Going to redo the sketch to make it a bit easier to work with:
___1___2___|
Since all collisions are elastic, the following velocity formulas apply:
v1 = (u1(m1-m2) + 2m2u2)/(m1+m2)
v2 = (u2(m2-m1) + 2m1u1)/(m1+m2)
where: m1,m2 are the masses, u1,u2 are the initial velocities, and v1,v2 are the final velocities.
The initial velocity of mass 2 is 0, so we can simplify the above expressions to get the following:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)
There are 3 cases here:
Case 1: m1=m2
If both masses are the same, the velocities after the first collision are:
v1 = 0
v2 = u1
In other words, there is a complete trade of momentum between the two masses. After rebounding off the wall, the velocity of m2 gets reversed so we have:
v1 = 0
v2 = -u1
After the second collision, we have:
v1 = -u1
v2 = 0
And m2 is stopped dead in its tracks again. Therefore, when both masses are the same there will only be 2 collisions.
Case 2: m2>m1
If m2 is greater than m1, the velocities after the first collision are:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)
Since m2>m1, (m1-m2) is negative which means m1 reverses direction. Comparing the magnitude of these velocities, we have:
|v2|/|v1| = |2m1u1|/|u1(m1-m2)| = |2m1/(m1-m2)| = 2m1/(m2-m1)
There are two sub-cases here. Letting |v2|/|v1| = 1 to find the point where both velocities are the same, we find:
2m1/(m2-m1) = 1
2m1 = m2-m1
3m1 = m2
m2/m1 = 3
Case 2(a)
m2 is at least 3 times as massive as m1. Plugging this back into the equation above, we see that v2 is less than (or equal to) v1. Therefore, m2 will never catch up with m1, so there is only 1 collision.
Case 2(b)
m2 is greater than m1, but less than 3 times as massive. Plugging this back into the equation above, we see that v2 is greater than v1. Therefore, m2 will eventually catch up with m1. My first guess is to say that there will only be 2 collisions total, but I'll have to double-check (more work on this in the next post!).
Case 3: m1>m2
If m1 is greater than m2, the velocities after the first collision are:
v1 = u1(m1-m2)/(m1+m2)
v2 = 2m1u1/(m1+m2)
Since m1>m2, (m1-m2) is positive which means m1 continues in the same direction. Comparing the magnitude of these velocities, we have:
|v2|/|v1| = |2m1u1|/|u1(m1-m2)| = |2m1/(m1-m2)| = 2m1/(m1-m2)
This is strictly greater than 1, which means m2 will increase in speed after each collision (I think...need to double check that). There will be a finite number of collision until eventually the momentum from the quick moving m2 reverses the direction of m1, after which m2 should start slowing down resulting in a finite number of collisions akin to Case 2 (small object hitting a large object).
As in Case 2, when I get more time I'll double check the math and try to find an expression for the number of collisions in each unanswered case.