Skip to content
Exercises · 6.7

Q.Two particles, each of mass mm and speed vv, travel in opposite directions along parallel lines separated by a distance dd. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

Madhya Pradesh MpbseTextbookSubjective· 3mImportance★★★★★est
33% · 19/57 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For a system whose total linear momentum is zero the angular momentum is the same about every point. Here the two equal-and-opposite momenta cancel, so L\mathbf{L} is independent of the reference point, with magnitude mvdmvd.

Shift-of-origin identity. Moving the reference point from OO to O′O' (let R\mathbf{R} be the vector from O′O' to OO) changes the total angular momentum by

LO′=∑i(ri−R)×pi=LO−R×Ptotal,Ptotal=∑ipi.\mathbf{L}_{O'} = \sum_i(\mathbf{r}_i - \mathbf{R})\times\mathbf{p}_i = \mathbf{L}_O - \mathbf{R}\times\mathbf{P}_{\text{total}}, \qquad \mathbf{P}_{\text{total}} = \sum_i \mathbf{p}_i.

Total momentum is zero. The two particles carry momenta p1=mv x^\mathbf{p}_1 = mv\,\hat{\mathbf{x}} and p2=−mv x^\mathbf{p}_2 = -mv\,\hat{\mathbf{x}}, so

Ptotal=mv x^−mv x^=0.\mathbf{P}_{\text{total}} = mv\,\hat{\mathbf{x}} - mv\,\hat{\mathbf{x}} = 0.

Hence R×Ptotal=0\mathbf{R}\times\mathbf{P}_{\text{total}} = 0 for every R\mathbf{R}, giving LO′=LO\mathbf{L}_{O'} = \mathbf{L}_O: the angular momentum is the same about any point.

Its value. Taking the two parallel lines as y=+d/2y = +d/2 and y=−d/2y = -d/2, …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.