Finding Coefficient of Restitution for 2D Collision

[bobred]

Linear momentum 2d collision

Homework Statement

Two particles A and B of mass m and 3m respectively, A collides with B. Find the coefficient of restitution e if \textbf{v}_{A} is purely in the \textbf{j}-direction.

Velocity of each particle before collision.
\dot{\textbf{r}}_{A}=9 \textbf{i}+5\textbf{j}
\dot{\textbf{r}}_{B}=2 \textbf{i}+2\textbf{j}

The x and y velocity components before collision
\dot{x}_{A}, \dot{y}_{A}, \dot{x}_{B} and \dot{y}_{B}

The x and y velocity components after collision
\dot{X}_{A}, \dot{Y}_{A}, \dot{X}_{B} and \dot{Y}_{B}

Homework Equations

The common tangent is the \textbf{j} axis.

\dot{y}_{A}\textbf{j}=\dot{Y}_{A}\textbf{j} and \dot{y}_{B}\textbf{j}=\dot{Y}_{B}\textbf{j}

(\dot{X}_{A}-\dot{X}_{B})\textbf{i}=-e(\dot{x}_{A}-\dot{x}_{B})\textbf{i}

m\dot{\textbf{r}}=m\textbf{v}

The Attempt at a Solution

With the values above I find the velocities after collision

\textbf{v}_{A}=(-\frac{21}{4}e+\frac{15}{4})\textbf{i}+5\textbf{j}
\textbf{v}_{B}=(\frac{7}{4}e+\frac{15}{4})\textbf{i}+2\textbf{j}

How do I find e if \textbf{v}_{A} is purely in the \textbf{j}-direction?

If I use(which is in the i-direction)

(\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})

I get e=0, a totally inelastic collision.

Thanks in advance.

 

[kuruman]

What do you think is the correct expression for the coefficient of restitution when you have a collision in 2-d? How did you conclude that the y-component of each particle's momentum is the same before and after the collision? Finally, what is a "common tangent" and why is it the y-axis?

 

[bobred]

Hi

The common tangent is the plane perpendicular to the common normal plane. We are told the common normal is in the i-direction. In the course text we are told

'In an (instantaneous) collision between two smooth non-rotating objects, where the area of contact at the moment of impact lies on a common tangent plane, the velocities parallel to the tangent plane remain unchanged before and after impact.'

Thanks

 

[kuruman]

Hi

The common tangent is the plane perpendicular to the common normal plane. We are told the common normal is in the i-direction. In the course text we are told

'In an (instantaneous) collision between two smooth non-rotating objects, where the area of contact at the moment of impact lies on a common tangent plane, the velocities parallel to the tangent plane remain unchanged before and after impact.'

Thanks

OK, then, the coefficient of restitution is given by

e=\left| \frac{\dot{X_A}-\dot{X}_B}{\dot{x_A}-\dot{x}_B} \right|
You know both terms in the denominator and it is given that

\dot{X}_A=0

Use momentum conservation in the x-direction to find the remaining term.

 

[bobred]

Hi

Thanks for the reply, in explaining to you the coordinate system it got me thinking again and obviously there is no \dot{X}_A so obviously I need to solve

<br /> (\dot{X}_{A}-\dot{X}_{B})=-e(\dot{x}_{A}-\dot{x}_{B})<br />

for e

Thanks

 

[kuruman]

I hope you can finish from this point.

 

[bobred]

Yes, all done, thanks again