When an in-motion object collides with a stationary object does that necessarily mean that the in-motion object will become stationary after collision When an in-motion object collides with a stationary object, does that necessarily mean that the in-motion object will always become stationary after the collision because of the equation:
$$ v_1=(m_1-m_2)u_2 + \frac{2 m_2 u_2}{m_1+m_2}$$
Where $v_1$ = velocity of the in-motion object after the collision, $m_1$ = mass of the in-motion object, $m_2$ =mass of the stationary object, $u_2$ = velocity of the stationary object.
And because $u_2 = 0$, doesn't that mean that $v_1$ will always be $0$?
 A: If the masses of the two colliding bodies are equal and the moving body collides with the stationary body, they will exchange their velocities according to the equation you stated. It is a consequence of the elastic nature of the collision. However, this equation does not apply if the collision is not perfectly elastic, i.e. the kinetic energy isn't conserved.
I would like to add that the equation you mentioned has a small mistake. The correct equation is:
$$v_1 = \frac{(m_1 - m_2) u_1}{m_1 + m_2} + \frac{2 m_2 u_2}{m_1 + m_2}$$
You miswrote the first fraction. Not that it changes anything in this situation as ($m_1 - m_2$) evaluates to zero.
A: 
When an in-motion object collides with a stationary object, does that necessarily mean that the in-motion object will always become stationary after the collision ?

Not necessarily. It depends on the relative masses of the objects and the coefficient of restitution of the collision.
Suppose the in-motion object has mass $m_1$ and initial velocity $u_1$ and becomes stationary after the collision. Suppose the initially stationary object has mass $m_2$ and velocity $v_2$ after the collision. Then by conservation of momentum we have
$$m_1u_1 = m_2 v_2$$
If the coefficient of restitution is $e$ then we also have
$$v_2 = eu_1
\\ \Rightarrow m_1u_1 = em_2u_1$$
But $u_1$ is non-zero (no collision can take place if both objects are stationary) so we can divide both sides by $u_1$ to get a necessary and sufficient condition:
$$m_1 = em_2$$
In a perfectly elastic collision we have $e=1$, so the condition becomes $m_1=m_2$.
A: Head to head elastic collision
$$v_{1f}={\frac { \left( m_{{1}}-m_{{2}} \right) { v_{1i}}}{m_{{2}}+m_{{1}}}}+2
\,{\frac {m_{{2}}{\,v_{2i}}}{m_{{2}}+m_{{1}}}}$$
$$v_{2f}=2\,{\frac {m_{{1}}{\,v_{1i}}}{m_{{2}}+m_{{1}}}}+{\frac { \left( m_{{2}}
-m_{{1}} \right) {v_{2i}}}{m_{{2}}+m_{{1}}}}$$

*

*$v~$ velocity

*index i intial

*index f final

you can check that with those equations the conservation of the energy
$$\frac 12\,m_1\,v_{1f}^2+\frac 12\,m_2\,v_{2f}^2=
\frac 12\,m_1\,v_{1i}^2+\frac 12\,m_2\,v_{2i}^2$$
and the conservation of the linear momentum
$$m_1\,v_{1f}+m_2\,v_{2f}=m_1\,v_{1i}+m_2\,v_{2i}$$
is fulfilled.
Those:
if $~m_1=m_2~$ and $~v_{2i}=0~$ then $~v_{1f}=0~$ and
$v_{2f}=v_{1i}$
