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Topic review (newest first)

jimi70
2012-11-08 23:53:19

Hi bob

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

(2de^2)/(1+e^2)

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

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

A LITTLE LATER:

Curious.  I'm getting



which agrees with your numeric version.

Bob

jimi70
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:
Let:
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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