Does the wall exert more energy to bring Car 2 to a stop, because Car
1 only has momentum, but Car 2 has both momentum AND force?
I assume by "exert more energy", you mean does Car 2 absorb more energy than Car 1 when colliding with a perfectly rigid wall. I believe the answer is yes, but only by a relatively tiny amount
An expert in car collision physics (which I am decidedly not) would be the best person to answer this, but consider the following:
Acceleration Force:
To separate the energy absorbed by the acceleration force from the energy absorbed by the average impact force associated with the change in momentum, visualize the car having no velocity and its front bumper in contact with the rigid wall. We then apply a force to the rear of the car equal to the maximum possible accelerating force and measure the compression of the bumper shock absorber spring to calculate the theoretical absorbed energy due to the acceleration force only.
The force that causes the acceleration of the car is the static friction force, $f_s$, of the road acting forward on the drive wheel(s). That force is limited to the maximum possible static friction force, $f_{s-max}$ where
$$f_{s-max}=\mu_{s}N$$
Where $N$ is the normal force of the road acting on the tire and $\mu_s$ is the coefficient of static friction between the road and tire. For simplicity we can take $N$ to equal the weight of the car, $mg$. Thus
$$f_{s-max}=\mu_{s}mg$$
For ordinary (non race car) tires on dry asphalt pavement, $\mu_s\lt 1$. Thus, taking for your 500 kg car and letting $\mu_{s}=1$ the maximum accelerating force becomes
$f_{s-max}=$ 4900 N
If we let the shock absorber spring constant be 410 kN/m $^1$, the compression of the spring would be 0.012 m, and the absorbed energy ($\frac{1}{2}kx^2$) due to the 4900 N force would be 29.3 J
Impact Force:
Now consider the average impact force for stopping the cars in your example. Per the work energy theorem, where the net work done on an object equals its change in kinetic energy.
$$W_{net}=F_{ave}d=\frac{1}{2}mv^2$$ where $F_{ave}$ is the average impact force acting on the car, and $d$ is the stopping distance.
Taking $v$ = 10 m/s (about 22 mph), d= 0.25 m (about 50% of the car crumple zone distance $^2$), and 500 kg for the mass, we obtain an average impact force will be about 25,000 N and absorbed kinetic energy of 6,250 J.
If we were to increase the speed to 27 m/s (about 60 mph) and the average impact force becomes 182,250 N with absorbed energy of 45,562 J.
Hope this helps.
Notes:
Value frequently associated with problems solved on internet, though I could not find the source.
https://www.sciencedirect.com/topics/engineering/crumple-zone#:~:text=In%20frontal%20and%20rear%20impacts,it%20is%20around%20150%20mm