 2, each of mass $m,$ moving in opposite directions with same speed $v$ along parallel lines. At a particular instant,${\overrightarrow{r}}_1$ and ${\overrightarrow{r}}_2$ are their respective position vectors drawn from point $A$ which is in the plane of the parallel lines.


Choose the correct options:

• A. Total angular momentum of the system about $A$ is
  
  $\overrightarrow{l}=|mv({\overrightarrow{r}}_1+{\overrightarrow{r}}_2)|⨀$
• B. Total angular momentum of the system about $A$ is
  
  $\overrightarrow{l}=mv(d_2-d_1)⨂$
• C. Angular momentum ${\overrightarrow{l}}_2$ of particle $2$ about $A$ is ${\overrightarrow{l}}_2=mv{r}_2⨀$
• D. Angular momentum ${\overrightarrow{l}}_1$ of particle $1$ about $A$ is ${\overrightarrow{l}}_1=mv\left(d_1\right)⨀$

Answer 1

Answer: (B), (D)

Angular momentum ${\overrightarrow{l}}_1$ of particle $1$ about $A$ is ${\overrightarrow{l}}_1=mv\left(d_1\right)⨀$

Formula used: $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$
Draw vector diagram showing angular momentum of the two masses.


Find angular momentum of the particle $1$ about point $A.$

Given, position vector for particle $1,{\overrightarrow{r}}_1$

And speed of the particle $1,=v$

So, angular momentum ${\overrightarrow{L}}_1={\overrightarrow{r}}_1\times {\overrightarrow{p}}_1$

$l_1=m(r_1\times v)$

${\overrightarrow{l}}_1=m{r}_{1}v\sin {\theta }_1⨀$

From figure $r_1\sin {\theta }_1=d_1$

${\overrightarrow{l}}_1=mv\left(d_1\right)⨀$

Find angular momentum of the particle $2$ about point $A.$

Given, position vector for particle $2,{\overrightarrow{r}}_2$

And speed of the particle $2,=v$

So, angular momentum ${\overrightarrow{l}}_2={\overrightarrow{r}}_2\times {\overrightarrow{p}}_2$

$l_2=m(r_2\times v)$

${\overrightarrow{l}}_2=mv{r}_2⨂$

From figure $r_2\sin {\theta }_2=d_2$

${\overrightarrow{l}}_2=mv\left(d_2\right)⨂$

Find total angular momentum of the system about $A.$

Total angular momentum $\overrightarrow{l}={\overrightarrow{l}}_1+{\overrightarrow{l}}_2$

$\overrightarrow{l}=mv{d}_1⨀+mv{d}_2⨂$

$\overrightarrow{l}=-mv{d}_1⨂+mv{d}_2⨂$

$\overrightarrow{l}=mv(d_2-d_1)⨂(\because d_2>d_1)$

Final Answer: Option (a),(d)