Chapter 14, Problem 14.28P

(a)

To determine

Prove that ${\xi }_{\text{lab}}=\frac{\left(\pi -{\theta }_{\text{cm}}\right)}{2}$

Answer to Problem 14.28P

The angle is ${\xi }_{\text{lab}}=\frac{\left(\pi -{\theta }_{\text{cm}}\right)}{2}$.

Explanation of Solution

Consider particle 1 with initial momentum $p_{\text{lab}\text{1}}$ and particle 2 with zero initial momentum. Final momentum of particle 1 is $p{\text{'}}_{\text{lab}1}$ and final momentum of the particle 2 is $p{\text{'}}_{\text{lab}2}$.

According to law of conservation of momentum

    $\begin{array}{c}{p}_{\text{lab}1}+0=p{\text{'}}_{\text{lab}1}+p{\text{'}}_{\text{lab}2}\\ p{\text{'}}_{\text{lab}2}=p_{\text{lab}1}-p{\text{'}}_{\text{lab}1}\end{array}$

From the figure, $AC$ denotes $p_{\text{lab}\text{1}}$, $AD$ represents $p{\text{'}}_{\text{lab}1}$ and $p{\text{'}}_{\text{lab}2}$ is represented by $AC-AD$

From the triangular law of vectors, $AC-AD=DC$

$BD$ and $BC$ is the radius of the circle. Thus the angles opposite to these sides are equal.

From the triangle $\Delta BCD$,

    $\begin{array}{c}\angle BCD=\angle BDC\\ ={\xi }_{\text{lab}}\end{array}$

And,

    $\angle DBC={\theta }_{\text{cm}}$

The sum of all three angles of the triangle is equal to $\pi \text{rad}$.

Therefore for $\Delta BCD$

    $\begin{array}{c}{\xi }_{lab}+{\xi }_{lab}+{\theta }_{cm}=\pi \\ 2{\xi }_{lab}+{\theta }_{cm}=\pi \\ {\xi }_{lab}=\frac{\pi -{\theta }_{cm}}{2}\end{array}$

Conclusion:

The angle is ${\xi }_{\text{lab}}=\frac{\left(\pi -{\theta }_{\text{cm}}\right)}{2}$.

(b)

To determine

Show that in the case of equal masses, the angle of these outgoing particles in elastic collision is $90°$.

Answer to Problem 14.28P

For the collision of equal masses, ${\xi }_{\text{lab}}+{\theta }_{\text{lab}}=\frac{\pi }{2}$

Explanation of Solution

Write down the expression connecting ${\theta }_{\text{lab}}$ and ${\theta }_{\text{cm}}$

    $\tan {\theta }_{lab}=\frac{\sin {\theta }_{cm}}{\lambda +\cos {\theta }_{cm}}$

$\lambda =1$ for $m_1=m_2$.

Then,

    $\begin{array}{c}\tan {\theta }_{lab}=\frac{\sin {\theta }_{cm}}{1+\cos {\theta }_{cm}}\\ =\frac{2\sin \frac{{\theta }_{\text{cm}}}{2}\cos \frac{{\theta }_{\text{cm}}}{2}}{2{\cos }^2\frac{{\theta }_{\text{cm}}}{2}}\\ =\tan \frac{{\theta }_{\text{cm}}}{2}\end{array}$

That is,

    $\begin{array}{l}{\theta }_{\text{lab}}=\frac{{\theta }_{\text{cm}}}{2}\\ {\theta }_{\text{cm}}=2{\theta }_{\text{lab}}\end{array}$

Substitute $\pi -2{\xi }_{\text{lab}}$ for ${\theta }_{\text{cm}}$

    $\begin{array}{c}{\xi }_{\text{lab}}=\frac{\pi -{\theta }_{\text{cm}}}{2}\\ {\xi }_{\text{lab}}=\frac{\pi }{2}-{\theta }_{\text{lab}}\\ \frac{\pi }{2}={\xi }_{\text{lab}}+{\theta }_{\text{lab}}\end{array}$

Hence proved.

Conclusion:

For the collision of equal masses, ${\xi }_{\text{lab}}+{\theta }_{\text{lab}}=\frac{\pi }{2}$

(c)

To determine

Prove ${\xi }_{\text{lab}}+{\theta }_{\text{lab}}=\frac{\pi }{2}$ using law of conservation of momentum.

Answer to Problem 14.28P

The expression ${\xi }_{\text{lab}}+{\theta }_{\text{lab}}=\frac{\pi }{2}$ has been proven using the law of conservation of momentum.

Explanation of Solution

Apply law of conservation of momentum in horizontal direction

    $p_{\text{lab1}}=p{\text{'}}_{\text{lab1}}\cos {\theta }_{\text{lab}}+p{\text{'}}_{\text{lab2}}\cos {\xi }_{\text{lab}}$

Along vertical direction,

  $\begin{array}{c}0=p\text{'}la{b}_1\sin {\theta }_{lab}-p{\text{'}}_{lab}\sin {\xi }_{\text{lab}}\\ p{\text{'}}_{lab1}\sin {\theta }_{lab}=-p{\text{'}}_{lab2}\sin {\xi }_{\text{lab}}\end{array}$

Apply the law of conservation of energy

    $\frac{p^{\text{2}}{}_{\text{lab1}}}{2m_1}=\frac{p{\text{'}}^{\text{2}}{}_{\text{lab1}}}{2m_1}+\frac{p{\text{'}}^{\text{2}}{}_{\text{lab2}}}{2m_2}$

Substitute $m_1=m_2$,

    $p^{\text{2}}{}_{\text{lab1}}=p{\text{'}}^{\text{2}}{}_{\text{lab1}}+p{\text{'}}^{\text{2}}{}_{\text{lab2}}$

Then,

    $\begin{array}{c}p{\text{'}}^{\text{2}}{}_{\text{lab1}}+p{\text{'}}^{\text{2}}{}_{\text{lab2}}={\left(p{\text{'}}_{\text{lab1}}\cos {\theta }_{lab}+p{\text{'}}_{lab2}\cos {\xi }_{lab}\right)}^2\\ p{\text{'}}^{\text{2}}{}_{\text{lab1}}+p{\text{'}}^{\text{2}}{}_{\text{lab2}}=p{\text{'}}^2{}_{\text{lab1}}{\cos }^2{\theta }_{lab}+p{\text{'}}^2{}_{lab2}{\cos }^2{\xi }_{lab}+2p{\text{'}}_{\text{lab1}}\cos {\theta }_{\text{lab}}p{\text{'}}_{\text{lab2}}\cos {\xi }_{\text{lab}}\\ 0=p{\text{'}}^2{}_{\text{lab1}}\left(1-{\cos }^2{\theta }_{lab}\right)+p{\text{'}}^2{}_{lab2}\left(1-{\cos }^2{\xi }_{lab}\right)-2p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\cos {\theta }_{\text{lab}}\cos {\xi }_{\text{lab}}\\ 0=p{\text{'}}^2{}_{\text{lab1}}{\sin }^2{\theta }_{lab}+p{\text{'}}^2{}_{lab2}{\sin }^2{\cos }^2{\xi }_{lab}-p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\cos {\theta }_{\text{lab}}\cos {\xi }_{\text{lab}}\end{array}$

On further simplification,

    $\begin{array}{c}p{\text{'}}^2{}_{\text{lab1}}{\sin }^2{\theta }_{lab}+p{\text{'}}^2{}_{\text{lab2}}{\sin }^2{\cos }^2{\xi }_{\text{lab}}=p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\cos {\theta }_{\text{lab}}\cos {\xi }_{\text{lab}}\\ {\left(p{\text{'}}^2{}_{\text{lab1}}{\sin }^2{\theta }_{lab}+p{\text{'}}^2{}_{\text{lab2}}{\sin }^2{\cos }^2{\xi }_{\text{lab}}\right)}^2=2p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\left(\cos {\xi }_{\text{lab}}\cos {\theta }_{\text{lab}}+\cos {\xi }_{\text{lab}}\cos {\theta }_{\text{lab}}\right)\\ {\left(p{\text{'}}^2{}_{\text{lab1}}{\sin }^2{\theta }_{lab}-p{\text{'}}^2{}_{\text{lab2}}{\sin }^2{\cos }^2{\xi }_{\text{lab}}\right)}^2=2p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\left(\cos {\xi }_{\text{lab}}\cos {\theta }_{\text{lab}}+\cos {\xi }_{\text{lab}}\cos {\theta }_{\text{lab}}\right)\\ 0=2p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\left(\cos {\xi }_{\text{lab}}+\cos {\theta }_{\text{lab}}\right)\end{array}$

Here $2p{\text{'}}_{\text{lab1}}p{\text{'}}_{\text{lab2}}\ne 0$. Therefore $\cos {\xi }_{\text{lab}}+\cos {\theta }_{\text{lab}}=0$

Take cosine inverse on both sides

    $\begin{array}{c}{\xi }_{lab}+{\theta }_{lab}={\cos }^{-1}0\\ =\frac{\pi }{2}\end{array}$

Conclusion:

The expression ${\xi }_{\text{lab}}+{\theta }_{\text{lab}}=\frac{\pi }{2}$ has been proven using the law of conservation of momentum.