# Thread: Mechanics, collisions help

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 > $\frac{3}{5}$
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 $A \rightarrow C$ as positive, and I'm assuming that the velocities of $A$ and $B$ are $v_a$ and $v_b$ 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):

$3 - 2 = v_a + v_b$

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

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

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

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

Add (1) and (2):

$2v_b = 1 + 5e$

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

$\Rightarrow 1 + 5e > 4$

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