Chapter 6: Problem 16
Two small balls, each of mass \(m=0.2 \mathrm{~kg}\), hang from strings oflength \(L=1.5\) \(\mathrm{m}\). The left ball is released from rest with\(\theta=-45^{\circ}\). As a result of the initial collision the right ballswings through a maximum angle of \(30^{\circ}\). Determine the coefficient ofrestitution between the two balls.
Short Answer
Expert verified
The coefficient of restitution is approximately 0.66.
Step by step solution
01
Energy Analysis of the Left Ball
Calculate the potential energy stored in the left ball at the angle \(\theta = -45^\circ\). At its highest point, all the energy is potential:\[U_1 = m g L (1 - \cos \theta_1)\]where \(g = 9.8 \, \text{m/s}^2\). Substituting the given values:\[U_1 = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.5 \, \text{m} \times (1 - \cos(-45^\circ)) \approx 0.618 \, \text{J}\]
02
Velocity of the Left Ball Before Collision
Convert the potential energy to kinetic energy just before the collision:\[K_1 = \frac{1}{2}mv^2 = U_1\]Solve for velocity \(v\):\[v = \sqrt{2gL(1 - \cos \theta_1)}\]Substitute the known values:\[v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.5 \, \text{m} \times (1 - \cos(-45^\circ))} \approx 3.08 \, \text{m/s}\]
03
Energy Analysis of the Right Ball
Determine the potential energy at the maximum height of the right ball when it reaches \(\theta_2 = 30^\circ\):\[U_2 = m g L (1 - \cos \theta_2)\]Substitute the known values:\[U_2 = 0.2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.5 \, \text{m} \times (1 - \cos(30^\circ)) \approx 0.264 \, \text{J}\]
04
Velocity of the Right Ball After Collision
Convert the potential energy of the right ball at its highest point back to kinetic energy at the lowest point right after collision:\[K_2 = U_2 = \frac{1}{2}mv_{2f}^2\]Solve for velocity \(v_{2f}\):\[v_{2f} = \sqrt{2gL(1 - \cos \theta_2)}\]Substitute the known values:\[v_{2f} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.5 \, \text{m} \times (1 - \cos(30^\circ))} \approx 2.04 \, \text{m/s}\]
05
System of Equations and Coefficient of Restitution
Apply the conservation of momentum and coefficient of restitution (e) as two equations. For linear momentum:\[mv_1 = mv_{1f} + mv_{2f}\]And for the coefficient of restitution:\[e = \frac{v_{2f} - v_{1f}}{v_1}\]From previous steps, we know \(v_1\) and \(v_{2f}\), and \(v_{1f} = 0\) since the left ball stops:\[0.2 \times v_1 = 0.2 \times v_{2f}\]Thus, solve for \(e\):\[e = \frac{v_{2f}}{v_1} = \frac{2.04}{3.08} \approx 0.66\]
06
Conclusion
After solving the equations, the coefficient of restitution between the two balls is \(e \approx 0.66\), indicating a moderately elastic collision.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Energy Conservation
In this problem, energy conservation plays a key role in the analysis of the movement of the balls. Energy conservation is the principle that states that the total energy in an isolated system remains constant over time, as long as no external forces are acting on it. This concept is often divided into potential and kinetic energy components, which can transform into one another depending on the state of the system.
In the case of this exercise, the left ball starts with purely potential energy at its highest point due to its vertical height relative to the lowest point. When released, the potential energy is converted to kinetic energy as it descends. Just before the collision, all this energy has shifted from potential to kinetic - the ball's velocity at this point can be determined using this energy transformation.
After the collision, the situation reverses partly – the kinetic energy of the right ball is converted back into potential energy as it rises to its maximum height. Understanding how energy transitions from potential to kinetic and vice versa is crucial to solving problems involving motion and collisions.
Momentum Conservation
Momentum conservation is another fundamental principle that aids in resolving this problem. Momentum is a vector quantity defined as the product of an object's mass and velocity. In an isolated system, where no external forces are acting, the total momentum before any interaction is equal to the total momentum afterward.
During the collision of the two balls, momentum conservation helps us set up an equation that describes the relationship between their velocities before and after the collision. The equation used: \[mv_1 = mv_{1f} + mv_{2f}\] describes the transfer of momentum between the left and right ball. This equation allows us to track how the speed of each ball changes through the collision.
By combining this equation with the coefficient of restitution, which describes how "bouncy" a collision is, we derive the final result. Hence, conservation of momentum helps to quantify how the motion's energy is shared between the interacting objects.
Elastic Collision
An elastic collision is a concept where two objects collide and then separate without any loss of kinetic energy in the system. Practically, it means objects bounce back from each other without getting deformed or generating heat. The coefficient of restitution, denoted by \(e\), measures the elasticity of a collision.
In this problem, after releasing the left ball, it collides with the stationary right ball. The coefficient of restitution is calculated using the velocities before and after the collision:\[e = \frac{v_{2f}}{v_1}\]The calculation shows a coefficient of about 0.66, indicating a moderately elastic collision. This implies that while some kinetic energy is conserved, a portion of it could have been transformed to other forms, like potential energy in the aftermath.
Understanding whether a collision is fully elastic, partially elastic, or completely inelastic helps in forecasting how the objects behave post-collision, especially in terms that involve mechanically reversible processes.
Potential Energy
Potential energy is a type of stored energy related to an object's position within a force field, typically a gravitational field. In simple terms, it can be thought of as the energy an object has due to its height above a reference point. For the balls in this problem:\[U = mgL(1 - \cos \theta)\]This formula calculates the potential energy based on the gravitational pull and angular displacement of the object.
Initially, the left ball has maximum potential energy when it is pulled back to a 45-degree angle. Once it's released, this energy converts into kinetic energy.
Similarly, after the collision, the potential energy of the right ball reaches its maximum when it swings up to a 30-degree angle. Knowing the initial and final potential energies helps deduce other quantities, such as velocity and height, crucial for solving dynamic problems.
Kinetic Energy
Kinetic energy is the energy of motion, essential in problems involving moving objects. With the formula:\[K = \frac{1}{2}mv^2\]Kinetic energy is proportional to the mass of the object and the square of its velocity.
In the exercise, as the left ball is released and starts moving, its potential energy gradually shifts to kinetic energy, which reaches a maximum just before the collision. This transformed energy determines the velocity of the ball – a critical step in finding how it will interact with the right ball.
Post-collision, the kinetic energy of the right ball is used to calculate how high and fast it will go, as some of this energy shifts back into potential energy. Altogether, these energy transformations play a central role in analyzing motions and collisions in physics.
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