Problem 73 - University Physics with Modern Physics

Question

CP A 10.0 .0 -kg mass is traveling to the right with a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second 10.0 -kg mass that is initially at rest but is attached to a light spring with force constant 110.0 \(\mathrm{N} / \mathrm{m}\) . (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

Step-by-Step Solution

Answer

Frequency: \(0.373 \, \text{Hz}\), Amplitude: \(0.43 \, \text{m}\), Period: \(2.68 \, \text{s}\), Return time: \(1.34 \, \text{s}\).

Step 1: Determine the Velocity after Collision

Since the two masses collide and stick together, we can use the conservation of momentum to find the velocity of the combined mass immediately after the collision. The initial momentum is the product of the mass and velocity of the moving mass: \[ p = m_1 v_1 = 10.0 \times 2.00 = 20.0 \, \text{kg m/s} \]. The final momentum for the system is the product of the combined mass and its velocity \( v \): \[ (m_1 + m_2) v = 20.0 \, \text{kg m/s} \].The combined mass is \( 20.0 \, \text{kg} \), so:\[ 20.0 v = 20.0 \, \text{kg m/s} \Rightarrow v = 1.00 \, \text{m/s} \].Thus, the combined mass moves at \( 1.00 \, \text{m/s} \).

Step 2: Find the Angular Frequency

The angular frequency \( \omega \) of a mass-spring system is given by the formula:\[ \omega = \sqrt{\frac{k}{m}} \].Here \( k = 110.0 \, \text{N/m} \) and \( m = m_1 + m_2 = 20.0 \, \text{kg} \):\[ \omega = \sqrt{\frac{110.0}{20.0}} = \sqrt{5.5} \, \text{rad/s}. \]

Step 3: Calculate the Frequency

The frequency \( f \) is related to the angular frequency by \( \omega = 2\pi f \). Hence,\[ f = \frac{\omega}{2\pi} = \frac{\sqrt{5.5}}{2\pi} \approx 0.373 \, \text{Hz}. \]

Step 4: Determine the Period of Oscillation

The period \( T \) is the reciprocal of the frequency:\[ T = \frac{1}{f} \approx \frac{1}{0.373} \approx 2.68 \, \text{s}. \]

Step 5: Calculate Amplitude of Oscillation

The amplitude of oscillation can be determined from the initial conditions using energy conservation. Maximum kinetic energy immediately after collision becomes the maximum potential energy in the spring:\[ \frac{1}{2}mv^2 = \frac{1}{2}kA^2 \].Using \(m = 20.0\, \text{kg}, v = 1.00\, \text{m/s}, k = 110.0\, \text{N/m}\):\[ 20.0 \times (1.00)^2 = 110.0 \times A^2 \Rightarrow A^2 = \frac{20.0}{110.0} \Rightarrow A \approx 0.43 \, \text{m}. \]

Step 6: Calculate Time to Return to Initial Position

This time corresponds to half the period of one complete cycle since the system starts from a point and returns to it after half a cycle. Thus, the time is:\[ \frac{T}{2} \approx \frac{2.68}{2} \approx 1.34 \, \text{s}. \]

Key Concepts

Conservation of Momentum

The conservation of momentum is a fundamental principle of physics. It tells us that the total momentum of a closed system remains constant unless acted upon by an external force. In this exercise, we start with a 10 kg mass moving at 2 m/s, colliding with another stationary 10 kg mass. Post-collision, the two masses stick together. To see how momentum conservation works, calculate the system's initial momentum. We use the formula:\[ p = m_1 v_1 + m_2 v_2 \]where:

• \( m_1 = 10 \, \text{kg}, \) the moving mass,
• \( v_1 = 2 \, \text{m/s}, \) the velocity of the moving mass,
• \( m_2 = 10 \, \text{kg}, \) the resting mass,
• \( v_2 = 0 \, \text{m/s}. \) the velocity of the resting mass.

This results in an initial momentum of 20 kg·m/s. After they collide and stick together, their combined mass (20 kg) has this same total momentum. We thus calculate their velocity post-collision by rearranging \( (m_1 + m_2) v = 20 \, \text{kg·m/s} \) to find a velocity of 1 m/s for the combined mass.

Mass-Spring System

A mass-spring system is a classic model used to describe oscillatory motion. It consists of a mass attached to a spring that can compress and extend. In this scenario, the masses stick together and attach to a spring with a force constant \( k = 110 \, \text{N/m}. \) The properties of this system govern its oscillatory behavior.The mass-spring system follows Hooke's law, which states:\[ F = -kx \]where \( F \) is the force applied to the spring and \( x \) is the displacement from equilibrium. The negative sign indicates that the force is restorative, aiming to bring the system back to equilibrium.The main parameters describing motion in the system include:

• Force constant \( k \): Indicates how "stiff" the spring is. A larger \( k \) means a firmer spring.
• Mass \( m \): Combined mass after collision. Here, it is 20 kg.
• Displacement/Amplitude \( A \): The maximum extent from equilibrium.

These parameters work together to define the system's oscillations, frequency, and period.

Angular Frequency

Angular frequency \( \omega \) characterizes the speed of oscillation, specifying how fast the system oscillates back and forth. It's part of what's known as harmonic motion and relates to the stiffness of the spring and the mass attached.In mass-spring systems, angular frequency is given by:\[ \omega = \sqrt{\frac{k}{m}} \]where \( k \) is the force constant (110 N/m) and \( m \) is the total mass (20 kg). Here, substituting the given values yields:\[ \omega = \sqrt{\frac{110}{20}} = \sqrt{5.5} \, \text{rad/s} \]Angular frequency helps to determine other vital characteristics of the system:

• It's directly related to frequency (how many cycles occur in a second).
• It helps to compute the period, the time it takes to complete one cycle.

Period of Oscillation

The period of oscillation \( T \) is the time taken for the mass-spring system to complete one full cycle of motion. Understanding this helps predict when the mass returns to its starting position.Using the relation between frequency and period:\[ T = \frac{1}{f} \]where \( f \) is the frequency calculated from:\[ f = \frac{\omega}{2\pi} \]From the previous angular frequency computation \( \omega = \sqrt{5.5} \), we find that:\[ f \approx 0.373 \, \text{Hz} \] Thus, the period is calculated as:\[ T \approx \frac{1}{0.373} \approx 2.68 \, \text{s} \]The period tells us the oscillation's rhythm and allows us to determine how long it will take for the system to return to its original position, which is half a period or approximately 1.34 seconds.