Sphere collision

I came across a problem in physics and I'm struggling a bit with getting the right intuition behind it. Below is the problem, my answer attempt and my question.

Problem: Two small spheres $$P$$ and $$Q$$ of equal mass lie touching each other in a smooth circular horizontal groove of radius $$a$$. Sphere $$P$$ is projected away from $$Q$$ with speed $$U$$. Assuming no resistance to motion in the groove (so that between collisions the speeds of the balls are constant), and a coefficient of restitution $$e$$ of each collision, find the time that elapses before the $$n$$th collision.

Assume that the speeds of $$P$$ and $$Q$$ after the $$n$$th collision satisfy $$V_n^P-V_n^Q=(-e)^nU,\\ V_n^Q+V_n^P=U.$$ My attempt: The time $$\Delta t_1$$ elapsed before the first collision is given by $$\Delta t_1=2\pi a/U$$. For the time between first and second collision, $$\Delta t_2$$, we first notice that, for some $$x$$, $$V_1^P=\frac{x}{\Delta t_2}\,\,\text{ and }\,\,V_1^Q=\frac{2\pi a+x}{\Delta t_2}.$$ Thus $$\Delta t_2=\left|\frac{2\pi a}{V_1^Q-V_1^P}\right|=\frac{2\pi a}{eU}.$$ The same argument could be used to calculate the remaining terms up to $$\Delta t_{n}$$. Then, simply sum these terms.

Question: My answer is done assuming that the movement of the two spheres is always kept in the same direction after a collision, thus the $$2\pi a +x$$. If the spheres were to bounce in opposite directions after every collision, I'd need to to use $$2\pi a-x$$ instead, which would yield a different result. How can I distiguinsh the two cases, and how would I know they would keep the same movement direction from the way the question is formulated?

Any insight is appreciated.

• If spheres are of equal mass it seems, they should "exchange velocities", e.g. when they collide one starts moving, while the other stops. You can look for any video on "Newton's pendulum" to see the effect. Thus, n-th collision should appear after $2n\pi a/ U$ Oct 24, 2019 at 15:11
• I understand that perspective, but that is not what the solution tells me. The solution says the time elapsed is $\frac{2\pi a}{U}\frac{1-e^n}{e^{n-1}-e^n}$, which is what you get following my attempt. I just don't understand it fully, that "exchange" must be somewhat different in this case. Oct 24, 2019 at 15:39
• I guess, I missed the point on the restitution coefficient. The problem is a bit strange: as I understand, the energy conservation law is not obeyed, but for some reasons momentum is conserved... Thus you have $e (v_b^Q - v_b^P) = (v_a^P - v_a^Q)$ and $v_b^Q + v_b^P = v_a^P + v_a^Q$ (b = before collision, a = after). This leads to $v_a^P = (1+e)v_b^Q/2 + (1-e)v_b^P/2$ and $v_a^Q = (1-e)v_b^Q/2 + (1+e)v_b^P/2$, thus $v_a^P-v_a^Q = e(v_b^Q-v_b^P)$. The latter means, the velocity difference after k-th collision is $e^k U$. P.S. My appologies if I'm missing +- somewhere -- It's 3 AM here. Oct 24, 2019 at 23:56
• So is it reasonable to assume that, after a collision, both spheres continue their movement in the same direction? Oct 28, 2019 at 15:35
• That depends. Suppose, you know $v_b^P$ and $v_b^Q$ -- velocities before collision. This means $e(v_b^Q-v_b^P)=(v_a^P-v_a^Q)$ and $v_b^Q+v_b^P=v_a^P+v_a^Q$ are two linear equations with two unknown variables -- you are not free to choose signs of $v_a^P$ and $v_a^Q$ as they are fully determined by the initial conditons $e$, $v_b^P$, and $v_b^Q$. Oct 28, 2019 at 17:03

1 Answer

In the meantime, I think I've solved it. Here's my solution.

The reason both spheres move in the same direction (clockwise, for example) after a collision is due to the fact that the collision is inelastic ($$e<1$$). If it was perfectly elastic ($$e=1$$), sphere $$P$$ would stop and only $$Q$$ would move after the first collision (as seen in Newton's cradle). To see this, consider simply the case where two spheres, $$1$$ and $$2$$, of equal mass collide. Sphere $$2$$ is initially at rest ($$u_2=0$$), and sphere $$1$$ has initial velocity $$u_1$$. Then, conservation of momentum and restitution yields \begin{align*} \begin{cases} u_1=v_1+v_2\\ v_2-v_1=eu_1 \end{cases} \end{align*} Thus, solving for $$v_1$$, $$v_1=u_1-v_2=u_1-eu_1-v_1 \Rightarrow v_1=(1-e)u_1/2.$$ Notice that $$u_1,u_2,v_1$$ and $$v_2$$ are velocities, i.e, they are vectors. Hence, if $$e=1$$, we get $$v_1=0$$ and if $$e<1$$, since $$1-e>0$$, we have that $$v_1$$ has the same direction as $$u_1$$ and $$v_2$$.

• Looks good for me. Oct 28, 2019 at 21:58