Problem 617 - JEE Physics

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

Step-by-Step Solution

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 1: 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: \(p = m · v\)

Step 2: 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.

Step 3: 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: \(L_z = r_x p_y - r_y p_x\)

Step 4: 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: \(L_z = x(t) · 0 - 0 · (m · v_x)\) \(L_z = 0\)

Step 5: 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

Key Concepts

Cross Product

The cross product is a crucial concept in vector mathematics, especially in physics. It refers to an operation on two vectors in three-dimensional space. When we take the cross product of two vectors, the resulting vector is perpendicular to both of the original vectors. This is particularly useful in determining quantities that involve rotation, such as angular momentum.
In the context of angular momentum, the cross product allows us to calculate how much 'twist' or rotational potential a moving object has around a certain point. It involves two vectors: the position vector and the linear momentum vector. With mathematical expression, if \(\mathbf{r}\) is the position vector and \(\mathbf{p}\) is the linear momentum, then the angular momentum \(\mathbf{L}\) is calculated as the cross product:

• \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\)

The magnitude of the cross product vector depends on the angle between the original two vectors. Notably, if the vectors are parallel or anti-parallel (the angle is 0 or 180 degrees), the cross product will be zero.

Position Vector

The position vector is fundamental in describing the motion of a particle in space. It's a vector that points from the origin of the coordinate system to the location of the particle.
A position vector \(\mathbf{r}\) can be expressed in terms of its components along the x, y, and z axes. In this problem, it is given as \(\mathbf{r} = (x(t), 0, 0)\), with the particle moving parallel to the x-axis. The dynamic part here is \(x(t)\), which denotes that the specific component might change with time depending on the motion of the particle.
Even though this vector evolves as the particle moves, its importance lies in the fact that it designates where the particle is relative to a chosen point, usually the origin. In calculations involving the angular momentum, the position vector helps determine how far the mass is from the point giving a measure of its rotational effect.

Linear Momentum

Linear momentum is the measure of an object's motion. It is calculated as the product of the mass \( m \) and velocity \( \mathbf{v} \) of the object. In equation form, linear momentum \( \mathbf{p} \) is expressed as:

• \( \mathbf{p} = m \cdot \mathbf{v} \)

This concept not only considers how fast the object is moving but also how much mass is contributing to its movement.
For the exercise in question, since the mass is moving with a constant velocity, the linear momentum remains constant over time as well. Furthermore, because the motion is purely along the x-axis, we can say \( \mathbf{p} = (m \cdot v_x, 0, 0) \). The constancy of linear momentum underlies many fundamental principles in physics, significantly affecting how we calculate and understand related concepts like angular momentum.

Velocity Vector

In physics, the velocity vector is a fundamental concept to describe the rate of change of an object's position. In essence, it quantifies both the speed and the direction of motion. For an object moving along a path, the velocity vector \(\mathbf{v}\) gives important insights. It is usually expressed as:

• \(\mathbf{v} = (v_x, v_y, v_z)\)

In the current problem, where the mass is moving parallel to the x-axis, the velocity vector simplifies to \(\mathbf{v} = (v_x, 0, 0)\), indicating motion only in the x-direction. This is due to the fact that the mass is not changing position in the y or z directions.
Understanding the velocity vector helps in visualizing and calculating other movements or influences on the object, such as determining angular momentum. With constant velocity as given in the problem, it leads to straightforward calculations where the changes or dynamics over time don't complicate analysis.