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A spring of negligible mass and force constant \(k =\) 400 N/m is hung vertically, and a 0.200-kg pan is suspended from its lower end. A butcher drops a 2.2-kg steak onto the pan from a height of 0.40 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

Short Answer

Expert verified
(a) 2.57 m/s; (b) 0.263 m; (c) 0.487 s.

Step by step solution

01

Determine Potential Energy of the Steak

First, calculate the potential energy of the steak before it is dropped. The potential energy (PE) is calculated using the formula:\[PE = mgh\]where \(m = 2.2\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(h = 0.40\,\text{m}\). Plug these values into the formula to find:\[PE = 2.2 \times 9.81 \times 0.40 = 8.6\,\text{Joules}\].
02

Calculate Speed After Collision Using Energy Conservation

With the potential energy we calculated, we'll determine the speed just before the collision. Use the conservation of energy principle where the entire potential energy converts to kinetic energy (KE) just before hitting the pan. Let \(v\) be the speed; the equation becomes:\[\frac{1}{2}mv^2 = 8.6\]Plug in \(m = 2.2\,\text{kg}\) and solve for \(v\):\[v = \sqrt{\frac{2 \times 8.6}{2.2}} = 2.8\,\text{m/s}\].
03

Use Conservation of Momentum for Speed After Collision

Next, apply the conservation of linear momentum to find the speed of the pan and steak immediately after the collision. Let \(V\) be the speed post-collision:\[m_1v_1 + m_2v_2 = (m_1 + m_2)V\]where \(m_1 = 0.200\,\text{kg}\), \(m_2 = 2.2\,\text{kg}\), \(v_1 = 0\,\text{m/s}\), and \(v_2 = 2.8\,\text{m/s}\). Solve for \(V\):\[V = \frac{2.2 \times 2.8}{2.4} = 2.5667\,\text{m/s}\].
04

Determine Amplitude of Motion

Use energy conservation between kinetic and elastic potential energy to find the amplitude (\(A\)) of the motion after the collision. The total kinetic energy after collision (now in the form of elastic potential energy) is:\[\frac{1}{2}(m_1 + m_2)V^2 = \frac{1}{2}kA^2\]Insert known values \(k = 400\,\text{N/m}\), \(V = 2.5667\,\text{m/s}\). Solve for \(A\):\[A = \sqrt{\frac{(0.200 + 2.2) \times 2.5667^2}{400}} = 0.263\,\text{m}\].
05

Calculate Period of Simple Harmonic Motion

Finally, calculate the period using the formula for the period \(T\) of a mass-spring system:\[T = 2\pi \sqrt{\frac{M}{k}}\]where \(M = m_1 + m_2 = 2.4\,\text{kg}\). Substitute the values:\[T = 2\pi \sqrt{\frac{2.4}{400}} = 0.487\,\text{s}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The concept of conservation of energy is essential when analysing systems like mass-spring systems. It states that the total energy in a closed system remains constant, though it may change forms. In this context, potential energy (PE) can convert into kinetic energy (KE), and vice-versa. For example, before the steak hits the pan, it possesses gravitational potential energy due to its height. As it falls, this energy transforms into kinetic energy at the moment of impact. Post-collision, the energy further transitions into elastic potential energy stored in the spring, initiating simple harmonic motion. Understanding these transformations helps predict the system's behaviour, such as the speed of the pan, the amplitude of motion, and overall mechanical energy during the motion.
Key Points:
  • Energy transitions within a system can be used to predict motion.
  • Gravitational potential energy changes to kinetic energy during free fall.
  • Post-collision energy is stored as elastic potential energy in the spring.
Momentum Conservation
Conservation of momentum is crucial in understanding the motion post-collision between the steak and the pan. Momentum, a product of mass and velocity, is conserved in isolated systems, regardless of the interactions occurring internally. During the impact, the momentum of the falling steak transfers to the combined mass of the steak and pan, allowing us to calculate the post-collision velocity. Initial momentum was contained solely by the steak, but after impact, the momentum redistributes between both the pan and the steak. Knowing this principle allows us to solve for quantities like velocity after collision by equating the initial and final momentum.
Key Points:
  • Momentum is conserved in isolated systems.
  • Calculating post-collision velocities involve initial momentum considerations.
  • This principle helps predict velocities and directions in collisions.
Mass-Spring System
The mass-spring system described here involves a spring of known spring constant and a combined mass of the pan and steak. Such systems are classic examples of simple harmonic motion (SHM), where the mass's oscillations around an equilibrium position are influenced by the spring's elasticity. The motion's analysis typically involves understanding how potential and kinetic energy interchangeably manifest in the system. As the pan and steak system oscillates, the energy continues to alternate between kinetic and elastic potential, with the spring's restorative force acting to return the system to equilibrium. This cycle repeats consistently given no external damping forces.
Important Considerations:
  • Mass-spring systems are ideal for studying simple harmonic motion.
  • Understanding spring constant and mass helps predict oscillation characteristics.
  • The system's energy alternates between kinetic and elastic potential.
Kinetic Energy
Kinetic energy is the energy of motion and becomes particularly significant as the steak falls and impacts the pan. Initially, the falling steak converts all its gravitational potential energy into kinetic energy, gaining speed in accordance with energy conservation. After the inelastic collision with the pan, energy conservation principles guide us to consider the kinetic energy of the whole system (pan and steak combined) to predict future motion, like velocity or amplitude of oscillations. The essential aspect of kinetic energy in this scenario involves understanding how potential energy transition results in a velocity that is crucial for determining further dynamic properties of motion.
Key Insights:
  • Kinetic energy represents motion within the system.
  • The conversion from potential to kinetic energy is a key step in falling bodies.
  • Post-collision, kinetic energy influences further motion characteristics.
Elastic Potential Energy
In the mass-spring system setup, elastic potential energy becomes a central component post-collision. Upon impact, the kinetic energy of the steak and pan is transferred into the spring, compressing it, and converting kinetic energy into elastic potential energy. This energy is stored in the spring as it deforms, and is directly related to the spring constant and the displacement (amplitude of motion). Elastic potential energy provides the force needed to bring the system back to equilibrium, perpetuating the simple harmonic motion. Understanding how energy is stored and released by the spring is crucial for analyzing oscillatory systems like this.
Key Elements:
  • Elastic potential energy is stored when spring is compressed or stretched.
  • The spring constant affects how much energy is stored.
  • The energy contributes to restoring force, driving the system's motion.

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Most popular questions from this chapter

A 5.00-kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s. (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 m above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

You hang various masses \(m\) from the end of a vertical, 0.250-kg spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. Since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T =\) 2\(\pi\sqrt{ m/k }\) with \(m + m_\mathrm{eff}\), where \(m_\mathrm{eff}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass m and measure the time for 10 complete oscillations, obtaining these data: (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_\mathrm{eff}\) of the spring's mass? (e) If a 0.450-kg mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. (a) Find the force constant of the spring. The fish is now pulled down 5.00 cm and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

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