dav2008 said:
Is that a joke post? What do you mean the man has a force of 440N?
I`m sure by force he meant weight.
Momentum is conserved.
(I just googled for "mass of bullet" so I'm using those numbers)
http://hypertextbook.com/facts/2000/ShantayArmstrong.shtml
m_{bullet}=0.0042\ kg
v_{bullet}=900\ m/s
m_{man}=70 \kg
(.0042\ kg)(900\ m/s)+0\ kg\cdot m/s=(v_f)(70.0042\ kg)
v_f=.054\ m/s = .12\ mph
Ok, well let's see what I can do.
A 175 grain bullet (such as wolrams referenced 7.62) weighs in at 11.3 grams (took me a while to find this, and it may be wrong. But if you check it make sure youŕe looking for the right bullet, they`re not all the same). Let's say you`ve got a man at 80 kg (heck even I weigh more than that, and people still think I am malnourished). Now, 2010 joules doesn seem unreasonable for the amount of chemical energy released in the gunpowder charge (dav2008s numbers give about 1700 J). But not all of that energy goes into the bullet, some of it goes into recoil. So, let's assume a simple one-dimensional model, with conservation of energy and momentum.
The exit velocity of the bullet can be determined by
<br />
m_b v_b = m_1 v_1<br />
<br />
\frac{1}{2}m_b v_b^2 + \frac{1}{2}m_1 v_1^2 = 2010 J<br />
Where
m_b Mass of bullet, 11.3 g
m_1 Mass of shooter, 80 kg.
v_b Bullet exit velocity
v_1 One dimensional recoil velocity
We get
v_b = 596 m/s
Not 900 m/s. So either the bullet has a lot more energy in that charge(and I am no munitions expert), or something funny is going on (like momentum not being conserved funny). Incidently at this speed, the bullet has 2006.97 J of kinetic energy.
Now, on impact, the bullet does not impart all of its energy into the target. Nor is it an elastic collision either though. That said, let's solve for each case to get an idea of the limits.
For an elastic collision, we solve again:
<br />
m_b v_b = m_2 v_2 (where m2 =m1, just a different person)
\frac{1}{2} m_b v_b^2 + \frac{1}{2} m_2 v_2^2 = 2006.97
This gives
v_2 = 8.417 \times 10^-2 m/s
Or roughly 3.3 inches per second, 0.188 miles per hours, etc.
And if the bullet were to deposit all of its energy into the target in an inelastic collision (and not exit the target)
m_b v_b = (m_b+m_2) v_2
gives
v_2 = 8.423 \times 10¯2 m/s
Again a tiny 3.3 inches per second. The difference between elastic and inelastic is quite small, but the bullet cannot impart more speed than this to the person.
Now, certainly getting shot by such a large bullet is going to cause all sorts of problems fo the person, that will cause them to fall or stagger, even in the direction the bullet was traveling, but theyŕe not going to go flying or get floored, not by the impact anyway.
I am stupid. Why did I conserve energy in a collision.
