1. ## 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.

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?

Craig

2. ## 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)

$\displaystyle 2v_b = 1 + 5e$

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

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

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