Picture of question

I was able to make the sketch of this, but I wanted to find the formulas for the x,y velocity components for each ball after the collision. I let $v_{x1}, v_{y1}$ be the velocities of the first ball after the collision, and similarly $v_{x2}, v_{y2}$ for the second ball. This is 4 unknown variables. Then I wrote the conservation of momentum along x, y axis, and conservation of energy:

$mv - (2m)v = mv_{x1} + 2mv_{x2}$ along x axis

$0 = mv_{y1} +2mv_{y2}$ along y axis

$\frac{mv^2}{2} + \frac{2mv^2}{2} = \frac{m(v_{x1}^2 + v_{y1}^2)}{2} + \frac{2m(v_{x2}^2 + v_{y2}^2)}{2}$ for energy.

This is only 3 formulas, and 4 unknowns, so I can not solve for the velocity components without more equations. I think there is some information involved with the geometry of the question (that one of the balls' center aligns with the other's top or bottom), but I do not know how to write this out mathematically, especially since the masses of the two objects are different.

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    $\begingroup$ Does this question (and links therein) give you what you need? $\endgroup$ – Floris Mar 14 '18 at 2:42
  • $\begingroup$ Yes, I think that provides a generalized solution to my question (although I have different masses, but I think it will work out anyway) Thanks :) $\endgroup$ – A. Remorov Mar 14 '18 at 3:17
  • $\begingroup$ Possible duplicate of Elastic collision between two circles $\endgroup$ – sammy gerbil Mar 14 '18 at 12:34
  • $\begingroup$ Divide all equations by m, and also note that $v_{x2}$ = $-v_{x1}$, so a substitution can be made that eliminates one of the velocity variables. $\endgroup$ – David White Mar 24 '18 at 18:57

Shift your perspective into the rest frame of one of the masses. Then, you can use the geometry of the setup to determine which direction the at-rest ball goes after the collision, since it has to move in the direction of the force applied by the other ball. This will give you the fourth equation.

  • $\begingroup$ Doesn't that mathematically reduce to the law of conservation of momentum (since the $dt$ cancels out from acceleration and we are left with velocities)? $\endgroup$ – A. Remorov Mar 14 '18 at 3:20
  • $\begingroup$ @A.Remorov It's not the momentum that's important here, but the direction of the force, which is determined by the way the balls collide. In the rest frame of a ball, the final velocity of that ball will be in the same direction as the contact force. So, when you get an angle from the geometry of the collision, you'll have $\tan\theta = v'_{y1}/v'_{x1}$ (the primes indicate the velocity in the shifted frame). Rewrite the rest of your equations in this frame, then shift back to the original frame after solving. $\endgroup$ – Mark H Mar 14 '18 at 4:21

There is a hidden assumption that the spheres are smooth. This implies that they retain their initial components of velocity parallel to the plane surface of contact. Then there are only 2 unknown variables, the components perpendicular to this plane, which can be found using the 2 equations for conservation of kinetic energy and conservation of linear momentum perpendicular to the plane of contact.

If the spheres were rough you would need additional information : the coefficient of friction and the moments of inertia of the spheres. The impact would provide a torque on each sphere, causing rotational as well as translational motion.


I think the graphical nature of this problem is being missed. One: do not think about velocity. Think about momentum, in units of $p_0 = mv$. You identified that the initial (and hence final) momentum is $-p_0\hat x$. Moreover, the initial (and hence final) energy is:

$$ E_0 = \frac 3 {2m} p_0^2 $$

Now consider the final state, there are 2 momenta: $\vec p_1$ and $\vec p_2$, and their sum is fixed:

$$ \vec p_1 + \vec p_2 = -p_0\hat x$$

Draw this, per the problem's request. Note how the tail of $\vec p_1$ and the head of $\vec p_2$ look like the foci of an ellipse.

The foci are at:

$(0,0)$ and $(-p_0, 0)$.

Now add the energy constraint. Suppose $\vec p_1 = (x, y)$, then:

$$ \frac{(x^2 + y^2)}{2m} + \frac{(x+p_0)^2+y^2}{4m}=\frac 3 2 \frac{p_0^2}m$$

Set $p_0=1$ so the foci are at the origin and $(-1,0)$:

and the equation is:

$$ 3x^2+2x+3y^2=5$$

Complete that square and the ellipse give you the allowed value of $\vec p_1=(x, y)$ for a scattering angle, $\theta$ of

$$ \tan \theta = y/x $$.

From there you can compute $\vec p_2$ directly.


Let u = velocity of object A and -u = velocity of object B Allow me to assume that the co efficient of restitution p = 1/4 At the moment of impact , the line joining the two centers makes an angle z with the original direction of motion and $sin z = r/2r = 1/2 $ so $z = 30^0$ Let the line making 30 degrees be the new frame of reference then the principle of linear momentum will conserve velocity in the j direction as the normal reaction to the collision will be in the i direction
... velocity before impact ....mass .... velocity after impact
[A].. $ u cos 30 i + u sin 30 j $ .... $m$ .......$x i + u sin 30 j$ [pclm]

[B].. $ -u cos 30 i - u sin 30 j $....$m$ .......$y i - u sin 30 j$

In the i direction the ratios of the relative velocities before and after impact =p [ Newtons law of restitution] ,so

$[x-y]/[u cos 30 - (-u cos 30)] = -[1/4]$

$[x-y]/[2 u cos 30] =1/4$

$x-y = - [\sqrt 3]/4 $....... Eq 1

Next use pclm in the i direction mx + my = mu cos 30 +m[- u cos 30] = 0

x+y=0 ...... Eq 2

Solve for Eq 1 and Eq 2 then x= -[ sq rt 3]u/8 and y = [sq rt 3]u/8 So the new velocity v for A after impact is

v[A] = x i + u sin 30 j = -[sq rt 3] [u/8]i + u sin 30 j and the new velocity v for B after impact is

v[B] = y i - u sin 30 j = +[sq rt 3][u/8]i - u sin 30 j


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