Question

Two particles each of mass m and speed v, travel in opposite direction along parallel lines separated by a distance d. Show that the vector angular momentum of this system of particles is the same about any point taken as origin.

Answer 1

Consider, Two particles be at points P and Q at any instant.

Angular momentum of the system about point P:

Lp = mv × 0 + mv × d

LR = mvd …(1)

Angular momentum of the system about point Q:

LQ = mv × d + mv × 0

LR = mvd ….(2)

Consider a point R, which is at a distance y from point Q, i.e. QR = y

∴ PR = d – y

Angular momentum of the system about point R:

LR = mv × (d – y) + mv × y

LR = mvd ….(3)

Comparing equations (1), (2), and (3), we get:

LP = LQ = LR

Thus, we can say that the angular momentum of a system is same wherever taken.

Answer 2

Given: Mass of the particle = m

Speed of the particle = v

Distance = d

To Find: Similar vector angle

Solution:

Let two particles be at points = P and Q .

Angular momentum of system at P will be -

Lp = mv × 0 + mv × d

LR = mvd --- eq 1

Angular momentum of system at Q will be -

LQ = mv × d + mv × 0

LR = mvd --- eq 2

Let R, be a point at a distance y from Q,

Thus, QR = y and PR = d – y

Angular momentum of system at R will be -

LR = mv × (d – y) + mv × y

LR = mvd --- eq 3

From equations 1, 2 and 3

LP = LQ = LR

Answer : The angular momentum of a system is same with the similar vector angles.