When I push an open door I feel less force but when I push a wall I feel a greater force So I observed yesterday when I pushed a closed door I experience a lesser reaction force but when I push a wall I experience a greater force is it because of the fact while pushing an open door rate of change of momentum is less so less force is experienced by me and vice versa
 A: Two things going on here:

*

*The door moves and so the reaction force is less. In fact, if you push at a distance $\ell$ from the pivot, and the mass moment of inertia about the pivot is $I$ then the relationship between force $F$ and rotational acceleration $\alpha$ is $$ F = \tfrac{I}{\ell} \alpha$$


*The wall has a lot more mass than the door, so the force would be much higher, but since the wall is not moving, the reaction force is exactly however much you push.
A: I presume you are contrasting pushing a closed door open, with pushing on a wall? You only experience a greater reaction force from pushing on the wall than from pushing on the closed door to open it if you choose to push harder on the wall than on the door.  You can push gently on the wall and experience a less reactive force.  You can choose to push harder on the open door resulting in a faster opening of the door and a greater reactive force.
The force to open the door is ma where m is the door mass and a its acceleration (actually torque opens the door).  Its just that to open the door with a reasonable acceleration for the door a relatively small force is required, because the door is designed to be opened with a little effort (mass, and hinges and moment of inertia).  If you tried to open a heavy vault door, more force would be required. The force on the wall is just how hard you choose to push on the wall; the wall does not move (it slightly buckles) due to the forces holding the wall in place (assuming you do not break through the wall).
