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NCERT Exemplar · Q10

Q.Two identical particles 1 and 2, each of mass mm, move with equal speeds vv in opposite directions along two parallel straight lines. A point A lies in the plane containing the two lines. The perpendicular distance from A to the line of particle 1 is d1d_1, and to the line of particle 2 is d2d_2, with d2>d1d_2 > d_1 (A is nearer the line of particle 1). r⃗1\vec{r}_1 and r⃗2\vec{r}_2 are the position vectors of the two particles measured from A. Let ⊗\otimes denote a unit vector directed into the plane of the page and ⊙\odot a unit vector directed out of the page. Choose the correct option(s); more than one may be correct.

(a) The angular momentum l⃗1\vec{l}_1 of particle 1 about A has magnitude l1=mvd1l_1 = mvd_1
(b) The angular momentum l⃗2\vec{l}_2 of particle 2 about A is l⃗2=mv r⃗2\vec{l}_2 = mv\,\vec{r}_2
(c) The total angular momentum of the system about A is l⃗=mv(r⃗1+r⃗2)\vec{l} = mv(\vec{r}_1 + \vec{r}_2)
(d) The total angular momentum of the system about A is l⃗=mv(d2−d1) ⊗\vec{l} = mv(d_2 - d_1)\,\otimes
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The magnitude of angular momentum of a particle about A is (momentum)×\times(perpendicular distance), i.e. mvdmvd, not mv∣r⃗∣mv|\vec r|. So (A) is correct and (B) wrong. Because the particles move oppositely, their angular momenta about A point in opposite senses; the net is mv(d2−d1)mv(d_2-d_1) directed into the page, which is (D). Statement (C) is wrong.

Concept

For a single particle, l⃗=r⃗×mv⃗\vec{l} = \vec{r}\times m\vec{v}, and ∣l⃗∣=mv d|\vec{l}| = mv\,d where dd is the perpendicular distance from the reference point to the line of motion. Only the perpendicular distance matters, not the full length of r⃗\vec r.

Directions (take ⊗\otimes = into page, ⊙\odot = out of page)

  1. Particle 1: ∣l⃗1∣=mvd1|\vec l_1| = mvd_1; from r⃗1×mv⃗\vec r_1\times m\vec v its sense is out of the page (⊙\odot). Its magnitude mvd1mvd_1 makes (A) correct. …

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