Question

A mass \(\mathrm{m}\) is moving with a constant velocity along the line parallel to the \(\mathrm{x}\) -axis, away from the origin. Its angular momentum with respect to the origin \(\\{\mathrm{A}\\}\) Zero \\{B \\} remains constant \(\\{\mathrm{C}\\}\) goes on increasing \\{D\\} goes on decreasing

Short answer

The angular momentum L with respect to the origin can be calculated as \(L = r × p\). Using the given position vector \(r = (x(t), 0, 0)\), and the velocity vector \(v = (v_x, 0, 0)\), we obtain the angular momentum in the z-direction as \(L_z = r_x p_y - r_y p_x\). Plugging in the values, we get \(L_z = 0\). Therefore, the angular momentum remains constant with respect to the origin. The correct answer is (B) remains constant.

Step-by-step solution

Recall the Definition of Angular Momentum

The angular momentum L of a particle with respect to a point is given by the cross product of the position vector r and its linear momentum p, which can be defined as: \(L = r × p\) Note that linear momentum p is defined as the product of the mass (m) and velocity (v) of the particle: \(p = m · v\)

Analyze the Position and Velocity Vectors

In this problem, mass m is moving with a constant velocity parallel to the x-axis. We can express its position vector and velocity vector as: Position vector: \(r = (x(t), 0, 0)\), where x(t) is the distance from the origin. Velocity vector: \(v = (v_x, 0, 0)\) As the mass is moving along the x-axis, v_x is constant.

Calculate the Angular Momentum

Using the definition of angular momentum and the given position and velocity vectors, we can determine the angular momentum L: Since the cross product of parallel vectors is zero, we only need to obtain the cross product of the non-parallel components. Here, we have no components in the y or z direction, so we only need to calculate L in the z-direction: \(L_z = r_x p_y - r_y p_x\)

Analyze the Terms

From Step 3, we notice that r_x is same as x(t) and r_y = 0. Also, the mass is moving parallel to the x-axis, which means p_y = 0 and p_x = m*v_x. Thus, we can write L_z as: \(L_z = x(t) · 0 - 0 · (m · v_x)\) \(L_z = 0\)

Determine the Behavior of Angular Momentum

Since L_z and thus the entire angular momentum L is always equal to 0, it remains constant with respect to the origin. Hence, the correct option is: (B) remains constant