How Do Colliding Spheres Behave in a Newton's Cradle Setup?

[Hoppa]

A novelty toy has two spheres of equal mass suspended by strings of equal length and negligible mass, and arranged so that the spheres can collide head-on at the bottom of each string’s swing. One sphere is given an initial velocity u towards the other, which is at rest hanging vertically. A collision between the two spheres has a coefficient of restitution e, which is less than 1. After a collision the spheres swing on the strings until they collide again. You may assume that the amplitudes of the swings are small and that swinging is frictionless. Under these conditions the spheres will collide again
at the bottom of their swings with exactly the same speeds as they had just after the previous collision, though the direction of their velocities will have reversed.

1.Investigate the sequence of speeds at successive collision for each of the spheres, by tracking the calculations through a few collisions. You should be able to detect a pattern emerging. To simplify your calculations you might express velocities in units of u.

2.What is the ultimate velocity of each of the spheres?

3. How much of the the original kinetic energy is ultimately lost in the collisions?

 

[Hoppa]

At the point of collision the line of centres of the spheres is at 45° from the y axis. The collision alters the velocities along this direction; the components perpendicular to this direction are unchanged. The equations relating velocities along the line of centres are, first the conservation of momentum, of which mass is irrelevant as it is same for both spheres.

v1 + v2 = u
Ö 2

And secondly the relation between normal velocities of approach and recession.

v2 - v1 = e u
Ö 2

Where v1 and v2 are the final velocities of the initially swinging sphere and the sphere at rest, respectively, along the line of centres, and e is the coefficient of restituion.

v1 = 1 – e u
2Ö2

And

v2 = 1 + e u
2Ö2

Sphere 1:
The component of the initial velocity along the direction perpendicular to the line of centres at contact is u and this is unchanged.
Ö 2
The final component along the lines of centres is 1 – e
2Ö2

Taking components of both these along the x and y axis:

v1x = æ1 – e ö u
è2 Ö2 ø
= (0.5 - e ö u
Ö2 ø
And

v2x = æ1 + e ö u
è2 Ö2 ø
= (0.5 + e ö u
Ö2 ø


Sphere 2:
There is no initial velocity perpendicular to the line of centres. The final component along the lines of centres is 1 + e
2(root)2

In the second collision the direction of the spheres velocities has reversed.

 

[Hoppa]

ah the editing, those 0 symbol things are menat to be square root signs.

 

[whozum]

Just a few things, A novelty toy has two spheres of equal mass suspended by strings of equal length and negligible mass, and arranged so that the spheres can collide head-on at the bottom of each string’s swing says that the balls collide head on at the bottom of their swing, which means at the bottom of the swing (vertical string) of the moving ball, it will collide with the second ball along hte direction of its motion (horizontal), so the line between the center of the spheres is actually parallel to the ground.

Try to learn latex, because your work is almost unreadable.