1. The problem statement, all variables and given/known data There are two problems: (A) Consider two identical billiard balls (spheres), each of mass M and radius R. One is stationary (ball 2) and the other rolls on a horizontal surface without slipping, with a horizontal speed v (ball 1). Assume that all the frictional forces are small enough so as to be negligible over the time of the collision, and that the collision is completely elastic. What is the final velocity of each ball after the collision? i.e. when each ball is rolling without slipping again A link to the problem: http://physics.columbia.edu/files/physics/content/Quals2010Sec1.pdf , Problem 5. The solution is on page 20~21 (B) Two identical billiard balls of radius R and mass M rolling with velocities ±v collide elastically head-on. Assume that after the collision they have both reversed motion and are still rolling. (a) Find the impulse which the surface of the table must exert on each ball during its reversal of motion (b) What impulse is exerted by one ball on the other? A link to the problem: http://physics.columbia.edu/files/physics/content/Section 1 - Classical Mechanics with Solutions.pdf , Problem 1. The solution is on page 7 2. Relevant equations v=ωR L=Iω 3. The attempt at a solution I can mostly follow the solution to (A), except: - Why the solution claims "Just after the collision, v1'=0, v2'=v" before saying anything about angular momentum after the collision? I know this result is obviously correct when they is no rolling. But now he have rolling and therefore there are rotational kinetic energy terms in the conservation of energy equation, and it is not immediately obvious we can still use the familiar non-rolling result. Indeed, ω1 and ω2 did not change after the collision because the impact force vector goes through centers of the spheres and therefore there is no net torque on either ball but I just feel that something is a bit illogical with the solution..... For (B), I don't really understand what "impulse which the surface must exert on each ball during its reversal of motion" means. After the collision, the angular velocity of each ball is unchanged for the same reason in (A), however the velocities are switched. Therefore friction by surface of the table is necessary to get the balls rolling without slipping again. And is this where the "impulse" in part (a) of this problem is coming from? Another interpretation is that the surface gives the ball an impulse during the collision. But I assume the only force involved in the collision is the impact between the two balls and the ground should not give the ball an impulse. Why would the surface want to give an extra lift to the ball? Anyway, let's do something with this problem. Take the left ball as example. Before the collision, v1=+Rω, ω1=ω (into the page). Immediately after the collision, v1'=-Rω, ω1'=ω(into the page). The angular momentum of the ball relative to a fixed point on the ground immediately after the collision is therefore L=MRv1'+Iω1'=MωR^2(out)+(2/5) MωR^2 (in) = (3/5)MωR^2 (out). Relative to the fixed point on the ground, the angular momentum of the ball is conserved. So (3/5)MωR^2 (out)=MRv_final+Iω_final, and this gives v_final=-3Rω/7 And what to do next? By the way, I can't understand even a single word in the solution to problem (B)..... Is there a source that details the general theory of elastic collisions of rolling objects?