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 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 \(L\) of a particle with to origin is to \(L =r \times p\) where \(r\) is the position vector of the particle and \(p\) is the linear momentum. The direction of \(L\) is perpendicular to \(dr\) and \(p\) 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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