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## Homework Statement

Particle A of mass m has initial velocity v0. After colliding with particle B of mass 2m initially at rest, the particles follow the paths shown in the sketch (see attachment). Find ##\theta##

## Homework Equations

collisions

## The Attempt at a Solution

The momentum before and after collision is

##

\left\{

\begin{array}{}

\vec P_i = mv_0 \hat \imath \\

\vec P_f = (mv_0' \cos{\theta} + \sqrt{2}m v_1')\ \hat\imath + (-m \sin(\theta) v_0' +\sqrt{2}mv_1')\ \hat\jmath

\end{array}

\right.

##

By conservation of momentum, ##v_0'## and ##v_1'## can be expressed in terms of ##v_0## and ##\theta## :

##

\left\{

\begin{array}{}

v_0' = \frac{v_0}{\cos(\theta) +\sin(\theta)} \\

v_1' = \frac{v_0}{\sqrt{2}} \frac{\sin(\theta)}{\cos(\theta) +\sin(\theta)}

\end{array}

\right.

##

Now in terms of energy:

## K_f = \frac{1}{2}mv_0'^2 + m v_1'^2 = \frac{1+2t^2}{(1+t)^2} K_i##,

where ## t = \tan(\theta)##.

The ratio between energy lost in the collision and initial kinetic energy is:

## \alpha = \frac{Q}{K_i} = 1 - \frac{K_f}{K_i} = \frac{t(2-t)}{(1+t)^2} ##

which can be re-written :

## (\alpha+1) t^2 + 2(\alpha -1) t + \alpha = 0 ##

The roots are real numbers only if ##\alpha \le 1/3## which is the same as saying that no more than one third of initial energy can be lost in the collision. In that case:

## t = \frac{1-\alpha}{1+\alpha} \pm \frac{\sqrt{1-3\alpha}}{1+\alpha} =

\frac{K_i - Q}{K_i+Q} \pm \frac{\sqrt{K_i}\sqrt{(K_i - 3Q)}}{K_i+Q}##

The answer will be ##\theta = \arctan(t)##, but which root should we keep ?