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Chapter 14: Problem 83

A rifle bullet with mass \(8.00 \mathrm{~g}\) and initial horizontal velocity \(280 \mathrm{~m} / \mathrm{s}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a friction less surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of \(15.0 \mathrm{~cm} .\) After the impact, the block moves in SHM. Calculate the period of this motion.

Short Answer

Expert verified
The period of the motion is approximately 1.09 s.

Step by step solution

01

Apply conservation of momentum

Before the bullet hits, the momentum of the block is zero as it's at rest. So, the total momentum before the bullet hits is equals to the momentum of the bullet, which is \(0.008 \ \mathrm{kg} \times 280 \ \mathrm{m/s} = 2.24 \ \mathrm{kg \ m/s}\). After the impact, the bullet and the block move together hence their total momentum is still \(2.24 \ \mathrm{kg \ m/s}\). Now, the mass of the block together with the bullet is \(1 \ \mathrm{kg}\) (since 0.992 kg + 0.008 kg = 1kg). So, we can solve for final velocity, \(v_f\), of the block after the bullet hits by equating the momentum before and after the impact: \(1 \ \mathrm{kg} \times v_f = 2.24 \ \mathrm{kg \ m/s}\). Solving for \(v_f\) gives \(v_f = 2.24 \ \mathrm{m/s}\).
02

Apply Hooke's Law to find the spring constant

In the case of maximum compression, the block comes momentarily to rest, thus all its kinetic energy is transferred into potential energy stored in the spring. So, set \(0.5 \cdot m \cdot v_f^2 = 0.5 \cdot k \cdot x^2\), where \(x\) is the maximum compression, \(15.0 \ \mathrm{cm} = 0.15 \ \mathrm{m}\). Solving for \(k\) gives \(k= \frac{m \cdot v_f^2}{x^2}= \frac{1 \ \mathrm{kg} \cdot (2.24 \ \mathrm{m/s})^2}{(0.15 \ \mathrm{m})^2}= 334.222 \ \mathrm{N/m}\).
03

Calculate the period of the motion

For a mass-spring system with no damping, the period \(T\) of the motion is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). Substituting the values of \(m\) and \(k\) obtained earlier, the period becomes \(T = 2\pi\sqrt{\frac{1 \ \mathrm{kg}}{334.222 \ \mathrm{N/m}}} = 1.09 \ \mathrm{s}\).

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Key Concepts

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

Momentum Conservation
When a bullet strikes a block and they move together, the law of momentum conservation plays a crucial role. Initially, the bullet has momentum because it's moving, while the block has none since it is at rest. Momentum is calculated by the formula \( p = m imes v \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
  • Total initial momentum = bullet鈥檚 momentum.
  • After collision, both bullet and block move as one unit, thus conserving the momentum.
This principle equates the initial momentum of the bullet to the combined momentum of the bullet-block system post-collision. Thus, the final velocity can be calculated using \( m_1\times v_1 = (m_1 + m_2)\times v_f \). Understanding momentum conservation is key in predicting the velocity of the combined system.
Hooke's Law
Hooke's Law is fundamental when dealing with springs in physics. It describes the relationship between the forces applied to spring and the displacement it causes. According to Hooke's Law, the force exerted by a spring is proportional to its displacement from equilibrium position. This can be represented as \( F = -k \times x \), where \( k \) is the spring constant and \( x \) is the displacement.
  • A larger \( k \) value indicates a stiffer spring.
  • Negative sign shows the force is always in the opposite direction of displacement.
In our scenario, the spring compresses when the bullet-block system hits it, temporarily storing all kinetic energy as potential energy, showcasing Hooke's forces at work.
Spring Constant
The spring constant, \( k \), is a pivotal value that determines the stiffness of a spring. A spring with a higher spring constant is stiffer and requires a larger force to achieve the same displacement as a softer spring. The constant \( k \) is determined from the energy equation where the system's kinetic energy equals the potential energy stored in the spring.
This relationship is represented by \( \)\( 0.5 \times k \times x^2 = 0.5 \times m \times v^2 \). Solving for \( k \) following an impact enables us to determine how the spring reacts when external forces, like a bullet, contact a connected block. This constant helps us understand system dynamics in simple harmonic motions.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, calculated as \( KE = 0.5 \times m \times v^2 \), where \( m \) is mass and \( v \) is velocity. In our problem, the bullet moving towards the block possesses significant kinetic energy and transfers this energy upon impact.
  • Initial kinetic energy stems solely from the bullet.
  • After collision, kinetic energy is redistributed throughout the bullet-block system.
After the collision, as the block and bullet move together compressing the spring, their kinetic energy is transformed into potential energy, illustrating the dynamic energy conservation in systems exhibiting simple harmonic motion.
Potential Energy
Potential energy, specifically elastic potential energy in this scenario, arises when the spring is compressed or extended from its equilibrium state. It can be defined using \( PE = 0.5 \times k \times x^2 \), where \( x \) is the displacement of the spring from its rest position and \( k \) is the spring constant. At maximum spring compression, all of the block-bullet system's kinetic energy is converted into potential energy, exemplifying the energy transformation characteristics of harmonic motion.
  • The stored potential energy in the spring influences the system's return to equilibrium.
  • This stored energy determines the oscillatory motion observed in a mass-spring system.
Understanding potential energy is instrumental in determining how energy is stored and redistributed within mechanical systems, especially those showing periodic motion.

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Most popular questions from this chapter

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\). (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

You are watching an object that is moving in SHM. When the object is displaced \(0.600 \mathrm{~m}\) to the right of its equilibrium position, it has a velocity of \(2.20 \mathrm{~m} / \mathrm{s}\) to the right and an acceleration of \(8.40 \mathrm{~m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta / 2\) from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

An unhappy \(0.300 \mathrm{~kg}\) rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{~N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\). (a) If the constant \(b\) has the value \(0.900 \mathrm{~kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is its period on the surface of Mars, where \(g=3.71 \mathrm{~m} / \mathrm{s}^{2} ?\)
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