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:

diagram

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.

 

 

Advertisements