A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s in the x direction along the line y = 5.What is its angular momentum (in kg.m²/s)relative to the origin? A)-30 $$\hat { k }$$ B)30 $\hat { k }$ C)-15 $\hat { k }$ D)15 $\hat { k }$ E)45 $\hat { k }$

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Quizplus · Accepted Answer

The angular momentum L is given by L = r x p, where r is the position vector relative to the origin and p is the linear momentum. Since the particle is moving only in the x direction, its linear momentum is p = mv = 2 kg x 3 m/s = 6 kg m/s in the positive x direction. To find the position vector, we can choose any point on the line y = 5 and the corresponding position vector will be parallel to the y-axis. Let's choose the point (0,5). The position vector r from the origin to this point is r = <0,5>. Taking the cross product, we get: L = r x p = <0,5,0> x <6,0,0> = <0,0,-30> The magnitude of L is given by |L| = |r| |p| sin(theta), where theta is the angle between r and p. Since r and p are perpendicular, theta = 90 degrees and sin(theta) = 1. Therefore, |L| = |r| |p| = 5 m x 6 kg m/s = 30 kg m^2/s. The direction of L is given by the right-hand rule: if you curl your fingers in the direction from r to p, your thumb points in the direction of L, which is out of the plane of the paper (negative z direction). So, the angular momentum of the particle relative to the origin is L = -30 kg m^2/s in the negative z direction. The correct answer is therefore A) -30.

A Particle Whose Mass Is 2 Kg Moves in the Xy