Question

A system consists of two particles. At \(t\) = 0 one particle is at the origin; the other, which has a mass of 0.50 kg,is on the \(y\)-axis at \(y\) = 6.0 m. At \(t\) = 0 the center of mass of the system is on the \(y\)-axis at \(y\) = 2.4 m. The velocity of the center of mass is given by \((0.75 m/s^3)t^2\hat{\imath}\). (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time t. (c) Find the net external force acting on the system at \(t\) = 3.0 s.

Short answer

The total mass is 1.25 kg, acceleration is \(1.5t \text{ m/s}^2\), and force at \(t=3.0\text{ s}\) is 5.625 N.

Step-by-step solution

Understanding the Problem

We need to find the total mass of a system with two particles, the acceleration of the center of mass, and the net external force on the system at a specific time. We use given positions and velocities of the center of mass and individual particles.

Calculate Total Mass of the System

Let the mass of the particle at the origin be denoted as \( m_1 \), and the mass of the second particle on the y-axis be \( m_2 = 0.50 \text{ kg} \). The position of the center of mass in the y-direction is given by:\[ y_{cm} = \frac{m_1 \cdot 0 + m_2 \cdot 6.0}{m_1 + m_2} = 2.4\]Solving for \( m_1 \):\[ 2.4(m_1 + 0.5) = 0.5 \times 6 \implies 2.4m_1 + 1.2 = 3 \ 2.4m_1 = 1.8 \implies m_1 = 0.75 \text{ kg} \\]Thus, the total mass \( M = m_1 + m_2 = 0.75 + 0.50 = 1.25 \text{ kg} \).

Calculate Acceleration of the Center of Mass

The velocity of the center of mass is given by:\[v_{cm} = (0.75 \text{ m/s}^3)t^2\hat{\imath}\]To find the acceleration, take the derivative of velocity with respect to time:\[a_{cm} = \frac{d}{dt}((0.75 \text{ m/s}^3)t^2) = 2(0.75)t = 1.5t \text{ m/s}^2 \hat{\imath}\]

Calculate Net External Force at t = 3.0 s

Using the formula for the net external force \( F_{net} = M \cdot a_{cm} \), where \( M = 1.25 \text{ kg} \) and at \( t = 3.0 \text{ s} \), the acceleration is \( 1.5 \times 3.0 = 4.5 \text{ m/s}^2 \):\[F_{net} = 1.25 \times 4.5 \hat{\imath} = 5.625 \text{ N} \hat{\imath}\]

Completion

Now we have all the required values: the total mass of the system is 1.25 kg, the acceleration of the center of mass is \(1.5t \text{ m/s}^2\), and the net external force at \(t = 3.0\text{ s}\) is \(5.625 \text{ N}\).

Key concepts

Total Mass of the System

The total mass of a system is a fundamental concept when dealing with multiple particles. It is simply the sum of the masses of all the particles within the system. In this exercise, we have two particles. One particle, denoted as \( m_2 \), has a mass of 0.50 kg and is located on the y-axis at 6.0 m. The other particle is at the origin. The position of the center of mass along the y-axis is given as 2.4 m.
To find the mass \( m_1 \) of the particle at the origin, we use the equation for the y-coordinate of the center of mass:

• \( y_{cm} = \frac{m_1 \cdot 0 + m_2 \cdot 6.0}{m_1 + m_2} = 2.4 \)

Solving this equation, we find \( m_1 \) to be 0.75 kg. Therefore, the total mass \( M \) of the system is:

• \( M = m_1 + m_2 = 0.75 \text{ kg} + 0.50 \text{ kg} = 1.25 \text{ kg} \)

Velocity of Center of Mass

The velocity of the center of mass is an important concept as it describes how the entire system moves through space over time. In this exercise, the velocity of the center of mass is given by the equation:

• \( v_{cm} = (0.75 \text{ m/s}^3)t^2 \hat{\imath} \)

This equation tells us that the velocity changes with time and its magnitude depends on the square of time \( t \). At \( t = 0 \), the velocity of the center of mass is zero, meaning the system is initially at rest along the x-axis.
This concept is crucial for understanding how forces applied to the system's particles affect the motion of the center of mass, integrating both position and time into the system's movement.

Acceleration of Center of Mass

Acceleration represents how quickly the velocity of an object changes over time. For the center of mass, we derive its acceleration by differentiating the velocity function with respect to time. Given the velocity:

• \( v_{cm} = (0.75 \text{ m/s}^3)t^2 \hat{\imath} \)

The acceleration of the center of mass can be calculated by taking its derivative:

• \( a_{cm} = \frac{d}{dt}((0.75 \text{ m/s}^3)t^2) = 2(0.75)t = 1.5t \text{ m/s}^2 \hat{\imath} \)

This indicates that the acceleration changes linearly with time, increasing as time progresses. Understanding acceleration helps in analyzing how the system's movement evolves and is especially useful when calculating the net force acting upon it.

Net External Force

The net external force acting on a system is crucial as it causes changes in the motion of the system's center of mass. It can be defined as the product of the total mass and the acceleration of the center of mass using Newton's second law:

• \( F_{net} = M \cdot a_{cm} \)

In this exercise, at \( t = 3.0 \) s, the acceleration is calculated as:

• \( a_{cm} = 1.5 \times 3.0 = 4.5 \text{ m/s}^2 \)

With the total mass \( M = 1.25 \text{ kg} \), the net external force is:

• \( F_{net} = 1.25 \times 4.5 \hat{\imath} = 5.625 \text{ N} \hat{\imath} \)

This force is what drives the system's center of mass to change its velocity, demonstrating the direct relationship between applied force, mass, and resulting acceleration.