If you stop a pendulum at the bottom of its cycle, so that all of its energy is kinetic energy of the (point) mass itself, then the equations are the same as for a mass sliding horizontally without friction. In short, if the mass M has a velocity v, then the impulse to stop it is the integral of force F(t) times time dt = M v. The energy lost is just (1/2) M v2.
If the mass hits a rigid (infinite mass) stop and bounces off the stop with 100% coefficient of restitution, the direction of pendulum moton is changed, no momentum is transferred to the stop (because the integral F(x,t) dx is zero), and the pendulum loses no kinetic energy. In this case, the integral F(x,t) dt is doubled.
If the pendulum mass is finite size and therefore has angular momentum, then the mass has to hit the stop at the center of its percussion so that both the linear kinetic energy and the rotational kinetic energy are completely absorbed.
For a simple pendulum, you could put a mass equal to the pendulum mass at the bottom of the swing, and if the collision between the pendulum and the mass is completely elastic, all the momentum (and energy) would be transferred to the mass, and the pendulum would stop.