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Q6.7:
Two particles, each of mass (m) and speed (v), travel in opposite directions along parallel lines separated by a distance (d). 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.

Solution :

Step 1: System Definition and Coordinate Geometry

Consider a Cartesian coordinate system where the two particles, $P_1$ and $P_2$, each of mass $m$, move with velocity vectors $\vec{v}_1 = v\hat{i}$ and $\vec{v}_2 = -v\hat{i}$. The particles travel along lines parallel to the x-axis, separated by a distance $d$ along the y-axis.

  • Position of $P_1$: $\vec{r}_1 = x_1\hat{i} + y_1\hat{j}$
  • Position of $P_2$: $\vec{r}_2 = x_2\hat{i} + (y_1 + d)\hat{j}$
P1 (v) P2 (-v) d

Step 2: Angular Momentum about an Arbitrary Point O

Let an arbitrary point $O$ be the origin $(0,0,0)$. The angular momentum $\vec{L}$ of the system is the sum of the angular momenta of the individual particles [Per the Principle of Superposition of Angular Momentum]:

$\vec{L} = \vec{r}_1 \times \vec{p}_1 + \vec{r}_2 \times \vec{p}_2$

Substituting $\vec{p}_1 = mv\hat{i}$ and $\vec{p}_2 = -mv\hat{i}$:

$\vec{L} = (x_1\hat{i} + y_1\hat{j}) \times (mv\hat{i}) + (x_2\hat{i} + (y_1 + d)\hat{j}) \times (-mv\hat{i})$

Step 3: Algebraic Expansion and Vector Cross Product

Utilizing the properties of cross products ($\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$):

$\vec{L} = [x_1mv(\hat{i} \times \hat{i}) + y_1mv(\hat{j} \times \hat{i})] + [x_2(-mv)(\hat{i} \times \hat{i}) + (y_1 + d)(-mv)(\hat{j} \times \hat{i})]$

$\vec{L} = 0 - y_1mv\hat{k} + 0 + (y_1 + d)mv\hat{k}$

$\vec{L} = (-y_1mv + y_1mv + dmv)\hat{k}$

$\vec{L} = dmv\hat{k}$

Step 4: Analysis of Invariance

The resulting expression $\vec{L} = dmv\hat{k}$ is independent of the coordinates $(x_1, y_1)$ and $(x_2, y_1+d)$. This demonstrates that the choice of the origin (the point about which angular momentum is calculated) does not affect the total angular momentum of the system.

Parameter Value
Net Angular Momentum $mv d$
Direction Perpendicular to the plane of motion

Final Solution: Since the coordinates $(x, y)$ of the particles cancel out during the vector summation, the angular momentum $\vec{L} = dmv\hat{k}$ is invariant and remains constant regardless of the reference point O.


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