(a) Three identical smooth spheres lie at rest on a smooth horizontal table with their centres in a straight line - Leaving Cert Applied Maths - Question 5 - 2008

Let the mass of each sphere be $m$.

Initially, sphere 1 has a speed of $u_1 = 2$ m/s and sphere 2 is at rest ($u_2 = 0$): Using the principle of conservation of momentum (PCM) and the law of restitution (NEL), we can set up the equations:

1. For collision between sphere 1 and sphere 2:

   $m(2) + m(0) = mv_1 + mv_2$

   Applying NEL, we have: $v_2 - v_1 = e(0 - 2)$ Rearranging gives: $v_2 = - e(2) + v_1$

   From the momentum equation, we have: $2 = v_1 + v_2$ Substituting $v_2$ gives: $2 = v_1 + (- e(2) + v_1)$

   Thus: $2 = 2v_1 - e(2)$

   Finally, solving for $v_1$ yields: v_1 = rac{2 + e(2)}{2}

   And for $v_2$: v_2 = e(2) + rac{2 + e(2)}{2}

2. For the second collision (between sphere 2 and sphere 3), a similar analysis gives:

   v_3 = rac{1 + e}{ 1 + e + e^2}

   The post-collision speeds can thus be expressed generally, and we find: velocities ext{ after 2nd impact}: ext{(1 - e)} imesrac{(1 - e^2)}{(1 + e)}.