Theoretical Mechanics IPSP

Jürgen Vollmer, Universität Leipzig

User Tools

Site Tools

Trace:
book:chap5:5.2_collisions_of_particles

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Next revision		Previous revision
book:chap5:5.2_collisions_of_particles [2022/01/03 14:30] – created jvbook:chap5:5.2_collisions_of_particles [2022/01/04 05:12] (current) – abril
Line 1: Line 1:
- 
- 
 ===== 5.2  Collisions of hard-ball particles  ===== ===== 5.2  Collisions of hard-ball particles  =====
 +<WRAP id=section_particleScattering />
  
-\label{section:particleScattering} 
 We consider two spherical particles We consider two spherical particles
 and denote their radii and masses  and denote their radii and masses 
Line 12: Line 10:
 \begin{align*}  \begin{align*} 
   \mathbf q_i (t) = \mathbf q_i ( t_0 ) + v_i \: ( t - t_0 ) \, ,   \mathbf q_i (t) = \mathbf q_i ( t_0 ) + v_i \: ( t - t_0 ) \, ,
-  \quad\text{for} i \in \{1,2\}+  \quad\text{for}\quad i \in \{1,2\}
 \end{align*} \end{align*}
  
Line 21: Line 19:
 and a relative motion $\mathbf r (t)$. and a relative motion $\mathbf r (t)$.
 Introducing the notion $M = m_1 + m_2$ the former amounts to Introducing the notion $M = m_1 + m_2$ the former amounts to
-\begin{align}\label{eq:EXT-MQ}+ 
 +<wrap #eq_EXT-MQ></wrap> 
 +\begin{align}
   M \; \mathbf Q (t)   M \; \mathbf Q (t)
   = m_1 \: \mathbf q_1 (t) + m_2 \: \mathbf q_2 (t)   = m_1 \: \mathbf q_1 (t) + m_2 \: \mathbf q_2 (t)
-  = M \: \mathbf Q (t_0) + \dot{\mathbf Q} ( t_0 ) \: ( t - t_0 )+  = M \: \mathbf Q (t_0) + \dot{\mathbf Q} ( t_0 ) \: ( t - t_0 ) \tag{5.2.1}
 \end{align} \end{align}
 Since there are not external forces the total momentum Since there are not external forces the total momentum
-$M \, \dot{\mathbf Q} (t)$ is conserved (cf.\Thm{Newton-conserveP}) +$M \, \dot{\mathbf Q} (t)$ is conserved (cf. [[book:chap3:3.4_constants_of_motion_cm #Thm_Newton-conserveP |Theorem 3.5]]) 
-such that \cref{eq:EXT-MQ} applies for all times -- even when the particles collide.+such that [[#eq_EXT-MQ |Equation 5.2.1]] applies for all times -- even when the particles collide.
 A collision will therefore only impact the evolution relative to the center of mass. A collision will therefore only impact the evolution relative to the center of mass.
-\cref{eq:EXT-MQ} holds for all times.+[[#eq_EXT-MQ |Equation 5.2.1]] holds for all times.
  
  
Line 42: Line 42:
 $  \mathbf L = \mathbf r \times \mathbf p $, $  \mathbf L = \mathbf r \times \mathbf p $,
 and it is conserved when the collision force is acting along the line connecting the centers of the particles and it is conserved when the collision force is acting along the line connecting the centers of the particles
-(cf.\Thm{Newton-conserveL} and the discussion of Kepler's problem in \cref{sec:Kepler}).+(cf. [[book:chap3:3.4_constants_of_motion_cm #Thm_Newton-conserveL|Theorem 3.6]] and the discussion of Kepler's problem in [[book:chap4:4.7_worked_example_the_kepler_problem|Section 4.7]]).
 Moreover, $\mathbf r(t)$ is the only time-dependent quantity in this equation Moreover, $\mathbf r(t)$ is the only time-dependent quantity in this equation
 because $\mathbf L$ and $\mathbf p$ are preserved. because $\mathbf L$ and $\mathbf p$ are preserved.
Line 77: Line 77:
 that acts in the direction of the line $\mathbf r (t_c)$ connecting the particles. that acts in the direction of the line $\mathbf r (t_c)$ connecting the particles.
 Hence, \\ Hence, \\
-1. the momentum component in the $\hat{\boldsymbol \alpha}$ direction is preserved during the collision +  - The momentum component in the $\hat{\boldsymbol \alpha}$ direction is preserved during the collision because there is no force acting in this direction. 
-because there is no force acting in this direction +  - The collision in $\hat{\boldsymbol \beta}$ direction proceeds like a one-dimensional collision, [[book:chap3:3.4_constants_of_motion_cm #Example_One-dimensional |Example 3.12]], with the exception that one must retrace the argument using the center-of-mass frame, as discussed in [[book:chap4:4.10_problems #quest_ODE-1dCollision |Problem 4.29]].\\ 
-\\ +
-2. the collision in $\hat{\boldsymbol \beta}$ direction proceeds like a one-dimensional collision, +
-\Example{1dCollision}, with the exception +
-that one must retrace the argument using the center-of-mass frame, +
-as discussed in \cref{quest:ODE-1dCollision}. +
-\\+
 Consequently, we obtain the following momentum $\mathbf p'$ after the collision Consequently, we obtain the following momentum $\mathbf p'$ after the collision
 \[ \[
Line 97: Line 92:
 ==== 5.2.4 Self Test ==== ==== 5.2.4 Self Test ====
  
-\begin{marginfigure}  +<WRAP 140pt right #fig_HardBallScatteringAngle> 
-\centering +{{EOM_HardBallScattering.png}} 
-{{./Sage/EOM_HardBallScattering__TrajectoryShape.pdf}} + 
-{{./Sage/EOM_HardBallScattering__ScatteringAngle.pdf}} +Figure 5.7: Collision of two hard-ball particles with radii $R_1$ and $R_2$:
-\caption{ Collision of two hard-ball particles with radii $R_1$ and $R_2$:+
 (top) Trajectory shape. The labels denote the ratios $(\mathbf p \cdot \hat{\boldsymbol \alpha}) / (\mathbf p \cdot \hat{\boldsymbol \beta})$. (top) Trajectory shape. The labels denote the ratios $(\mathbf p \cdot \hat{\boldsymbol \alpha}) / (\mathbf p \cdot \hat{\boldsymbol \beta})$.
-(bottom) Scattering angle $\theta$. } +(bottom) Scattering angle $\theta$.
-\label{fig:HardBallScatteringAngle}+
 </WRAP> </WRAP>
- 
-----  
  
 <wrap #quest_HardBallScatteringAngle > Problem 5.2: </wrap>** Scattering angle for hard-ball particles ** <wrap #quest_HardBallScatteringAngle > Problem 5.2: </wrap>** Scattering angle for hard-ball particles **
 \\ \\
-In\cref{fig:HardBallScatteringAngle} we show shows the trajectory shape and the scattering angle for hard-ball scattering.+In [[#fig_HardBallScatteringAngle |Figure 5.7]] we show shows the trajectory shape and the scattering angle for hard-ball scattering.
  
-  -  What is the dimensionless length scale adopted to plot the trajectory shapes? +**a)** What is the dimensionless length scale adopted to plot the trajectory shapes?\\ 
-  -  What is the impact of the angular momentum on the trajectory shape? + 
-\\ +**b)** What is the impact of the angular momentum on the trajectory shape? What is the impact of the energy?\\ 
-What is the impact of the energy? + 
-  -  Verify that  +**c)** Verify that  
-\begin{align}  \label{eq:HardBallScatteringAngle} +<wrap #eq_HardBallScatteringAngle></wrap> 
-      \sin^2\theta = \frac{ L^2 }{2\mu \, E \, (R_1+R_2)^2}+\begin{align}  
 +      \sin^2\theta = \frac{ L^2 }{2\mu \, E \, (R_1+R_2)^2} \tag{5.2.2}
 \end{align} \end{align}
-and that this dependence is plotted in the lower panel of\cref{fig:HardBallScatteringAngle}. +and that this dependence is plotted in the lower panel of\cref{fig:HardBallScatteringAngle}.\\ 
-  -  What happens when $L^2 > 2\mu \, E \, (R_1+R_2)^2$? \\ + 
-Which angle $\theta$ will one observe in that case? +**d)** What happens when $L^2 > 2\mu \, E \, (R_1+R_2)^2$? Which angle $\theta$ will one observe in that case?\\ 
-  -  **$\star$** Show that\cref{eq:HardBallScatteringAngle,eq:CoulombScatteringAngle} agree when one identifies + 
-the length scale $R_1+R_2$ of the hard-ball system with the distance $R_{\text{eff}}$ of symmetry point of the cone section from the origin, +**e)** :!: Show that [[#eq_HardBallScatteringAngle |Equation 5.2.2]] and [[book:chap5:5.1_motivation_and_outline #eq_CoulombScatteringAngle |Equation 5.1.1]] agree when one identifies 
-i.e., with the mean value of the two intersection points with the $\hat x$-axis+the length scale $R_1+R_2$ of the hard-ball system with the distance $R_{\text{eff}}$ of symmetry point of the cone section from the origin, i.e., with the mean value of the two intersection points with the $\hat x$-axis
 \begin{align*}  \begin{align*} 
       R_{\text{eff}}       R_{\text{eff}}
Line 144: Line 136:
  
 ----  ---- 
 +
 +<WRAP 120pt right>
 +{{02_billiard_A1.png}}
 +</WRAP>
  
 <wrap #quest_Conservation-05 > Problem 5.4: </wrap> <wrap #quest_Conservation-05 > Problem 5.4: </wrap>
Line 149: Line 145:
 \\ \\
 The sketch to the right shows a billiard table. The sketch to the right shows a billiard table.
-The white ball should be kicked (i.e. set into motion with velocity $\mathbf v$), +The white ball should be kicked (i.e. set into motion with velocity $\mathbf v$), and hit the black ball such that it ends up in pocket to the top right. What is tricky about the sketched track?What might be a better alternative? 
-and hit the black ball such that it ends up in pocket to the top right. +
-\\ +
-What is tricky about the sketched track? +
-\\ +
-What might be a better alternative? +
-\begin{marginfigure}[-30mm] +
-{{./Sketch/02_billiard_A1.png}} +
-</WRAP>+
  
  
-~~ DISCUSSION ~~ +~~DISCUSSION~~ 
  
book/chap5/5.2_collisions_of_particles.1641216623.txt.gz · Last modified: 2022/01/03 14:30 by jv