If a bullet were to hit a target point blank, would the force upon impact be the same force as the recoil? When a bullet fires from a gun the force exerted on the bullet by the blast is equal and opposite to what you feel as the recoil. I know that acceleration on both the gun and bullet is different due to their different masses, but I was wondering. If I shot something point blank would that force, the bullet produces, be equal to what I felt in the recoil, but just centered in a small area?
 A: 
If a bullet were to hit a target point blank, would the force upon impact be the same force as the recoil?

The answer is no.
From the conservation of momentum we have:
$m_g*v_g = m_b*v_b$ (1), where $g$ and $b$ come from gun and bullet.
Using the formula for the kinetic energy and the conservation of momentum (1) we get:
$E_b=m_b*v_b^2/2$ (2)
$E_g=m_g*v_g^2/2 = (m_b/m_g)*E_b$ (3)
As $m_g>>m_b$ it immediately follows from (3) that $E_g/E_b<<1$
We also know that: 
$E_g = F_g *d_g$ (4)
$E_b = F_b *d_b$ (5)
where $F_b$ is the average resistance force the bullet faces inside the target and $F_g$ the average force the gun exerts during the recoil till it stops. $d_{b,g}$ are the distances  traveled by the bullet inside the target and the gun, respectively.
From (3), (4), (5) it follows that:
$F_g *d_g / (F_b *d_b) <<1$
which means that:
$F_g <<F_b$ if we consider $d_g = d_b$ (similar traveled distances that is a realistic case for a bulled hitting a relatively tough target situated close to the gun).
The opposite case $F_g = F_b$ is not realistic because it leads to $d_g << d_b$ which means a bullet traveling a quite soft target for a long distance, hundreds of times greater, more precisely $m_g/m_b$ greater, than the distance traveled by the gun back.
A: In a perfectly ideal system, the entire linear momentum of the bullet (and angular momentum if the bullet is spinning, but this is not really relevant to the question) will be conserved. An ideal system implies that all the potential energy from the explosion of gun powder is transferred to the bullet so that $PE = 1/2mv^2$. If the pass of the bullet is constant, velocity will increase with relation to $PE$. If no friction or other forces exist, then the bullet could theoretically travel forever at the same speed (Newton's First Law) and so it wouldn't matter if the gun was at point blank range or not. If it did eventually come into contact with another object, the bullet would exert a force on that object proportional to its change in momentum.
If the system is not closed and outside forces such as friction and gravity can act on the bullet, then the further the bullet is travelling, the more energy it loses. As soon as the bullet is shot, its acceleration decreases because forces are acting against it $\Sigma F = -F_{air friction}-F_{...}$.
Considering that $F=ma$, knowing the mass and acceleration of the bullet when it strikes an object could give you the force, and if you want to know how much pressure is exerted due to the relatively small surface area of the bullet, you could use $P=F/A$ where A is the area of the head of the bullet.
A: In this scenario momentum is conserved; force is not. Think of your using a hammer on a nail. The force your arm applies to the hammer is not the same as the force the hammer applies to the nail (otherwise you could just push the hammer against the nail). However, the momentum your arm puts into the hammer is the same as the momentum the hammer puts into the nail (ignoring any bounce of the hammer).
The difference between the arm/hammer and hammer/nail forces is due to the differing distances over which the force is applied. In the same manner, the force between your gun and your bullet is applied over the barrel length of the gun, while the force between the bullet and the target is applied over the depth of the bullet's penetration into the target. The latter depends strongly on the characteristics of the target's material; harder targets decrease the penetration depth, increasing the force applied. That's why harder targets distort impinging bullets more than soft targets.
