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Chapter 13: Problem 77

A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a \(0.200-\mathrm{kg}\) pan is suspended from its lower end. A butcher drops a 2.2\(\cdot \mathrm{kg}\) 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 imto 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) 0.1987 m; (c) 0.49 s.

Step by step solution

01

Determine the final speed after collision

To find the speed of the pan and steak immediately after the collision, use the conservation of momentum principle: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]where \( m_1 = 0.200\, \text{kg} \) is the mass of the pan and \( v_1 = 0 \) (since it was initially at rest), and \( m_2 = 2.2\, \text{kg} \) is the mass of the steak. The steak's velocity \( v_2 \) can be found using the equation for gravitational potential energy converted to kinetic energy as it falls: \[ v_2 = \sqrt{2gh} = \sqrt{2 \times 9.8\, \text{m/s}^2 \times 0.4\, \text{m}} = 2.8\, \text{m/s} \].Plugging these values into the momentum equation gives:\[ 0.200 \times 0 + 2.2 \times 2.8 = (0.200 + 2.2) v_f \]\[ v_f = \frac{2.2 \times 2.8}{2.4} \approx 2.57\, \text{m/s} \].
02

Find the amplitude of the motion

The amplitude of the subsequent motion can be determined using the maximum kinetic energy right after the collision being converted into potential energy in the spring.The kinetic energy just after the collision is:\[ KE = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 2.4 \times (2.57)^2 \approx 7.86\, \text{J} \].This equals the spring potential energy at the amplitude:\[ PE_s = \frac{1}{2} k A^2 \].Setting these equal:\[ 7.86 = \frac{1}{2} \times 400 \times A^2 \].Solve for \( A \):\[ A = \sqrt{\frac{7.86 \times 2}{400}} \approx 0.1987\, \text{m} \].
03

Calculate the period of the motion

The period \( T \) of simple harmonic motion for a mass on a spring is given by the equation:\[ T = 2\pi \sqrt{\frac{m}{k}} \].Here \( m = m_1 + m_2 = 2.4\, \text{kg} \) and \( k = 400\, \text{N/m} \).Substitute these values into the formula:\[ T = 2\pi \sqrt{\frac{2.4}{400}} \approx 0.49\, \text{s} \].

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Key Concepts

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

Conservation of Momentum
In physics, the conservation of momentum is a crucial principle describing how the total momentum of a closed system remains constant, provided no external forces are acting on it.

When a 2.2 kg steak is dropped onto a 0.200 kg pan, connected to a spring, the principle of conservation of momentum allows us to calculate the speed of the system right after the steak collides with the pan.
  • The momentum before the collision is the sum of the individual momenta: the pan (initially at rest) has zero momentum, while the steak gains momentum as it falls.
  • Once the steak hits the pan, the two masses move together as one combined object. We calculate the final speed of this system using the equation: \( m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \) where \( v_f \) is the speed right after the collision.

In this case, the speed of the pan and steak system is 2.57 m/s. The conservation of momentum provides a powerful tool to understand collisions and similar interactions.
Spring Constant
The spring constant, denoted as \( k \), measures how stiff or flexible a spring is. In a simple harmonic motion scenario, this constant plays an essential role.
  • The spring in this exercise has a constant of 400 N/m, indicating the stiffness required to compress or extend the spring by 1 meter.
  • This constant directly affects how the system behaves: a stiffer spring (higher \( k \)) leads to a quicker oscillation and a greater force requirement for a given displacement.

Understanding the spring constant is vital because it helps determine other characteristics of the system, such as the amplitude and the period of the oscillation. By using potential energy stored in the spring ( \( PE_s = \frac{1}{2} k A^2 \)) and equating it to the kinetic energy right after the collision, you can solve for the amplitude of the system's motion.
Kinetic and Potential Energy
In the context of simple harmonic motion, kinetic and potential energy are exchanged continually. Right after the steak collides with the pan, the kinetic energy ( \( KE = \frac{1}{2} mv^2 \)) is at its maximum as the system moves with speed \( v_f \).
  • This kinetic energy, calculated to be 7.86 J, is then converted into potential energy as the spring stretches or compresses.
  • The energy stored in the spring at maximum displacement is the potential energy ( \( PE_s \)), calculated using \( PE_s = \frac{1}{2} k A^2 \).

As the system continues its oscillations, energy shifts back and forth. This continuous exchange between kinetic and potential energy is what sustains the motion, illustrating why this behavior is called "simple harmonic motion." Calculating these energy forms provides insights into the amplitude and motion characteristics of the system.
Period of Oscillation
The period of oscillation is the time it takes to complete one full cycle of the motion. For a mass-spring system in simple harmonic motion, the period depends on both the mass ( \( m \)) of the system and the spring constant ( \( k \)).
  • The equation \( T = 2\pi \sqrt{\frac{m}{k}} \) describes this relationship. Here, a smaller mass or a stiffer spring leads to a shorter period, implying faster oscillations.
  • In this example, with a total mass of 2.4 kg and a spring constant of 400 N/m, the period is calculated to be approximately 0.49 seconds.

Understanding the period is key to predicting the motion's timing, frequency, and overall dynamic behavior. It provides an essential characteristic of the oscillating system, enabling precise modeling and predictions of future states.

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Most popular questions from this chapter

A uniform, solid metal disk of mass 6.50 \(\mathrm{kg}\) and diameter 24.0 \(\mathrm{cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.23 \(\mathrm{N}\) tangent to the rim of the disk to turn it by \(3.34^{\circ},\) thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

(a) What is the change \(\Delta T\) in the period of a simple penduIum when the acceleration of gravity \(g\) changes by \(\Delta g ?\) (Hint: The new period \(T+\Delta T\) is obtained by substituting \(g+\Delta g\) for \(g :\) $$\boldsymbol{T}+\Delta \boldsymbol{T}=2 \pi \sqrt{\frac{\boldsymbol{L}}{\boldsymbol{g}+\Delta \boldsymbol{g}}}$$ To obtain an approximate expression, expand the factor \((g+\Delta g)^{-1 / 2}\) using the binomial theorem (Appendix B) and keep only the first two terms: $$(g+\Delta g)^{-1 / 2}=g^{-1 / 2}-\frac{1}{2} g^{-3 / 2} \Delta g+\cdots$$ The other terms contain higher powers of \(\Delta g\) and are very small if \(\Delta g\) is small.) Express your result as the fractional change in period \(\Delta T / T\) in terms of the fractional change \(\Delta g / g .\) (b) A pendulum clock keeps correct time at a point where \(g=9.8000 \mathrm{m} / \mathrm{s}^{2},\) but is found to lose 4.0 \(\mathrm{s}\) each day at a higher elevation. Use the result of part (a) to find the approximate value of \(g\) at this new location.

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and reased, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A \(0.400-\mathrm{kg}\) object undergoing \(\mathrm{SHM}\) has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m}\) . What is the time for one oscillation?

Weighing Astronauts. This procedure has actually been used to "weigh" astronauts in space. A \(42.5-\mathrm{kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?
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