A particle P with a mass 2.0 kg has position vector r = 3.0 m and velocity V = 4.0 m/s as shown. It is accelerated by the force F = 2.0N. All three vectors lie in a common plane. The angular momentum vector is
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Angular momentum L is given by the cross product of the position vector r and the linear momentum p (where p = mv). The magnitude is L = r * p * sinθ. From the diagram, the angle between r and v is 90°, so sin90° = 1.
Calculate: m = 2.0 kg, r = 3.0 m, v = 4.0 m/s. Therefore, L = (3.0 m) * (2.0 kg * 4.0 m/s) * 1 = .
The direction is found using the right-hand rule for the cross product r × v. Curling fingers from r to v, the thumb points out of the plane.
Final Answer: 24 kg m²/s out of the plane of the figure