Mechanics, collisions help

Hi, not been online in some time but back again now with some more probably simple questions ;)

I can't get my head around this question involving collisions, it's a bit wordy to please bear with me.

Quote:

3 smooth identical spheres each of mas 1kg move on a straight line on a smooth horizontal floor. B lies between A and C.

Spheres A and B are projected towards each other with speeds 3 m/s and 2 m/s respectivly, and C is projected away from sphere A at a speed of 2m/s.

The coefficient of restitution between any two spheres is *e*, show that sphere B will only collide with C if *e* > $\displaystyle \frac{3}{5}$

Obviously the final velocity of B needs to be greater than 2, but I'm having trouble making this into an equation?

Thanks in advance

Craig

Coefficient of Restitution

Hello Craig

There are two principles involved here. They are *momentum *and *restitution*. In the equations that follow, I am taking the direction $\displaystyle A \rightarrow C$ as positive, and I'm assuming that the velocities of $\displaystyle A$ and $\displaystyle B$ are $\displaystyle v_a$ and $\displaystyle v_b$ respectively, after the impact. So:

1 *The momentum of the system is conserved at the impact. *So (noting that the masses are all 1 kg):

$\displaystyle 3 - 2 = v_a + v_b$

$\displaystyle \Rightarrow v_a + v_b = 1$ (1)

2 The restitution equation is: *the velocity of separation = -e times the velocity of approach*.

In other words, the (velocity of $\displaystyle B -$ velocity of $\displaystyle A$) after impact $\displaystyle = -e \times$ (velocity of $\displaystyle B -$ velocity of $\displaystyle A$) before impact. So (remembering that we are taking $\displaystyle A \rightarrow C$ as the positive direction):

$\displaystyle v_b - v_a = -e(-2 - 3) = 5e$ (2)

Add (1) and (2):

$\displaystyle 2v_b = 1 + 5e $

and $\displaystyle v_b > 2$ if $\displaystyle B$ collides with $\displaystyle C$

$\displaystyle \Rightarrow 1 + 5e > 4$

$\displaystyle \Rightarrow e > \frac{3}{5}$

Grandad