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Chapter 16: Problem 12

In this project we address a collision between two identical atoms. You will learn how to determine external and internal motion of a diatomic molecule after a collision using a combination of analytical techniques, such as conservation laws, and numerical methods to determine the motion of the molecule. We want to address a collision between two identical atoms of mass \(m\), and we assume that we may consider the atoms to be point particles. The atoms are not affected by any external forces. Here, we will first analyze a simplified model for the collision-a one dimensional model-before we analyze the full collision process. First, we address a simplified model. The system we consider consists of two atoms: atom A moves along the \(x\)-axis with the velocity \(v_{0}\), and atom B is at rest in the origin as illustrated in Fig. 16.37. The atoms do not interact before they hit each other. After the collision they are stuck to each other. (a) Find the velocity of the center of mass for the system before the collision. (b) Find the velocity of the center of mass of the system after the collision. (c) What is the change in the system's kinetic energy through the collision. Let us make the model slightly more realstic by introducing a simplified model for the interactions between the two atoms. We will here not use a full model for the interatomic interaction, but instead assume that we can model the interatomic interaction using a spring force model. When atom A reaches a distance \(b\) from atom B, the two atoms become attached by a massless spring with spring constant \(k\) and equilibrium length \(b\). The atoms remain attached with this spring throughout the collision and the subsequent motion. (d) What is the velcoity of the center of mass immediately after the atoms are attached with the spring, that is, when atom A is at the distance \(b\) from atom B? What is the change in kinetic energy for the system before and immediately after the collision? Let us now address the motion of the atoms after the attachment. The positions of the atoms are \(x_{A}\) and \(x_{B}\). (e) Show that the force on atom A is: \(F=k\left(\left(x_{A}-x_{B}\right)-b\right)\), and find a corresponding expression for the force on atom B. (f) Find expressions for the acceleration for atom A and B, and formulate the differential equations you need to solve to find the motion of the atoms, including the initial conditions. (g) Write a program to determine the positions and velocities of atom A and atom B as a function of time. Assume \(m=0.1, k=20, b=0.2, v_{0}=1.0\), and \(\Delta t=0.001\). (h) Plot the position as a function of time for the center of mass of the system and for each of the atoms. (i) What is the maximum distance between the two atoms? The collision we have addressed so far is a special case-the case of a central colilsion. Let us now address a non-central collision. First, we address a simplified model for a non-central collision, as illustrated in Fig. 16.38. Atom A moves in the \(x\)-direction along the line \(y=b\) with velocity \(v_{0}\), and atom \(\mathrm{B}\) is at rest at the origin. The atoms do not interacti until they hit each other, which occurs when atom A reaches \(x=0\). After the collision, the atoms form a diatomic molecule, and the atoms remain attached at a fixed distance \(b\) from each other. (We are not studying a model without the spring forcee, but with a non-central collision. We will add the spring force again further on to get a complete, but still simplified model). (j) What is the velocity of the center of mass and the angular velocity around the center of mass immediately after the collision? Let us now introduce a more advanced model for this collision: When atom \(\mathrm{A}\) is in the position \(x=0, y=b\), and atom \(\mathrm{B}\) is in the position \(x=0\) and \(y=0\), the two atoms become attached with a massless spring with spring constant \(k\) and equilibrium length \(b\). The atoms remain attached throughouth the subsequent motion. (k) Rewrite your program to model the motion of the atoms in this case. (1) Plot the motion of the atoms and the center of mass after the collision. \((\mathbf{m})\) Discuss the motion of the angular velocity for the rotation about the center of mass for the motion after the collision.

Short Answer

Expert verified
Velocity of center of mass before and after collision: \(\frac{v_0}{2}\), change in kinetic energy: \(- \frac{1}{4} mv_0^2\). Forces derived using spring models, updated differential equations, simulation program needed.

Step by step solution

01

(a) Compute Velocity of Center of Mass Before Collision

Center of mass velocity is computed using the formula: \[ v_{cm} = \frac{m_A v_A + m_B v_B}{m_A + m_B} \]Given that atom A moves with velocity \(v_0\) and atom B is at rest, plug in the values: \[ v_{cm} = \frac{mv_0 + m \cdot 0}{m + m} = \frac{mv_0}{2m} = \frac{v_0}{2} \]
02

(b) Compute Velocity of Center of Mass After Collision

After the collision, the two atoms are stuck together and move as one entity with combined mass 2m, at velocity v’. By conservation of momentum:\[ (m_A + m_B)v_{cm} = m_A v_A + m_B v_B \rightarrow 2mv’ = mv_0 + 0m \rightarrow v’ = \frac{v_0}{2} \]Hence, the velocity of the center of mass after the collision is also \( \frac{v_0}{2}\).
03

(c) Calculate Change in Kinetic Energy Through the Collision

Before collision kinetic energy: \[ KE_{initial} = \frac{1}{2} mv_0^2 \]After collision kinetic energy: \[ KE_{final} = \frac{1}{2}(2m)\left(\frac{v_0}{2}\right)^2 = \frac{1}{2}(2m)\left(\frac{v_0^2}{4}\right) = \frac{1}{4}mv_0^2 \]Change in kinetic energy: \[ \Delta KE = KE_{final} - KE_{initial} = \frac{1}{4} mv_0^2 - \frac{1}{2} mv_0^2 = - \frac{1}{4} mv_0^2 \]
04

(d) Compute Velocity of Center of Mass and Change in Kinetic Energy with Spring

When atom A and B are attached via spring at distance \(b\), center of mass velocity remains \( \frac{v_0}{2} \) since no external forces act. Kinetic energy change: \[ KE_{initial} = \frac{1}{2}mv_0^2 \]Spring attachment (still combined mass 2m at \( \frac{v_0}{2} \)): \[ KE_{immediate} = \frac{1}{2}(2m)\left(\frac{v_0}{2}\right)^2 = \frac{1}{4}mv_0^2 \]Change: \[ \Delta KE = KE_{immediate} - KE_{initial} = - \frac{1}{4} mv_0^2 \]
05

(e) Show the Force on Atom A

Spring force model for atoms A and B distance: \[ F = k\left((x_A - x_B) - b\right) \]
06

(f) Express Accelerations and Differential Equations

Using Newton's second law: \[ F_A = m\frac{d^2 x_A}{dt^2} = k((x_A - x_B) - b) \] \[ F_B = m\frac{d^2 x_B}{dt^2} = -k((x_A - x_B) - b) \]
07

(g) Write Program for Atomic Motion

Write code for solving differential equations: Initialize values for m=0.1, k=20, b=0.2, v_0=1.0, \(\Delta t = 0.001\)
08

(h) Plot Positions

Plot functions of x_A, x_B and center of mass versus time.
09

(i) Find Maximum Distance Between Atoms

Use program to find peak value of \((x_A - x_B)\) as they oscillate after collision.
10

(j) Calculate Center of Mass Velocity and Angular Velocity (Non-Central)

Center of mass velocity remains \(\frac{v_0}{2}\). Angular velocity, calculated as: \[ \omega = \frac{v_0}{b} \]
11

(k) Update Program for Non-Central Collision

Adjust code to account for y-component in position data and solve modified differential equations.
12

(l) Plot New Motion

Lock initial conditions and compute positions x_A(t), y_A(t), x_B(t), y_B(t), plot respective pathways over time.
13

(m) Discuss Angular Velocity Post-Collision

Analyze angular velocity variations and discuss any expected damping or conservation qualities in the new motion path of the spring-coupled system.

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Key Concepts

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

collision mechanics
Collision mechanics is the study of how objects interact during impacts. To understand collisions between identical atoms, we look at both physical laws and properties such as momentum and kinetic energy. This problem involves atoms treated as point particles moving along straight lines. When two atoms collide and stick together, the conservation laws of momentum and energy play crucial roles.

  • Conservation of Momentum: The total momentum before and after the collision must be equal. This helps determine the velocities post-collision.
  • Conservation of Energy: While kinetic energy might not be conserved if the collision is inelastic, the overall energy (kinetic + potential) will still be conserved.

This exercise starts with a simplified one-dimensional collision to establish a basic understanding, before adding complexities such as spring forces to simulate more realistic interactions.
center of mass
The center of mass is a point representing the combined position of a system's entire mass. For two atoms of mass \(m\), it is crucial to understand its motion both before and after collision.

The center of mass velocity can be computed using the formula:

\[ v_{\text{cm}} = \frac{m_A v_A + m_B v_B}{m_A + m_B} \]
In our example, since atom B is initially at rest, this simplifies to:

\[ v_{\text{cm}} = \frac{mv_0 + m \cdot 0}{m + m} = \frac{mv_0}{2m} = \frac{v_0}{2} \]

After the collision, both atoms stick together and form a single entity, thus the center of mass moves with the system’s combined mass. Understanding the center of mass also helps in analyzing more complex motion, such as post-collision dynamics when spring forces are introduced.
kinetic energy
Kinetic energy is the energy of motion. In a collision, it is significant to track the changes in kinetic energy to understand if the collision is elastic (where kinetic energy is conserved) or inelastic (where some kinetic energy is converted into other forms of energy, like heat or sound).

For our collision, calculate the kinetic energy before and after the impact as follows:

Before the collision:

\[ KE_{\text{initial}} = \frac{1}{2} mv_0^2 \]
After the collision:

\[ KE_{\text{final}} = \frac{1}{2}(2m)\left(\frac{v_0}{2}\right)^2 = \frac{1}{4} mv_0^2 \]

The change in kinetic energy can be calculated by:

\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = - \frac{1}{4} mv_0^2 \]

This drop in kinetic energy indicates an inelastic collision where atoms lose kinetic energy, which is not uncommon in real-world scenarios.
spring force model
The spring force model introduces a spring with spring constant \(k\) and equilibrium length \(b\) connecting the two atoms after they collide. This model helps simulate more realistic post-collision behavior where the atoms remain 'attached' and oscillate around the equilibrium distance instead of sticking statically.

When attached by a spring, the forces and motion are governed by Hooke's law:

For atom A:

\[ F_A = k((x_A - x_B) - b) \]
For atom B:

\[ F_B = -k((x_A - x_B) - b) \]

These forces influence both atoms post-collision, leading to a new calculation of their respective accelerations and equations of motion using Newton's second law.

The velocity of the center of mass remains \( \frac{v_0}{2} \) immediately after the spring attachment. However, kinetic energy will change again due to the introduction of potential energy in the spring.
analytical and numerical techniques
Analytical and numerical techniques are essential for understanding and predicting the system's behavior both before and after the collision.

  • Analytical Techniques: Involve using mathematical equations and conservation laws to solve variables like velocity and energy changes. For example, using momentum conservation to find velocities and analyzing forces to form differential equations.
  • Numerical Techniques: Employed when problems become too complex for simple equations. Writing and running computer programs to simulate the motion of atoms under forces, solving differential equations numerically.

In this exercise, one program is created to model the positions and velocities of atoms over time. This involves initial values (\(m = 0.1\), \(k = 20\), \(b = 0.2\), \(v_0 = 1.0\), \(\Delta t = 0.001\)) and visualizing the motion and maximum distance between atoms with plots, giving deeper insight into the dynamic system post-collision.

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Most popular questions from this chapter

In this project you will apply your knowledge of linear and angular momentum to study the aggregation of small droplets of ice to form large grains of snow. As snow crystals form in clouds they start falling through the cloud. Due to air resistance, larger particles fall faster than smaller particles. A large particle will therefore overtake smaller particles. When a smaller particle is overtaken, it will stick to the larger particle, adding further to the size. This process forms aggregate snowflakes, which is one of the most common types of snowflakes. \({ }^{3}\) This mechanism is often called differential sedimentation, and is a process important for pattern formation in many natural systems, and it is also a process important for many industrial processes. An example of a complex aggregate formed by a related aggregation process called Diffusion Limited Aggregation in Fig. \(16.35\) shows the complex geometries typically found in aggregate grains. In this project, we will study the aggregation process in detail. We will study an approximately spherical grain of ice of mass \(M\) and radius \(R\), hitting and sticking to an identical grain of ice. First, let us address why large particles fall faster than small particles. The mass of an ice grain of radius \(R\) and mass density \(\rho_{m}\) is $$ M=\rho_{m} \frac{4 \pi}{3} R^{3} $$ We will assume that air resistance can be modelled using the approximation: $$ \mathbf{F}_{v}=-k_{v} \mathbf{v} $$ where $$ k_{v} \simeq 20.4 R \eta $$ is a constant depending on the viscocity \(\eta\) of the fluid. (a) Find the forces acting on an ice grain with radius \(R\), and write down Newton's second law of motion for the grain. (b) Show that the acceleration of the grain is $$ \mathbf{a}=\mathbf{g}-\frac{20.4 \eta}{\rho_{m} \frac{4 \pi}{3} R^{2}} \mathbf{v} $$ where \(\mathbf{g}=-g \mathbf{j}\) and \(g\) is the acceleration of gravity. Can you now explain why larger grains fall faster than smaller grains? We will now study a collision between two identical ice grains. One grain is at rest relative to the reference system and the other grain has a velocity \(v_{0}\) downwards. When the two grains collide, they stick together at the point of contact, and remain stuck together. We call this combination of two grains a compound grain. (c) The moment of inertia of one ice grain around its center is \(I_{c}=\frac{2}{5} M R^{2}\). Show that the moment of inertia, \(I\), around the center of mass for a compound grain consisting of two grains sticking together is \(I=(14 / 5) M R^{2}\) First, we consider a linear collision where the upper grain hits the lower grain directly in the center, as illustrated in Fig. 16.36a. We assume the collision to be instantaneous, so you can ignore the effect of air resistance and gravity during the collision. (d) What is the velocity, \(\mathbf{v}_{1}\), of the center of mass the compound grain after the collision? (e) What is the angular velocity, \(\omega_{1}\), around the center of mass of the compound grain after the collision? Let us now consider the more general case illustrated in Fig. 16.36b. When the two grains touch, the line between the centers of the two grains forms the angle \(\theta\) with the horizontal. The upper grain still has the initial velocity \(v_{0}\) downwards before the collision, and the lower grain is at rest. (f) What is the velocity, \(\mathbf{v}_{1}\), of the center of mass of the compound grain after the collision? (g) What is the angular velocity, \(\omega_{1}\), around the center of mass of the compound grain after the collision? (h) What is the loss of energy in the collision? Let us now address the motion of the compound grain after the collision. Initially, it is rotating with the angular velocity \(\omega_{1}\). (i) If we ignore air resistance, find \(\omega(t)\) as a function of time for the subsequent motion. In the following we will not ignore air resistance, but rather develop a simplified model for the air resistance. In order to determine the force acting on the compound object due to air resistance, we either need to perform experiments on such objects, or we can use numerical simulations of the fluid flow around the object to determine the forces. Here, we will use a strong simplification: We assume that we may consider the compound object to consist of two separate spheres. The force on each of the spheres due to air resistance is described by \((16.176)\), where the corresponding velocity, \(v\), in (16.176) is the velocity of the center of the sphere, and the force acts in the center of the sphere. The compound object has velocity \(\mathbf{v}_{\mathrm{cm}}\) and angular velocity \(\omega\). (j) Argue that the velocities, \(\mathbf{v}_{A}\) and \(\mathbf{v}_{B}\), of each of the ice grains \(A\) and \(B\) are \(\mathbf{v}_{A}=\mathbf{v}_{c m}+\omega \times \mathbf{r}\) and \(\mathbf{v}_{B}=\mathbf{v}_{c m}-\omega \times \mathbf{r}\), where \(\mathbf{r}\) describes the position of grain \(A\) relative to the center of mass of the compound grain. (k) Show that the net force on the center of mass of the compound object is \(\sum \mathbf{F}=\) \(2 M \mathbf{g}-2 k_{v} \mathbf{v}_{c m}\), where \(\mathbf{g}=-g \mathbf{j}\) and \(g\) is the acceleration of gravity. (l) Show that the torque around the center of mass of the compound object due to air resistance is \(\tau=-2 k_{v} \omega R^{2}\) (Hint: Use Lagrange's formula). (m) Show that the angular acceleration \(\alpha\) of the compound object around its center of mass can be written as \(\alpha=d \omega / d t=-\left(1 / t_{0}\right) \omega\), and find the characteristic time \(t_{0}\). (n) Describe (with words) the motion of the compound object. (o) Sketch the time development of the velocity \(v_{c m}\) and the angular velocity \(\omega\) of the compound object, and discuss how the behavior would change if you changed the radius, \(R\), of the grains. (p) How would our argument change if we instead studied large particles, where the air resistance force depends on the square of the velocity? Final comment: Notice that the result above for the net force on the compound grain indicates that small and large grains have the same acceleration, which is not consistent with our initial result. This is due to our (incorrect) simplification of adding the air resistance force for each of the grains together to get the air resistance force for the compound grain. For a real ice crystal formed by aggregation, the dependence of the air resistance on the size of the compound grain is more complicated, and will also depend on the complex geometry attained by a compound grain after a few hundred collisions with smaller grains.

In this project we will study a ball that bounces on a flat surface. We will only look at a single collision between the ball and the surface, but we will use different models for the interactions during the collision. Both the ball and the surface are deformed during the collision, but you can assume that this deformation is small, this means that the forces from the surface on the ball will only act in a single point on the ball throughout the collision, and that the distance from this point to the center of mass of the ball does not change. The ball slips against the surface during the collision and the coefficient of dynamic friction between the ball and the surface is the constant \(\mu .\) You can neglect air resistance. The ball has a mass \(m\) and a radius \(R\), the acceleration of gravity is \(g\), the moment of inertia of the ball about its center of mass is \(I\). You throw the ball from a height \(h\) with only a horizontal velocity. The velocity immediately before the collision with the surface is \(\mathbf{v}\left(t_{0}\right)=v_{0 x} \mathbf{i}+v_{0 y} \mathbf{j}\), where \(v_{0 x}\) is positive and \(v_{0 y}\) is negative. (a) Draw a free-body diagram for the ball while it is in contact with the surface. Identify the forces. Let us first assume that the normal force from the surface on the ball is constant, \(N_{0}\). (b) Find the vertical component of the velocity, \(v_{y}(t)\), and the vertical position, \(y(t)\), of the ball while it is in contact with the surface. (c) How long is the ball in contact with the surface? (d) Find the horizontal component of the velocity of the ball as a function of time, \(v_{x}(t)\), while it is in contact with the surface. What is the horizontal component of the velocity of the ball, \(v_{1 x}\), immediately after the collision? (e) Find the angular velocity as a function of time, \(\omega(t)\), as well as the angular velocity, \(\omega_{1}\), of the ball immediately after the collision. Describe the motion of the ball after the collision. (f) Is the energy of the ball conserved during the collision? Does the ball bounce back to the height \(h\) after the collision? Justify your answers. Now assume that the force from the surface on the ball is \(N=k(R-y)^{3 / 2}\) when the ball is in contact with the surface, i.e. when \(y
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