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bobred
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Linear momentum 2d collision
Two particles A and B of mass [tex]m[/tex] and [tex]3m[/tex] respectively, A collides with B. Find the coefficient of restitution [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction.
Velocity of each particle before collision.
[tex]\dot{\textbf{r}}_{A}=9 \textbf{i}+5\textbf{j}[/tex]
[tex]\dot{\textbf{r}}_{B}=2 \textbf{i}+2\textbf{j}[/tex]
The x and y velocity components before collision
[tex]\dot{x}_{A}[/tex], [tex]\dot{y}_{A}[/tex], [tex]\dot{x}_{B}[/tex] and [tex]\dot{y}_{B}[/tex]
The x and y velocity components after collision
[tex]\dot{X}_{A}[/tex], [tex]\dot{Y}_{A}[/tex], [tex]\dot{X}_{B}[/tex] and [tex]\dot{Y}_{B}[/tex]
The common tangent is the [tex]\textbf{j}[/tex] axis.
[tex]\dot{y}_{A}\textbf{j}=\dot{Y}_{A}\textbf{j}[/tex] and [tex]\dot{y}_{B}\textbf{j}=\dot{Y}_{B}\textbf{j}[/tex]
[tex](\dot{X}_{A}-\dot{X}_{B})\textbf{i}=-e(\dot{x}_{A}-\dot{x}_{B})\textbf{i}[/tex]
[tex]m\dot{\textbf{r}}=m\textbf{v}[/tex]
With the values above I find the velocities after collision
[tex]\textbf{v}_{A}=(-\frac{21}{4}e+\frac{15}{4})\textbf{i}+5\textbf{j}[/tex]
[tex]\textbf{v}_{B}=(\frac{7}{4}e+\frac{15}{4})\textbf{i}+2\textbf{j}[/tex]
How do I find [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction?
If I use(which is in the i-direction)
[tex](\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})[/tex]
I get [tex]e=0[/tex], a totally inelastic collision.
Thanks in advance.
Homework Statement
Two particles A and B of mass [tex]m[/tex] and [tex]3m[/tex] respectively, A collides with B. Find the coefficient of restitution [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction.
Velocity of each particle before collision.
[tex]\dot{\textbf{r}}_{A}=9 \textbf{i}+5\textbf{j}[/tex]
[tex]\dot{\textbf{r}}_{B}=2 \textbf{i}+2\textbf{j}[/tex]
The x and y velocity components before collision
[tex]\dot{x}_{A}[/tex], [tex]\dot{y}_{A}[/tex], [tex]\dot{x}_{B}[/tex] and [tex]\dot{y}_{B}[/tex]
The x and y velocity components after collision
[tex]\dot{X}_{A}[/tex], [tex]\dot{Y}_{A}[/tex], [tex]\dot{X}_{B}[/tex] and [tex]\dot{Y}_{B}[/tex]
Homework Equations
The common tangent is the [tex]\textbf{j}[/tex] axis.
[tex]\dot{y}_{A}\textbf{j}=\dot{Y}_{A}\textbf{j}[/tex] and [tex]\dot{y}_{B}\textbf{j}=\dot{Y}_{B}\textbf{j}[/tex]
[tex](\dot{X}_{A}-\dot{X}_{B})\textbf{i}=-e(\dot{x}_{A}-\dot{x}_{B})\textbf{i}[/tex]
[tex]m\dot{\textbf{r}}=m\textbf{v}[/tex]
The Attempt at a Solution
With the values above I find the velocities after collision
[tex]\textbf{v}_{A}=(-\frac{21}{4}e+\frac{15}{4})\textbf{i}+5\textbf{j}[/tex]
[tex]\textbf{v}_{B}=(\frac{7}{4}e+\frac{15}{4})\textbf{i}+2\textbf{j}[/tex]
How do I find [tex]e[/tex] if [tex]\textbf{v}_{A}[/tex] is purely in the [tex]\textbf{j}[/tex]-direction?
If I use(which is in the i-direction)
[tex](\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})[/tex]
I get [tex]e=0[/tex], a totally inelastic collision.
Thanks in advance.
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