The following question gave a little grief:

Two smooth spheres of masses $2m$ and $3m$ respectively lie on a smooth horizontal table.

The spheres are projected towards each other with speeds $4u$ and $u$ respectively.

i. Find the speed of each sphere after the collision in terms of $e$, the coefficient of restitution

ii. Show that the spheres will move in opposite directions after the collision if $e>\frac13$.

My contention is that the question erred in not specifying that the answers to part i. were to be in terms of $e$ and $u$.

Solution:

i. The following is the situation: Let $v_1$ and $v_2$ be the velocities of the smaller respectively larger sphere after collision. Note that the initial velocity of the larger sphere is minus $u$.

Using conservation of momentum, $m_1u_1+m_2u_2=m_1v_1+m_2v_2$ $\Rightarrow 2m(4u)+3m(-u)=2mv_1+3mv_2$ $\Rightarrow 5u=2v_1+3v_2$.

Using: $\displaystyle e(u_1-u_2)=-(v_1-v_2)\Rightarrow \frac{v_1-v_2}{u_1-u_2}=-e$,

Therefore, $\displaystyle \frac{v_1-v_2}{4u-(-u)}=-e$ $\Rightarrow v_1-v_2=-5ue\Rightarrow v_1=v_2-5ue$,

and so $5u=2(v_2-5ue)+3v_2\Rightarrow 5u=2v_2-10ue+3v_2$, $\Rightarrow 5v_2=5u+10ue\Rightarrow v_2=u(1+2e)$. $\Rightarrow v_1=u(1+2e)-5ue=u+2ue-5ue=u(1-3e)$.

ii. $v_2>0$. If $e>\frac13\Rightarrow 3e>1\Rightarrow 1-3e<0$ and so $v_1=u(1-3e)<0$;

that is the particles move in opposite directions.