A particle of mass m moves with a constant velocity v in the positive x-direction. At a particular instant, the particle is located at the position vector r = (0, y₀, 0) relative to the origin, where y₀ is a positive constant. What is the angular momentum of the particle about the origin at this instant?
Correct: B
The angular momentum (L) of a particle about a point is given by the cross product of its position vector (r) relative to that point and its linear momentum (p).
L = r × p
The given information:
Mass of particle = m
Velocity vector v = (v, 0, 0) (constant velocity in the positive x-direction)
Linear momentum vector p = mv = (mv, 0, 0)
Position vector r = (0, y₀, 0) (at this instant)
Now, calculate the cross product:
L = r × p = (0, y₀, 0) × (mv, 0, 0)
Using the determinant form for the cross product:
L = | i j k |
| 0 y₀ 0 |
| mv 0 0 |
L = i (y₀ * 0 - 0 * 0) - j (0 * 0 - 0 * mv) + k (0 * 0 - y₀ * mv)
L = i (0) - j (0) + k (-y₀ * mv)
L = -y₀mv k
So, the magnitude of the angular momentum is mv y₀, and its direction is in the negative z-direction (k is the unit vector in the z-direction).
The correct choice is B.