Is it possible to determine the outcome of any impact knowing only the ratio of masses? [duplicate]

In elastic collisions in 2-D if two balls $A$, $B$ ($m_A = m_B$, $R = 1$) have equal mass we can determine in advance the outcome of the collision.

If cue-ball $A$ impacts object-ball $B$ (at rest) at an angle of +60° (with the x-axis), we can say in advance that ball $B$ will move at an angle of -30° with same axis, we do not need other equations or data.

How do you determine the angle of ball $B$ if the ratio $R: m_A/ m_B \neq 1$? How do you determine the angle of ball $B$ in the above-cited case if, for example, $R = 1/3 \rightarrow \lambda = -46.1$ ?

If you think that impossible, considering the current state of the art, please state that clearly, even in a simple comment. This will help future readers looking for an answer.

I am adding a picture to visualize the problem:

• @RobJeffries, the usual notation is: A moves from -x to +x, impact at origin, and +60° clockwise from +x, -30° means anti-clockwise. The sum makes a right angle. – bobie Jan 17 '15 at 11:26
• I left it at the office, but I believe that you'll find a fairly complete treatment in Marion and Thornton. – dmckee --- ex-moderator kitten Jan 17 '15 at 15:41

I'll set up the problem in the following way. Let's assume the line of centers is along the x-axis. Initial speed of $m_1$ is $u_1$ at an angle $\theta$ to the x-axis. Final speed of $m_{1}$ is $v_1$ at an angle $\alpha$ to the x-axis. The final speed of $m_2$ is $v_2$ and is along the line of centers (the x-axis), such that $\alpha$ is the angle between the final velocities of the two balls.

Two conservation of momentum equations $$m_1 u_1 \cos \theta = m_1 v_1 \cos \alpha + m_2 v_2\ \ \ \ (1)$$
$$m_1 u_1 \sin \theta = m_1 v_1 \sin \alpha\ \ \ \ \ \ \ \ \ \ \ (2)$$

If $R= m_1/m_2$ then I can manipulate these to give $$v_2 = R u_1 \left(\cos \theta - \cos\alpha \frac{\sin\theta}{\sin \alpha}\right)\ \ \ \ \ \ (3)$$

Now using conservation of kinetic energy in an elastic collision $$m_1 u_1^{2} = m_1 v_1^{2} + m_2 v_2^{2}\ \ \ \ \ \ \ \ (4)$$

Substituting for $v_2$ from (3) and using $v_1 = u_1 \sin\theta /\sin \alpha$ from (2), after a bit of algebra, I get the following horrible expression

$$R = \frac{\sin^2 \alpha - \sin^2 \theta}{\cos^2 \theta \sin^2 \alpha - 2 \sin\theta \cos\theta \sin \alpha \cos\alpha + \sin^2\theta\cos^2\alpha}\ \ \ \ (5)$$

If $\alpha=\pi/2$, then indeed the solution is $R=1$ for any value of $\theta$.

From formula (5) above. Substitute $1/(1+\tan^2)$ for $\cos^2$, substitute $1 - 1/(1+\tan^2)$ for $\sin^2$ and the corresponding square roots for $\cos$ and $\sin$. This will give an equation exclusively in terms of $\tan\theta$ and $\tan \alpha$. $$R = \frac{\tan^{2}\alpha(1+\tan^2\theta) -\tan^2\theta(1+\tan^2\alpha)}{\tan^2\alpha - 2\tan\theta\tan\alpha + \tan^2\theta}$$ Rearrange this into a quadratic in $\tan^2\alpha$ thus: $$(R-1)\tan^2\alpha -2R\tan\theta\tan\alpha +(R+1)\tan^2\theta = 0$$ Thus $$\tan\alpha = \tan\theta \left(\frac{R \pm 1}{R-1}\right)$$

Only one of these roots makes sense $$\tan\alpha = \tan\theta \left(\frac{R+1}{R-1}\right),$$ and of course a negative value of $\tan\alpha$ just means that $\alpha>90^{\circ}$.

EDIT: As a test I plugged this into a graphical tool with $\theta=60^{\circ}$ (red curve). The plot below shows $R$ vs $\alpha$ (recall this is the angle between the final velocity vectors). Notice that the minimum value of $\alpha$ is $60^{\circ}$ when $R$ is very large. As $R$ diminishes, $\alpha$ gets larger, reaching about $73^{\circ}$ (i.e. $13^{\circ}$ from the x-axis on your diagram) for $R=3$, then $90^{\circ}$ for $R=1$, and about $106^{\circ}$ ($46^{\circ}$ to the x-axis in your diagram) for $R=1/3$. There is a unique curve for each $\theta$ (the blue curve shows $\theta=30^{\circ}$ for comparison). Note that the red and blue curves only cross at $R=1$. This value of $R$ is the only one that uniquely determines the separation angle, for all other values of $R$ it also depends on $\theta$.

$R$ vs $\alpha$ for two values of $\theta$.

• Hi Rob, the graph looks OK, but do you think that this procedure is simpler than the standard one? Can you show how you get the exact values of $cos\lambda = 5 - 7 /\sqrt{52}$ respectively for R = 1/3 and 3? – bobie Jan 18 '15 at 11:32
• @bobie What do you mean "the standard procedure"? No, I cannot show what you want - though if you plug the numbers in they work. I surmise therefore there is a way of rearranging the equation I have shown, I just cannot see it. Please feel free, or ask on Maths SE. The Physics question is answered. – Rob Jeffries Jan 18 '15 at 11:35
• What software did you use for the curves? – John Alexiou Jan 18 '15 at 17:32
• @ja72 Geogebra geogebra.org – Rob Jeffries Jan 18 '15 at 18:26
• Rob, I will not accept any answer, to give you time to refine your idea, keep in mind that Phoenix87's formula is good, but there is an even better and simpler formula. – bobie Jan 19 '15 at 13:09

I would reason along these lines. Put the system in the centre of mass. Then, at collision, consider the plane through the centre of mass and perpendicular to the line through the centres of the balls. Since this plane is fixed in the frame of the centre of mass, this behaves like a wall the balls smash against, and therefore the velocity vectors after collision are determined by the usual reflection laws of elastic collision of a ball against a fixed wall.

It can be checked through a direct computation that this argument yields the expected results in the cases

1. of $m_1=m_2$;
2. linear collision, i.e. $\theta = 0$, and different masses.

EDIT: After a pressing demand for details here are some.

Suppose ball $A$ is moving with speed $v_1$ along the positive direction of the horizontal, i.e. $x$, axis, and the ball $B$ is stationary. Let $\theta$ be the angle between the direction of motion of $A$ and the line through the centres of the ball at collision. Denote by $\mu$ the ratio between the two masses, i.e. set $$\mu=\frac{m_B}{m_A}.$$ The velocity of the centre of mass is given by $$\mathbf v_{cm} = \frac1{1+\mu}v_A\mathbf i,$$ and hence in the frame of the centre of mass we have $$\mathbf u_A =\frac\mu{1+\mu}v_A\mathbf i,\qquad\mathbf u_B =-\frac1{1+\mu}v_A\mathbf i.$$ If in the picture above $A$ is the ball on the left and $B$ is the ball on the right, after the collision we have velocities \begin{align}\mathbf u_A' &=-\frac\mu{1+\mu}v_A\cos(2\theta)\mathbf i - \frac\mu{1+\mu}v_A\sin(2\theta)\mathbf j,\\\mathbf u_B'&=\frac1{1+\mu}v_A\mathbf\cos(2\theta)\mathbf i + \frac1{1+\mu}v_A\mathbf\sin(2\theta)\mathbf j.\end{align} The last step is to bring everything back to the original frame by adding the velocity of the centre of mass back to all the computed velocities, thus obtaining \begin{align}\mathbf v_A' &=\left(\frac1{1+\mu}-\frac\mu{1+\mu}\cos(2\theta)\right)v_A\mathbf i-\frac\mu{1+\mu}v_A\sin(2\theta)\mathbf j,\\\mathbf v_B'&=\frac1{1+\mu}\left(1+\cos(2\theta)\right)v_A\mathbf i +\frac1{1+\mu}v_A\mathbf\sin(2\theta)\mathbf j.\end{align} One easily sees that the angle of ball $B$ is the same as in the case $\mu=1$, since $$\tan(\theta_B) = \frac{\sin(2\theta)}{1+\cos(2\theta)} =\tan\theta,$$ while the angle of ball $A$ now depends on $\mu$ through $$\tan(\theta_A) = \frac{\sin(2\theta)}{\frac1\mu-\cos(2\theta)}.$$ One can check that, for $\mu=1$, the expected result is obtained back, namely $$\mu = 1\quad\Rightarrow\quad\tan(\theta_A) = \cot(\theta).$$

Remark It goes without saying that those angles must be corrected in order to fall into the right quadrant, so one has to take the signs of the component of the velocities into account as well, and this is not done in the above formulae.

The above steps show how passing to the centre of mass drastically simplifies the original problem.

Here is the solution of $\theta_A$ vs $\theta$ for the case $\mu=3$. At $\theta = 60^\circ$ the angle $\theta_A$ is roughly $46^\circ$, while at $\theta = 0$, back scattering is predicted.

• by carrying out the computations i guess – Phoenix87 Jan 18 '15 at 10:24
• As I was suggesting in the answer: refer to the frame of the centre of mass. Treat the problem as above, then transform it back to the original frame of reference – Phoenix87 Jan 18 '15 at 10:35
• sure, but it makes computations easier – Phoenix87 Jan 18 '15 at 10:52
• This is the right approach but I think you should actually do the math and add diagrams showing the impact of the transformations. – Floris Jan 18 '15 at 18:45
• Very nice improvement with your most recent edit. If I had not already upvoted, I would do so now. – Floris Jan 18 '15 at 20:13