Elastic collisions and conservation of momentum If you have an elastic collision between objects 1 and 2 and  where 'kinetic energy is conserved', does this mean object 1 will always have the same velocity it had before the collision? 
Or will object 2 have all the energy from object 1 transferred to it and used for velocity or will both the objects always 'join' and have the same common velocity?
 A: In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well as their rotations. If we assume that no exchange between these two forms of kinetic energy occurs, i.e. that both forms are separately conserved, we have
$$
m_1\boldsymbol{v}_1 + m_2\boldsymbol{v}_2 = m_1\boldsymbol{w}_1 + m_2\boldsymbol{w}_2
$$
and
$$
\tfrac{1}{2}m_1\boldsymbol{v}_1^2 + \tfrac{1}{2}m_2\boldsymbol{v}_2^2 = \tfrac{1}{2}m_1\boldsymbol{w}_1^2 + \tfrac{1}{2}m_2\boldsymbol{w}_2^2
$$
where $\boldsymbol{v}$ and $\boldsymbol{w}$ denote the velocities before and after the collision, respectively. Thus, we have 4 equations (3 components of momentum and the energy) for 6 unknowns (3+3 components of the post-collision velocities). Consequently, the above equations (on conjunction with $\boldsymbol{v}_{1,2}$) do not uniquely constrain the $\boldsymbol{w}_{1,2}$. Some other information is required to determine the relative direction. This depends on the properties of the objects and the point of impact.
The above system of equations is invariant under a Galilean transformation, i.e. a change of the origin of velocity. They become particularly simple in the frame in which the total momentum vanishes. Let a prime denote velocities in that frame, then
$$
\boldsymbol{v}' = \boldsymbol{v} - \boldsymbol{V},\qquad
\boldsymbol{V} = \frac{m_1\boldsymbol{v}_1+m_2\boldsymbol{v}_2}{m_1+m_2}
$$
and we have
$$
m_1 \boldsymbol{w}'_1+m_2 \boldsymbol{w}'_2=0,\quad
|\boldsymbol{w}'_1| = |\boldsymbol{v}'_1|,\quad
|\boldsymbol{w}'_2| = |\boldsymbol{v}'_2|
$$
(these equations are not independent, there still only 4 independent scalar constraints). In particular, the speeds remain the same in this frame.
In 1D we have 2 equations for 2 unknowns and hence the solution is completely determined (also, there is no rotation in 1D). Since a collision requires $w_{1,2}\neq v_{1,2}$, we have
$w_{1,2}=-v_{1,2}$. Transforming back to the original frame, this gives
$$
w_{1,2} = 2 V - v_{1,2}.
$$
Requiring the speeds to remain the same ($w_{1,2}=-v_{1,2}$) gives us $V=0$. Thus in 1D, the speeds reamin the same if and only if the total momentum vanishes.
A: 
If you have an elastic collision between objects 1 and 2 and where 'kinetic energy is conserved' ... will both the objects always 'join' and have the same common velocity?

Linear momentum will be conserved, so if the objects 'join'
 $$\vec{p}_1+\vec{p}_2=(m_1+m_2)\vec{V}\ $$ where $\vec{V}$ is the final common velocity.
The initial kinetic energy is $$K_i=\frac{p_1^2}{2m_1}+\frac{p_2^2}{2m_2} $$The final kinetic energy will be 
$$K_f=\frac{(m_1+m_2)V^2}{2} = \frac{p_1^2+p_2^2+2\vec{p}_1\cdot\vec{p}_2}{2(m_1+m_2)}$$
We can easily show that if the particles have oppositely directed momenta or erpendicular momenta, $K_f<K_i$. The maximum of K_f will occur for parallel momenta, and hence parallel velocity.
$$K_{fmax}=\frac{p_1^2+p_2^2+2|p_1||p_2|}{2(m_1+m_2)}=\frac{(|p_1|+|p_2|)^2}{2(m_1+m_2)}.$$
A brief exercise in algebra will show that $K_i\geq K_f$, with the equality happening only if $v_1=v_2$, in which case, the particle never collide.
Conclusion: If the particles collide and stick together the collision cannot be elastic.
