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Chapter 9: Problem 57

a ball of mass \(m=60 \mathrm{~g}\) is shot with speed \(v_{i}=\) \(22 \mathrm{~m} / \mathrm{s}\) into the barrel of a spring gun of mass \(M=240 \mathrm{~g}\) initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after the ball stops in the barrel? (b) What fraction of the initial kinetic energy of the ball is stored in the spring?

Short Answer

Expert verified
(a) The speed of the spring gun is 4.4 m/s. (b) 80% of the initial kinetic energy is stored in the spring.

Step by step solution

01

Understand the Conservation of Momentum

In a system with no external forces, the total momentum before an event must equal the total momentum after. Here, the momentum before the ball hits the gun is given by the ball's momentum, since the gun is at rest:\[ p_{ ext{initial}} = m \cdot v_i \]
02

Calculate System's Initial Momentum

The initial momentum of the system is the momentum of the ball, as the gun is initially at rest:\[ p_{ ext{initial}} = (0.06 \, \text{kg}) \times (22 \, \text{m/s}) = 1.32 \, \text{kg} \cdot \text{m/s} \]
03

Apply Conservation of Momentum

After the collision, the combined mass of the ball and gun moves with a common velocity \( v_f \). Using conservation of momentum:\[ m \cdot v_i = (m + M) \cdot v_f \]Solving for \( v_f \):\[ v_f = \frac{m \cdot v_i}{m + M} \]
04

Calculate Final Velocity of System

Substituting the masses and initial velocity into the equation:\[ v_f = \frac{0.06 \, \text{kg} \times 22 \, \text{m/s}}{0.06 \, \text{kg} + 0.24 \, \text{kg}} = \frac{1.32}{0.3} = 4.4 \, \text{m/s} \]
05

Determine Initial Kinetic Energy of Ball

The initial kinetic energy of the ball is given by:\[ KE_{ ext{initial}} = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 0.06 \, \text{kg} \times (22 \, \text{m/s})^2 = 14.52 \, \text{J} \]
06

Compute Final Kinetic Energy of System

The kinetic energy of the system after the collision is given by:\[ KE_{ ext{final}} = \frac{1}{2} (m + M) v_f^2 = \frac{1}{2} \times 0.3 \, \text{kg} \times (4.4 \, \text{m/s})^2 = 2.904 \, \text{J} \]
07

Determine Energy Stored in Spring

The energy stored in the spring is the difference between the initial and final kinetic energies:\[ E_{ ext{spring}} = KE_{ ext{initial}} - KE_{ ext{final}} = 14.52 \, \text{J} - 2.904 \, \text{J} = 11.616 \, \text{J} \]
08

Calculate Fraction of Initial Energy Stored in Spring

The fraction of the initial kinetic energy stored in the spring is:\[ \frac{E_{ ext{spring}}}{KE_{ ext{initial}}} = \frac{11.616 \, \text{J}}{14.52 \, \text{J}} \approx 0.800 \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It can be imagined as the amount of work required to bring an object from rest to its current speed.
  • The formula for kinetic energy is: \[ KE = \frac{1}{2} m v^2 \]where \( m \) represents mass and \( v \) is velocity.
  • In the original exercise, the initial kinetic energy of the ball is calculated using its mass of \( 60 \) grams (or \( 0.06 \) kg converted to kilograms) and its velocity \( 22 \) m/s.
This calculation yields an initial kinetic energy of \( 14.52 \) Joules. Understanding this concept helps in figuring out how much energy is transformed in a system.
Energy Conservation
The principle of energy conservation states that in a closed system, energy cannot be created or destroyed, only transformed from one form to another.
  • In the context of the problem, we see this conservation between different types of energy: kinetic energy and potential energy in the spring.
  • Initially, the ball's kinetic energy is transformed into other forms after it gets in contact with the spring gun, compressing the spring.
The total energy in the system stays the same before and after the event, illustrating how conservation principles apply even in seemingly complex interactions.
Spring Potential Energy
Spring potential energy is the stored energy in a system due to the position of the spring being compressed or extended.
  • This energy is governed by Hooke's law which states that the force required to compress or extend a spring by a certain distance is proportional to that distance.
  • In mathematical terms, the potential energy stored in a spring can be expressed as:\[ PE_{spring} = \frac{1}{2} k x^2 \]where \( k \) is the spring constant, and \( x \) is the displacement from equilibrium position.
In the exercise, the energy stored in the spring after the ball comes to rest is calculated as the difference between the initial kinetic energy of the ball and the final kinetic energy of the system, resulting in \( 11.616 \) Joules of potential energy stored in the spring.

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