Are these collisions equivalent? Similar to the question if two cars with a velocity of 50 mph each colliding is the same as one car colliding with wall at 100 mph, I was wondering if the same amount of energy is produced when hitting a stationary object at 50 mph as hitting an object that's moving away from you at 20 mph with your velocity being 70 mph?
 A: Yes, you have to consider what the frame of reference is for the colliding objects.
The relative velocity between the objects is given by $v = v_1 - v_2$ where $v_2$ is the velocity of the stationary object and $v_1$ is the velocity of the car.
In the first situation the relative velocity is is $(50 - 0)\ \rm{mph}$. Thus the first object will hit the standing object at $50\ \rm{mph}$.
In the second situation, the relative velocity is $(70 - 20)\ \rm{mph}$ so again the car will hit the moving object at $50\ \rm{mph}$. 
In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by $50\ \rm{mph}$ during the impact time with everything else kept equal.
Edit:
In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to $50\ \rm{mph}$ in both equations, you will find that the collisions are the same.
Note:
Kinetic energy depends on the frame of reference. 
If the ground is the frame of reference, then the total energy in the first situation is $\frac 12\cdot 50^2 = 1250\ \rm{units}$
In the second frame of reference the total energy is $\frac 12\cdot70^2 + \frac 12\cdot 20^2 =  2650\ \rm{units}$.
Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at $50\ \rm{mph}$ relative to the ground. The amount of kinetic energy the object gained was $1250\ \rm{units}$ relative to the ground. 
In the second situation, the impacted object now moves at $70\ \rm{mph}$, and gained $\frac 12\cdot70^2 - \frac 12\cdot20^2 = 2250\ \rm{units}$ of kinetic energy relative to the ground. 
From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions still experience the same change in momentum.
The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt). 
A: It depends on your meaning of "is produced", but yes, in some important sense the same energy is produced.
Suppose you have a bunch of masses $m_i$ moving with velocities $\vec v_i$. The overall momentum in this reference frame is $$\vec P = \sum_i m_i \vec v_i$$ and the overall kinetic energy is $$K = \sum_i~\frac12 m_i v_i^2.$$These quantities are well-known to be reference-frame dependent, meaning that if you were to add the same velocity $\vec u$ to all of the particles, neither would stay the same. The first would transform to $\vec P' = \vec P + M \vec u$ where $M = \sum_i m_i,$ and the second will transform to $$K' = \sum_i\frac12 m_i\left(v_i^2 + 2 \vec v_i\cdot \vec u + u^2\right) = K + \vec P\cdot \vec u + \frac12 M u^2.$$
However there are things which are reference-frame independent which come from comparing two configurations of these particles at two different times. So if you had some net momentum $\vec P_0$ and it later became some different net momentum $\vec P_1,$ and the total mass $M$ did not change, you would find that this momentum difference $\Delta\vec P = \vec P_1 - \vec P_0$ to be independent of $\vec u$ because you'd have $$\Delta \vec P' = \vec P_1' - \vec P_0' = (\vec P_1 + M \vec u) - (\vec P_0 + M \vec u) = \vec P_1 - \vec P_0 = \Delta \vec P.$$
On the other hand the kinetic energy difference is still reference-frame-dependent, working out to:$$\Delta K' = K_1' - K_0' = \Delta K + \Delta \vec P \cdot \vec u.$$
So it is only reference-frame independent when we are talking about situations where the net momentum doesn't change, $\Delta \vec P = \vec 0.$
But wait. The cases that you are talking about involve two cars interacting purely with each other, not with some "external" force. This means that they have to obey Newton's third law, which forces momentum to be conserved in the collision, $\Delta \vec P = \vec 0.$ Of course if they collide and then skid to a stop against some wall or the ground or so, then that violates this momentum conservation unless we include the whole Earth as a set of particles into our system, but for just the collision itself, yes, we can model that assuming conservation of momentum.
So if we look only at the collision between the cars and define $E=-\Delta K$, the (now positive) energy dissipated in the collision, as the "energy produced", then yes, that is the same. So if you rear-end a car going 10mph while you are going 40mph, that is going to be largely the same as rear-ending a car going 30mph while you are going 60mph, in terms of the damage you will see. It will be slightly different from what you may see from crashing into a parked car going 0mph while you are going 30mph, but only because that car's brakes are probably engaged from the start and so we have to consider these "external forces." For that matter when you rear-end someone there is a chance that they hit the brakes reflexively (or you both hit a wall or so) and then there will be extra damage from the influence of these external forces when moving at speed. All such examples must connect to the outside world to create collisions where momentum is not conserved -- for all cases where momentum is conserved, the energy change of the collision does not depend on any absolute velocities, only the particles' velocities relative to each other.
