Question

What is the magnitude of the angular momentum relative to the origin of the 200 g particle in (figure 1)? Express your answer in kilogram-meter square.

a) 0.02 kg m²
b) 0.04 kg m²
c) 0.06 kg m²
d) 0.08 kg m²

Answer 1

The magnitude of the angular momentum of a 0.2-kg particle moving along the line y = 2.0 m with a velocity of 5.0 m/s is found by taking the cross product of the position vector and the momentum. This results in 0.02 kg m², which corresponds to Option (a). So, the correct option is a) 0.02 kg m²

Step-by-step explanation:

The question asks about the magnitude of the angular momentum (L) of a 0.2-kg particle moving along the line y = 2.0 m with a velocity of 5.0 m/s relative to the origin. In physics, the angular momentum of a particle relative to a point (like the origin) is given by the cross product of the position vector (r) and the momentum of the particle (p = mv).

In this case, the position vector of the particle can be represented as r = (0, 2.0, 0)m and the velocity vector as v = (5.0, 0 , 0)m/s.

The momentum p of the particle is:

m * v = 0.2 kg * 5.0 m/s = 1 kg*m/s (in the x-direction).

Using the right hand rule for cross products:

L = r x p

L = (0, 2.0, 0) m x (1, 0, 0) kg*m/s

L = (0, 0, 2.0 kg*m/s).

The magnitude of the angular momentum L is simply the magnitude of this cross product, which is:

|L| = 2.0 kg*m²/s (since the cross product results in a vector perpendicular to r and p, its magnitude is given by multiplying the magnitudes of r and p).

Therefore, the correct answer is:

Option (a) 0.02 kg m².