Thanks for checking. I've managed to solve it and you are right; it should be:

(2*d**e*^2)/(1+*e*^2)

It must be a misprint in the book.

]]>If e = 1 the 2de^2/(1 - e^2) answer is crazy.

Whereas 2de^2/(1+e^2) is correct .... perfectly elastic spheres exchange velocity, so after the first impact A stops and B carries on with A's old velocity. At the wall it bounces back at the same speed, and it hits A at the original spot. ie D = d

Bob

]]>Cannot find the error at the moment but I'll keep working on it.

It's not too bad to work with algebra throughout though:

and after the wall impact

Express all in terms of u_a

Let D be the distance from the wall for second impact

Back later when I've found some more paper to work this out.

A LITTLE LATER:

Curious. I'm getting

which agrees with your numeric version.

Bob

]]> 2*de*^2/(1 - e^2)

from the wall, where *e* is the coefficient of restitution for all the impacts involved.'

Before I tried the problem, I tried the formula on an example:

Let:

U*a* be the initial speed of sphere *a*

U*b* be the initial speed of sphere *b*

V*a* be the speed of sphere *a* after the first collision with sphere *b*

V*b* be the speed of sphere *b* after the first collision

I let U*a* = 2 m/s, *d* = 15 m and *e* = 1/2 and these are my workings:

by restitution:

1/2 = (V*b* - V*a*)/2

V*b* - V*a* = 1

by conservation of momentum:

2 = V*a* + V*b*

so: V*b* = 3/2 m/s and V*a* = 1/2 m/s

This means that sphere *b* will take 10 seconds to hit the wall and, during this time, sphere *a* will have travelled 5 m towards the wall. After the impact with the wall, sphere *b* will rebound with speed 3/2 x 1/2 = 3/4 m/s. Since there is now 10 m between the spheres and they are heading towards each other: 1/2t + 3/4t = 10 where t is time in seconds. This means they will collide after another 8 seconds and this will be at a distance of 6 m from the wall. Using the formula, however, I get 10 m. What am I doing wrong?