Does the physical cross section of a projectile have an effect on the motion of a target? Why can a baseball (moving with kinetic energy) move a baseball glove (or how a hammer displaces a nail),. yet a bullet can't move a person? I know the laws of conservation of energy, deformation, etc. I'm referring to smaller rounds, because a hit by 20 mm caliber will move a human body. I know the equations except this one thing, now I have ideas but I want to make sure.
That doesn't mean it does not do damage, the bullet exerts a massive force, which can break bones, rip tissue, etc. They can move ballistic gel blocks and send clay flying. But that is by impact force causing a TSC to form and that force is like a spring which sends it flying.
I'm guessing it has to do with the time the bullet exerts its force on tissue?
I'm not sure...
Other questions
Im guessing if we have psuedo-rigid objects, where the bullet doesn't deform as much as it is a tougher material, assuming the bullet also doesn't rebound
using work-energy and the Mechanical energy, if a bullet with 850Joules was about to hit a box which µ=1 so we will say 800N's of opposing force or retarding friction force.
This we will apply work-energy or I guess in this case, work-kinetic energy
W=∆KE but I won't show the final KE, just show TMEf and displacement.
KE=TMEi, so TMEi + Wext = TMEf
our bullet has some energy taken away to do force=>work to the material of bullet, i.e deformation, so the box lets say is displaced by 1 meter, so the bullet will apply a force to the box, now we could get to acceleration and dynamic but we wont digress, we know the box applies 800N's by Ffric, so we must get a force to displace that.
(850Joules)+(-800N's*1meter)=TMEf; 50 Joules remaining which will have been used up in heat,deformation, etc. 
 A: If you fire your projectile on a rigid target it will move, no matter what the diameter of your bullet is. All that counts in this scenario is the momentum and kinetic energy. If the target is a soft target you do also move the part of the target that is directly hit by the bullet, but since this part has a smaller area when you use a smaller caliber the circumference which connects the hit part of the target with the rest of the target is smaller and will therefore not spread the impact energy as much as it would do with a bigger projectile diameter. Also the pressure is higher with same energy and impulse but smaller area. Then you have higher penetration, which also moves target material, an even breaks it.
A: The important point here, which I think has been missed so far, is that momentum goes like velocity while energy goes like the square of velocity:
$$p = mv$$
but
$$E = \frac{mv^2}{2}$$
So let's consider a small object with mass $m$ hitting, and sticking to, a much larger object with mass $M$, if the initial velocity of the small object (bullet or baseball) is $v_i$, then, from conservation of momentum, the final velocity, $v_f$, of the resulting lump of two objects is given by
$$v_f = v_i\frac{m}{M+m}$$
The initial energy, $E_i$ is
$$E_i = \frac{mv_i^2}{2}$$
and the final energy $E_f$ is
$$E_f = \frac{(M+m)v_f^2}{2} = \frac{m^2v_i^2}{2(M+m)}$$
So the energy lost in the collision, dumped into the objects, is
$$E = \frac{mv_i^2}{2}-\frac{m^2v_i^2}{2(M+m)}$$
And, since $m\ll M$,
$$E \approx \frac{mv_i^2}{2}$$
and
$$v_f \approx \frac{v_i m}{M}$$
So consider a baseball, with mass $0.145\textrm{kg}$, moving at $30\textrm{m}/\textrm{s}$, hitting someone of $80\textrm{kg}$.


*

*The baseball dumps about $0.145\times 30^2/160\textrm{J} \approx 65\textrm{J}$.

*The final velocity is about $30\times 0.145/80\textrm{m}/\textrm{s}$: about $0.05\textrm{m}/\textrm{s}$.


Now consider a bullet, mass $0.002\textrm{kg}$ and travelling at $350\textrm{m}/\textrm{s}$, hitting the same person:


*

*This dumps about $0.002 \times 350^2 / 2\textrm{J}$, which is $245\textrm{J}$.

*And the final velocity is $350\times 0.002/80\textrm{m}/\textrm{s}$ or about $0.009\textrm{m}/\textrm{s}$.


So you can see that lighter objects, moving much faster, can dump a lot of energy into a person, even though their momentum is much lower.
(Additionally, of course, the bullet is tiny compared to the baseball, so everything happens over a much smaller area which exceeds the physical strength of the stuff you're made of: a lot of the energy then gets dumped inside you, which is particularly nasty.)
A: 
yet a bullet cant move a person

This is wrong.
Momentum
A bullet hitting a person is a complicated phenomenon so it is best to ignore the detailed effects and imagine a 3-gram projectile hitting a stationary 80000 gram block and coming to rest in the block.
In that case you simply use the principle of conservation of momentum.
Before impact the 3 gram bullet has momentum, the "person" has none.
After impact, the combined 80003 gram human+bullet has the same total momentum.
The momentum in the system of bullet and "human" is unchanged (i.e. conserved)
This means the "human" is moving. However due to the difference in mass, they are moving slowly.
Forces
How does the momentum get transferred? By the bullet exerting a force on the "human" (to accellerate it) and the "human" exerting a force on the bullet (to decellerate it)
The force acts over a limited time - the time it takes for the bullet to come to rest in the frame of reference of the "human".
A: Some bullets like a .38 passes through flesh, while an M-16 bullets tumble through the air to maximize damage and they will knock you back considerably. Slower and larger bullets pass threw flesh slower allowing the kinetic energy to push in forward instead of pushing to the side in the case of a faster and pointier bullet. A good example is a razor blade vs. butter knife against flesh.
