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 is out of the plane of the and perpendicular to r, and v).
Step 2: Find the angular momentum of the system.
Similarity is into the plane of the paW and perpendicular to and —p Hence, total angular momentum
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