1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.