Ncert Exercises & Solutions

The figure below shows two identical particles

1

$1$ and

2

$2$ , each of mass

m,

$m,$ moving in opposite directions with the same speed

v

$v$ along parallel lines. At a particular instant,

r_{1}

$r_1$ and

r_{2}

$r_2$ are their respective position vectors drawn from point

A

$A$ , 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 from the given ones:

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

Hint: In angular momentum, only perpendicular distance is considered.

Step 1: Find the angular momentum of the particle $1.$

The angular momentum

L

$L$ of a particle with to origin is to

L = r \times p

$L =r \times p$ where

r

$r$ is the position vector of the particle and

p

$p$ is the linear momentum. The direction of

L

$L$ is perpendicular to

d r

$dr$ and

p

$p$ by the right-hand rule.

For particle $1,$ $L_1=r_1×mv$ is out of the plane of the perpendicular to $r_1$ and $v$ ).

Step 2: Find the angular momentum of the system.

Similarly $L_2=r_2×m(-v)$ is into the plane of perpendicular to $r_2$ and $p.$ Hence, total angular momentum

$L = L_1 +L_{2} = r_{1} \times mv + \left(\right. - r_{2} \times mv \left.\right) \\ \left|\right. L \left|\right. = \left(mvd\right)_{1} - \left(mvd\right)_{2}$
$(d_{2} > d_{1}) \text{total angular momentum will be inward}$

$L = mv \left(\right. d_{2} - d_{1} \left.\right) \bigotimes$

Hence, option (2) is the correct answer.

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