Condition for 2 bodies to move together The condition for 2 (or more) bodies to move such that they are always in contact, is that their accelerations (and velocities) along their common normal should be same. 
Can someone explain why this is so? 
If we talk about 2D motion, let's say there is a wedge, and a block is kept on it. The wedge is moved horizontally, and so the box moves right and down with respect to ground. So their accelerations along the x-axis have to be same. But how do we conclude that accelerations along common normal have to be same?
Thank you.
 A: Let's say their acceleration and velocities are not equal. Then we can define a relative acceleration $a_r$ and velocity $v_r$ between the two.
After a time $t$, the distance between them would be $$x= \frac {1}{2} a_r t^2 = v_r t$$
Since $x$ might not be equal to zero as $t$ tends to greater values, the bodies will stop being in contact.
A: Consider two bodies sliding past each other along a particular ramp direction.

What is the possible relationship between the velocity vectors $\vec{v}_1$ and $\vec{v}_2$ ? Take the ramp and designate two directions $\hat{n}$ for perpendicular and $\hat{t}$ for along the ramp.
Since the two bodies can slide past each other along $\hat{t}$ direction, their relative velocity has to be along this vector
$$ \vec{v}_2 - \vec{v}_1 = u \hat{t} $$
where $u$ is the slip speed. Re-arrange the above to get
$$ \vec{v}_2 = \vec{v}_1 + u \hat{t} $$
This is the first and simplest of the kinematic relationships between two interacting objects.
Since the direction $\hat{t}$ cannot change with time in this scenario, the time derivative of the above gives us the acceleration kinematics
$$ \vec{a}_2 = \vec{a}_1 + \dot{u} \hat{t} $$
Since the two bodies are rotationally locked, then you also need the complimentary equations for the rotational velocity and rotational acceleration
$$ \vec{\omega}_2 = \vec{\omega}_1 $$
$$ \vec{\alpha}_2 = \vec{\alpha}_1 $$
The next step would be to consider two pinned bodies. But I am leaving this for another question.
A: your question is only a special case of constrained motion,in ur case it may be a wedge block system when there are motions along common normal,but when u talk about motion along x axis only normal reaction constraint does not work,so the material you have picked up only talks about that particular case.ur question that it is always along the normal is not true,thanks.
