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Chapter 17: Problem 24

A \(4.00-\mathrm{kg}\) block is suspended from a spring with a force constant of \(5.00 \mathrm{~N} / \mathrm{cm}\). A \(50.0-\mathrm{g}\) bullet is fired into the block from below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and comes to rest in the block. (a) Find the amplitude of the resulting simple harmonic motion. (b) What fraction of the original kinetic energy of the bullet appears as mechanical energy in the oscillator?

Short Answer

Expert verified
The amplitude of the resulting simple harmonic motion is \(x\) and the fraction of the original kinetic energy of the bullet that appears as mechanical energy in the oscillator is \(frac{ME_o}{KE_b}\).

Step by step solution

01

Conservation of linear momentum

The momentum of the block and bullet before the bullet hits the block is equal to the momentum of the block and bullet after the bullet hits the block. This gives the equation \(m_b*v_b = (m_b + m)*v\) where \(m_b\) is the mass of the bullet, \(m\) is the mass of the block, \(v_b\) is the initial velocity of the bullet and \(v\) is the velocity of the block and bullet system after the bullet hits the block. We can solve this equation for \(v\).
02

Calculate the maximum height

The kinetic energy of the block-bullet system is converted into potential energy when the system reaches its maximum height. This gives the equation \(0.5*(m_b + m)*v^2 = (m_b + m)*g*h + 0.5*k*x^2\) where \(g\) is the acceleration due to gravity, \(h\) is the initial height, \(k\) is the spring constant and \(x\) is the maximum displacement of the system which equals the amplitude. Because the spring was initially at rest, \(h\) is zero and we can solve this equation for \(x\).
03

Calculate the fraction of original kinetic energy

The initial kinetic energy of the bullet is given by \(KE_b=0.5*m_b*v_b^2\) and the mechanical energy in the oscillator is \(ME_o=0.5*(m_b + m)*v^2\). The fraction of the original kinetic energy appearing as mechanical energy in the oscillator is given by \(frac{ME_o}{KE_b}\). This fraction can then be calculated.

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Most popular questions from this chapter

A particle of mass \(m\) moves in a fixed plane along the trajectory \(\overrightarrow{\mathbf{r}}=\hat{\mathbf{i}} A \cos \omega t+\hat{\mathbf{j}} A \cos 3 \omega t .\) (a) Sketch the trajectory of the particle. (b) Find the force acting on the particle. Also find (c) its potential energy and \((d\) ) its total energy as functions of time. \((e)\) Is the motion periodic? If so, find the period.

(a) When the displacement is one-half the amplitude \(x_{m}\), what fraction of the total energy is kinetic and what fraction is potential in simple harmonic motion? (b) At what displacement is the energy half kinetic and half potential?

In an electric shaver, the blade moves back and forth over a distance of \(2.00 \mathrm{~mm}\). The motion is simple harmonic, with frequency \(120 \mathrm{~Hz}\). Find \((a)\) the amplitude, \((b)\) the maximum blade speed, and ( \(c\) ) the maximum blade acceleration.

A \(12.3-\mathrm{kg}\) particle is undergoing simple harmonic motion with an amplitude of \(1.86 \mathrm{~mm}\). The maximum acceleration experienced by the particle is \(7.93 \mathrm{~km} / \mathrm{s}^{2} .(a)\) Find the period of the motion. (b) What is the maximum speed of the particle? ( \(c\) ) Calculate the total mechanical energy of this simple harmonic oscillator.

Two particles execute simple harmonic motion of the same amplitude and frequency along the same straight line. They pass one another when going in opposite directions each time their displacement is half their amplitude. Find the phase difference between them.
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