The Elastic Pi Problem

Transcription

1 The Elastic Pi Problem Jacob Bien August 8, The problem Two balls of mass m 1 and m are beside a wall. Mass m 1 is positioned between m and the wall and is at rest. Mass m is moving with velocity v towards m 1, and all collisions (both ball-ball and ball-wall are elastic. If m m 1, there are 3 collisions before m escapes; if m 100m 1, there are 31 collisions; if m 100 m 1, there are 314; and if m m 1, there are In other words, if m 100 n m 1, then m escapes after 10 n π collisions. Why π? Verbal description of what happens When m hits m 1, it transfers some of its momentum (in the wall direction to m 1. This sets m 1 in motion towards the wall. Upon hitting the wall, m 1 reverses direction (maintaining the same speed since the collision is elastic. It alternates between hitting m and the wall. Every time m 1 hits m, m loses velocity in the wall direction, until eventually it is moving away from the wall fast enough that m 1 cannot catch up with it. At this point, m escapes off to infinity, and no more collisions occur. 3 A Preliminary esult We begin by establishing a general result about the change in velocity when two moving masses collide. Claim 1. Suppose two masses, m 1 and m 1, collide with velocities v 1 and v. Then after the collision their velocities are ( ( m1 m m ṽ 1 v m m m m ( ( m1 m m 1 ṽ v. m m m m 1

2 m 1 m Figure 1: Schematic of situation. Proof. The center of mass of the two balls is moving at velocity v cm m 1 m v m m, so in the reference frame of the center of mass, the balls are moving at velocities v 1 v CM and v v CM. In this frame, the total momentum is zero and thus their momentums are equal and opposite. After the collision, by conservation of momentum, they must have equal and opposite momenta and since the collision is elastic, they must be traveling at the same speeds (but with directions reversed. Therefore in the center of mass frame, after the collision the balls are traveling with velocities v CM v 1 and v CM v respectively. In the reference frame of the wall, their final velocities are therefore ṽ 1 (v CM v 1 + v CM and ṽ (v CM v + v CM. Thus, ṽ 1 v CM v 1 (m ( ( 1 m v (m m v 1 m1 m m v m m m m m m ṽ v CM v (m ( ( 1 m v (m m v m1 m m 1 v. m m m m m m 4 Calculating the velocity as a function of number of collisions We detail in this section how the velocities change after each collision. denote the velocities of the balls after t collisions. To start we have v (0 1 0 and v (0 Let v (t 1 and v (t 1 v (we define away from the wall to be positive. The system alternates between two kinds of collisions (1 Ball-Ball: By Claim 1, we have ( v (t+1 m1 m 1 m m ( v (t+1 m1 m m v (t v (t ( m m m ( m m 1 m m v (t v (t.

3 ( Ball-Wall: When m 1 collides with the wall, it changes direction, but doesn t change speed (since the collision is elastic: Putting this together, we see that ( v (t+ m1 m 1 m m ( v (t+ m1 m m v (t+ 1 v (t+1 1 v (t+ v (t+1. ( v (t m 1 m m ( m m 1 v (t Define m /m 1. Then this can be written as ( v (t+ 1 1 v (t v (t+ ( 1 v (t m m ( ( 1 v (t v (t. This can be expressed more compactly in vector/matrix form as v (t v (t. v (t+ Av (t, (1 where. A By Equation (1, ( ( 1/( /( /( ( 1/( and v (t ( v (t 1 v (t v (t Av (t A v (t 4 A t v (0. Thus, after t collisions, the balls will have velocities given by ( 0 A t v Our goal is to calculate the total number of collisions. Notice that the balls will continue to collide as long as v 1 v (since this means that m 1 is able to overtake m. Thus, after the last collision, v 1 < v for the first time. With this equation for (v 1, v as a function of the number of collisions, we can compute T min{t : v (t 1 < v (t }. And doing so for m /m 1 1, 100, 100,..., we confirm that T 3, 31, 314,...,

4 5 A Useful Geometric View We know that initially v 1 0 > v v and that through the course of collisions, the path in v 1 v space eventually crosses the v v 1 boundary. Based on the equation for v (t above, we can numerically calculate the path in v 1 v space for any fixed m /m 1 (see Figure. A striking feature of these plots is that the vectors v (t always lie on an ellipse with major axis along the v 1 axis (and a factor of of the minor axis. 6 Why an ellipse? Since all collisions are elastic, kinetic energy is conserved and therefore K 1 m 1v (t 1 m v (t. And initially K 1 m v (where the one disadvantage of this choice is that the units look wrong. Since the magnitude of v does not matter, we can take it to be 1 so that K m /. This is advantageous because then the above equation can be written more simply as 1 v (t 1 / + v (t which is the equation for the ellipse observed in the figure. 7 From ellipse to circle Observing that v (t is confined to an ellipse, we make things simpler by transforming to a coordinate system in which this is a circle. Let u 1 v 1 / and u v. Then, 1 u (t u (t. We can express the recurrence equation for v (t in terms of u (t. Since u (t always lies on a circle, the matrix which gives u (t must be simply a rotation matrix. From Equation (1, we have or ( ( u (t+ v (t+ Av (t 0 A 0 1 u (t ( ( ( u (t+ 1/ 0 ( 1/( /( /( ( 1/( 0 1 ( ( 1/ 0 ( 1/( /( 0 1 u (t /( ( 1/( ( ( 1/( /( u (t /( ( 1/( u (t 4

5 v v v v Figure : v (t as a function of v (t 1. 5

6 We claim that this is a rotation matrix. Indeed, observing that ( ( , ( ( we can write the above recurrence in terms of θ arcsin ( cos θ sin θ u (t+ sin θ cos θ +1 u (t. Multiplying a vector by this matrix rotates it by theta (counterclockwise. This observation is useful since t rotations by θ is identical to a single rotation by tθ. That is, ( t ( cos θ sin θ cos θt sin θt u (t u (0 u (0. sin θ cos θ sin θt cos θt ( 0 Finally, since u (0, it follows that 1 u (t ( sin θt cos θt This gives an expression for v (t that conveniently does not require t matrix multiplications: ( v (t sin θt. cos θt. : 8 Answering the question... at long last T min{t : v (t 1 < v (t } min{t : sin θt/ < cos θt/ and cos θt/ < 0} min{t : tan θt/ < 1/ } min{t : tan(π θt/ < 1/ } min{t : π θt/ < arctan(1/ } min{t : t > (/θ(π arctan(1/ } (/θ(π arctan(1/ That is, T π arctan(1/ ( arcsin +1 6

7 Now, for small x, sin x x and tan x x. Thus, for large, π / π / T +1 π 1 π 7