Why do particles of equal mass (with one at rest) undergoing elastic collisions scatter at only right angles?

Question

This is from the Section 9.6, page 351 of "Classical Dynamics of Particles and Systems" by Thornton and Marion.

By setting a up a system where mass 1 has initial momentum $m_1 u_1$ and mass 2 is at rest.

If we let

• $\psi$ be the scatter of angle of particle 1 from the line connecting $m_1$ and $m_2$ prior to contact in the Lab reference frame

• $\xi$ is the scatter angle of particle 2 from the same reference line as above in the Lab reference frame

• $V$ be the speed of the center of mass in the Lab reference frame

• $v_1$, $v_2$ be the final speed of particles 1 and 2 respectively in the Lab reference frame

• $u_1'$, $u_2'$ be the initial speeds in the center of mass reference frame

• $v_1'$, $v_2'$ be the final speeds in the center of mass reference frame

• $\theta$ be the scatter angle of particle 1 in the center of mass reference frame

Then from the fact that:

• Conservation of momentum and kinetic energy implies $u_1'=v_1'$ and $u_2'=v_2'$

• $u_1 = u_1' + u_2'$ is the relative speed of the two particles before collision, which are the same in both frames

• $u_2' = V$ since vectorally, the vectors have the same magnitude but opposite directions

• $u_2 = 0$ since the 2nd particle is at rest in the Lab reference frame

Through algebraic manipulations we can write $v_1'$ and $v_2'$ in terms of $u_1$

and through geometric inspection, we can determine that: $$tan\psi = \frac{v_1' sin\theta}{v1' cos\theta + V}$$ which (since $v/v_1' = m_1/m_2$) simplifies to: $$tan\psi = \frac{sin\theta}{cos\theta + (m_1/m_2)}$$

and that: $$tan\xi = \frac{v_2' sin\theta}{V - v_2' cos\theta}$$ which (since $V = v_2'$) simplifies to: $$tan\xi = cot\frac{\theta}{2}$$ implying: $$2\xi = \pi - \theta$$

Then in the case where $m_1 = m_2$ we have: $$\psi = \frac{\theta}{2}$$ so that: $$\xi + \psi = \frac{\pi}{2}$$

Despite all of this fancy mathematical acrobatics, the result seems counterintuitive to what we expect. Why does the scatter angle between the two particles need to always be 90 degrees?

What if for instance, particle 1 strikes particle 2 directly so that particle 1 bounces straight back, while particle 2 is pushed straight forwards, creating a scatter angle of 180? Isn't that what we normally see in billiards (where the balls are approximately equal in mass).

Answer 1

The perfect head on collision is a special case where we don't need to worry about any relative angles. We can solve it using Physics 1 conservation of momentum and energy, all in the lab frame.

For equal mass particles

$$m u_1 = m v_1 + m v_2 \implies u_1 = v_1 + v_2,$$

and

$$\frac{1}{2} m {u_1}^2 = \frac{1}{2} m {v_1}^2 + \frac{1}{2} m {v_2}^2 \implies {u_1}^2 = {v_1}^2 + {v_2}^2 .$$

We can square the momentum equation and find

$${u_1}^2 = {v_1}^2 + {v_2}^2 + 2 v_1 v_2 = {v_1}^2 + {v_2}^2.$$

So

$$2 v_1 v_2 = 0.$$

This means one of the particles has zero velocity after the collision. Either particle 1 passed right through particle 2 with no effect (aphysical), or particle 1 stops and transfers all of its momentum and energy to particle 2.

In the center of mass frame this would be scattering at $180^\circ$. The fact that the first particle stops must break one of M&T's assumptions. In this case the scattering angle of particle 1 $\theta$ is ill defined. What is the direction of motion for a particle that is stopped?

In billiards you see a ball bounce backwards, because it was spinning backwards before the collision (while sliding along the table). It stops. Angular momentum is conserved so it keeps spinning. Friction catches the spin on the table and it starts to roll backwards. If the ball is rolling on the table it will continue to move forward after the collision for the same reason.

I don't have the reputation to comment, but setting $\theta=0$ is not the problem, as another answer suggests:

$$\tan\xi=\cot 0\rightarrow\infty \implies \xi=\frac{\pi}{2}$$

Answer 2

For a head on collision, if we take $\psi = 0$ then this implies $\theta = 0$ and $\zeta = 0$. But in the derivation of $\zeta + \psi = \pi/2$ it is implicitly assumed that $\theta \neq 0$, therefore, the head on collision case is automatically excluded in the derivation.

Answer 3

I think deriving that same result in the "head-on" case where one the scattering angles are $0$ and $\pi$ ends up dividing by $0$.

$$\cot \frac\theta 2 = \frac{1}{0}$$