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Chapter 14: Problem 85

CP SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) 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 \(\mathrm{m}\) . 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) Speed after collision: 2.57 m/s (b) Amplitude: 0.2 m (c) Period: 0.487 s

Step by step solution

01

Find the Speed Before Collision

Use the conservation of energy principle. The initial potential energy of the steak at a height of 0.40 m is converted into kinetic energy just before it hits the pan. So we have: \[ mgh = \frac{1}{2}mv^2 \]Substitute \(m = 2.2\, \text{kg}, \ g = 9.8\, \text{m/s}^2, \ h = 0.4\, \text{m}\) into the equation:\[ 2.2 \times 9.8 \times 0.4 = \frac{1}{2} \times 2.2 \times v^2 \]Solve for \(v\):\[ v^2 = \frac{2 \times 2.2 \times 9.8 \times 0.4}{2.2} \]\[ v = \sqrt{2 \times 9.8 \times 0.4} \approx 2.8\, \text{m/s} \]
02

Calculate Speed After Collision

In a totally inelastic collision, momentum is conserved. The total mass after collision becomes \(M = 2.2 + 0.2 = 2.4\, \text{kg}\). Using the conservation of momentum:\[ m_1v_1 = Mv' \]\( m_1 = 2.2\, \text{kg}, \ v_1 = 2.8\, \text{m/s}, \ M = 2.4\, \text{kg} \)\[ 2.2 \times 2.8 = 2.4 \times v' \]Solve for \(v'\):\[ v' = \frac{2.2 \times 2.8}{2.4} \approx 2.57\, \text{m/s} \]
03

Determine the Amplitude of Motion

Use the conservation of mechanical energy. The initial kinetic energy of the system is converted to potential energy at maximum displacement (amplitude):\[ \frac{1}{2}Mv'^2 = \frac{1}{2}kA^2 \]Substitute \(M = 2.4\, \text{kg}, \ k = 400\, \text{N/m}, \ v' = 2.57\, \text{m/s} \):\[ \frac{1}{2} \times 2.4 \times (2.57)^2 = \frac{1}{2} \times 400 \times A^2 \]\[ 2.4 \times 6.6049 = 400A^2 \]Solve for \(A\):\[ A^2 = \frac{15.85}{400} \]\[ A \approx 0.2 \text{ m} \]
04

Calculate the Period of SHM

The period \(T\) of a simple harmonic motion is given by:\[ T = 2\pi \sqrt{\frac{M}{k}} \]Substitute \(M = 2.4\, \text{kg}, \ k = 400\, \text{N/m} \):\[ T = 2\pi \sqrt{\frac{2.4}{400}} \]\[ T = 2\pi \sqrt{0.006} \]\[ T \approx 0.487\, \text{s} \]

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Key Concepts

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

Inelastic Collision
In an inelastic collision, the objects involved stick together after colliding, rather than bouncing apart. This specific kind of collision is characterized by a loss of kinetic energy, but crucially, momentum is conserved.
\[ \text{Initial momentum} = \text{Final momentum} \]
During the collision of the 2.2-kg steak with the 0.2-kg pan, the masses combine to form a single mass of 2.4 kg. The speed of the combined system directly after the collision can be determined using the conservation of momentum. It shows how, even if energy is "lost" in the form of heat or sound, momentum continues to play a key role in determining motion right after the impact.
  • A characteristic, distinguishing point is that the energy is not conserved during inelastic collisions, only momentum is.
  • Due to their nature, inelastic collisions usually result in some permanent deformation or generation of heat.
Conservation of Energy
The principle of conservation of energy is central in solving dynamics and mechanics problems. Energy can neither be created nor destroyed; it can only transform from one form to another. This is particularly helpful in analyzing mechanical systems involving potential and kinetic energy transitions.
In the given problem, the initial potential energy of the steak from a height of 0.4 m is transformed into kinetic energy as it falls.
  • The calculation of this transformation helps to find the steak's speed just before impact.
  • Similarly, after the collision, we analyze energy again to determine how the kinetic energy transforms to potential when the system reaches its maximum displacement (or amplitude).
In simple harmonic motion (SHM), this transformation back and forth between kinetic and potential forms is constant and predictable.
Amplitude of Motion
Amplitude in simple harmonic motion (SHM) refers to the maximum displacement from the equilibrium position. In essence, it's the farthest point the system moves to either side during oscillation.
After the inelastic collision in this exercise, the amplitude of motion is determined by equating the kinetic energy converted into the spring's potential energy at the amplitude's displacement.
  • The equation \( \frac{1}{2}Mv'^2 = \frac{1}{2}kA^2 \) is used to find the amplitude.
  • This shows us that a larger initial speed after the collision would lead to a greater amplitude.
Amplitude does not depend on factors such as phase or the system's mass, but rather the energy provided to it.
Momentum Conservation
Momentum, a product of an object's mass and velocity, is a core concept in physics. In any isolated system, the total momentum before and after a collision remains constant. This is the foundation of momentum conservation.
For the steak-pan problem, this concept is showcased during the inelastic collision step. Before the collision, only the steak has momentum; however, after the collision, this momentum is shared between the combined steak and pan mass.
  • The formula used is \( m_1v_1 = Mv' \), reflecting this principle of shared momentum.
  • This conservation law becomes particularly evident in systems where external forces (like friction or external presses) are negligible.
Understanding momentum conservation allows one to predict the motion of various bodies engaged in collisions and is pivotal in mechanics.

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Most popular questions from this chapter

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

BID (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note \(B\) flat, which has a fre- quency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angu- lar frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm} .\) (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding onto the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) 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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