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Topic review (newest first)

2012-11-08 23:53:19

Hi bob

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


It must be a misprint in the book.

bob bundy
2012-11-07 08:59:02

One further thought.

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 bundy
2012-11-07 03:44:00

hi jimi70

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.  smile


Curious.  I'm getting

which agrees with your numeric version.


2012-11-06 21:40:37

'A small smooth sphere moves on a horizontal table and strikes an identical sphere lying at rest on the table at a distance d from a vertical wall, the impact being along the line of centres and perpendicular to the wall. Prove that the next impact between the spheres will take place at a distance

           2de^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:
Ua be the initial speed of sphere a
Ub be the initial speed of sphere b
Va be the speed of sphere a after the first collision with sphere b
Vb be the speed of sphere b after the first collision
I let Ua = 2 m/s, d = 15 m and e = 1/2 and these are my workings:

by restitution:
1/2 = (Vb - Va)/2
Vb - Va = 1

by conservation of momentum:
2 = Va + Vb

so: Vb = 3/2 m/s and Va = 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?

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