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Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. Show that the vector angular momentum of the two particle system is the same whatever be the point about which the angular momentum is taken.
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Angular momentum about $A$,
$$
L_A=m v \times 0+m v \times d=m v d
$$
Angular momentum about $B$,
$$
L_B=m v \times d+m v \times 0=m v d
$$
Angular momentum about $C$,
$L_c=m v \times y+m v \times(d-y)=m v d$
In all the three cases, the direction of angular momentum is the same.
$$
\therefore \quad \vec{L}=\vec{L}_B=\vec{L}_C .
$$
$$
L_A=m v \times 0+m v \times d=m v d
$$
Angular momentum about $B$,
$$
L_B=m v \times d+m v \times 0=m v d
$$
Angular momentum about $C$,
$L_c=m v \times y+m v \times(d-y)=m v d$
In all the three cases, the direction of angular momentum is the same.
$$
\therefore \quad \vec{L}=\vec{L}_B=\vec{L}_C .
$$
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