Ncert Exercises & Solutions

The figure below shows two identical particles 1 and 2, each of mass $m,$ moving in opposite directions with the same speed $v$ along parallel lines. At a particular instant, $r_1$ and $r_2$ are their respective position vectors drawn from point A, which is in the plane of the parallel lines.

Consider the following statements.

a.	angular momentum $l_1$ of particle 1 about A is $l_1=mv(d_1)$ ⊙
b.	angular momentum $l_1$ of particle 2 about A is $l_1=mv(r_2)$ ⊙
c.	total angular momentum of the system about A is $l=mv(r_1+r_2)$ ⊙
d.	total angular momentum of the system about A is $l=mv(d_2-d_1)$ ⊗

Choose the correct option:

1.	(a, c)
2.	(a, d)
3.	(b, d)
4.	(b, c)

(a, d) Hint: In angular momentum, only perpendicular distance is considered.

Step 1: Find the angular momentum of particle 1.

The angular momentum L of a particle with to origin is to L —r x p where. r is the position vector of the particle and p is the linear momentum. The direction of L is perpendicular to d r and p by the right-hand rule.

For particle $1, I_{1} = r_{1} \times mv$ is out of the plane of the and perpendicular to r, and v).

Step 2: Find the angular momentum of the system.

Similarity $I_{2} = r_{2} \times m (- v)$ is into the plane of the paW and perpendicular to $r_{2}$ and —p Hence, total angular momentum

$\begin{matrix} l = h + l_{2} = r_{1} \times mv + (- r_{2} \times mv) \\ | l | = {mvd}_{1} - {mvd}_{2} as d_{2} > d_{1} total angular momentum will be inward \end{matrix}$

$Hence, I = mv (d_{2} - d_{1}) \otimes$

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a.	angular momentum \(l_1\) of particle 1 about A is \(l_1=mv(d_1)\) ⊙
b.	angular momentum \(l_1\) of particle 2 about A is \(l_1=mv(r_2)\) ⊙
c.	total angular momentum of the system about A is \(l=mv(r_1+r_2)\) ⊙
d.	total angular momentum of the system about A is \(l=mv(d_2-d_1)\) ⊗