Feynman’s Lectures Exercises 8.2

‌A moving particle collides perfectly elastically with an equally massive particle initially at rest. Show that the two particles move at right angles to one another after the collision.

From conservation of momentum, we write the equations for the $x$ and $y$ axes:

${\displaystyle \begin{array}{rl}{\displaystyle v_1}& {\displaystyle =v_1^{\mathrm{\prime }}\cos \alpha +v_2^{\mathrm{\prime }}\cos \beta }\\ {\displaystyle 0}& {\displaystyle =v_1^{\mathrm{\prime }}\sin \alpha -v_2^{\mathrm{\prime }}\sin \beta }\end{array}}$

From the second equation:

${\displaystyle v_1^{\mathrm{\prime }}\sin \alpha =v_2^{\mathrm{\prime }}\sin \beta }$

This gives:

${\displaystyle \frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}=\frac{\sin \beta }{\sin \alpha }}$

Now, from conservation of kinetic energy:

${\displaystyle \begin{array}{rl}{\displaystyle \frac{1}{2}m{v}_1^2}& {\displaystyle =\frac{1}{2}m{v}_1^{\mathrm{\prime }2}+\frac{1}{2}m{v}_2^{\mathrm{\prime }2}}\\ {\displaystyle v_1^2}& {\displaystyle =v_1^{\mathrm{\prime }2}+v_2^{\mathrm{\prime }2}}\end{array}}$

Square the momentum equation and substitute $v_1^2$:

${\displaystyle \begin{array}{rl}{\displaystyle (v_1^{\mathrm{\prime }}\cos \alpha +v_2^{\mathrm{\prime }}\cos \beta {)}^2}& {\displaystyle =v_1^2}\\ {\displaystyle v_1^{\mathrm{\prime }2}{\cos }^2\alpha +v_2^{\mathrm{\prime }2}{\cos }^2\beta +2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}+v_2^{\mathrm{\prime }2}}\end{array}}$

Rearranging:

${\displaystyle \begin{array}{rl}{\displaystyle 2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}(1-{\cos }^2\alpha )+v_2^{\mathrm{\prime }2}(1-{\cos }^2\beta )}\\ {\displaystyle 2v_1^{\mathrm{\prime }}{v}_2^{\mathrm{\prime }}\cos \alpha \cos \beta }& {\displaystyle =v_1^{\mathrm{\prime }2}{\sin }^2\alpha +v_2^{\mathrm{\prime }2}{\sin }^2\beta }\end{array}}$

Divide both sides by $v_1' v_2'$:

${\displaystyle 2\cos \alpha \cos \beta =\frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}{\sin }^2\alpha +\frac{v_2^{\mathrm{\prime }}}{v_1^{\mathrm{\prime }}}{\sin }^2\beta }$

Use the earlier ratio:

${\displaystyle \frac{v_1^{\mathrm{\prime }}}{v_2^{\mathrm{\prime }}}=\frac{\sin \beta }{\sin \alpha }}$

Substitute:

${\displaystyle 2\cos \alpha \cos \beta =2\sin \alpha \sin \beta }$

So:

${\displaystyle \cos \alpha \cos \beta -\sin \alpha \sin \beta =0}$

This gives:

${\displaystyle \cos (\alpha +\beta )=0}$

Therefore:

${\displaystyle \alpha +\beta =90^{\circ }}$

The two particles move at right angles after the collision.

A more elegant solution can be found in Feynman’s Lectures Exercises 8.2 via the Center-of-Mass Frame.