Bull charges you or you charge the bull? I'm asking about two specific scenarios.  Let's say a young man charged a stationary bull or a bull charges a stationary human.  Physically, are these the same scenarios?  The bull and human end up attach at the end due to the horns.  Their movement velocity could be 10 m/s let's say for simplicity.  I know the kinetic energy is higher after the collision in the charging bull case using conservation of momentum.  However, I believe the difference in energy from after and before the collision is the same.  
I'm just interested since I thought about this as I was thinking about personal experiences going up against more massive humans.  Please be aware that no acceleration happens before.  
 A: It all boils down to the motion of the centre of mass of the man and the bull.   
If there are no horizontal external forces then the momentum of the centre of mass of the man and bull system cannot change.
Without doing the sums, if the bull is carrying the momentum (the bull is moving and the man is stationary) then the velocity of the centre of mass of the bull and man system will be just smaller than the velocity of the bull because the mass of the bull is so much greater than the mass of the man.
Thus the amount of kinetic energy available to be lost in the inelastic collision compared with the kinetic energy that the bull had before the collision is very small as the bull does not slow down very much.  
If the man is carrying the momentum then the velocity of the centre of mass of the man and the bull will be much lower than the velocity of the man and so there is lots of kinetic energy available to be lost in an inelastic collision and the amount which can be lost is comparable to the kinetic energy of the man.  

Bull mass $M$ with speed $v$ and man mass $m$ at rest.
Kinetic energy of the bull $\dfrac 1 2 Mv^2$ and the speed of the centre of mass is $\dfrac {M}{M+m} v$.  
Kinetic energy associated with the centre of mass motion is $\dfrac 1 2 (M+m)\dfrac {M^2}{(M+m)^2}v^2= \dfrac 1 2 \dfrac {M^2}{M+m} v^2$  
So the amount of kinetic energy which can be lost is $\dfrac 1 2 Mv^2-\dfrac 1 2 \dfrac {M^2}{M+m} v^2 = \dfrac 1 2 \dfrac {Mm}{M+m} v^2$.
This works out to be the same amount if the man was carrying the momentum.
If $M\gg m$ then $\dfrac 1 2 \dfrac {Mm}{M+m} v^2 \approx \dfrac 1 2 m v^2$ and so in both cases the amount of kinetic energy which could be lost is the same but when the bull is moving it has much more kinetic energy to start with than when the man is moving.
A: This is a totally inelastic collision in which a moving object of mass $m_1$ collides with stationary object of mass $m_2$. 
The fraction of KE lost is the ratio of stationary to total mass :
$\frac{\Delta K_1}{K_1}=\frac{m_2}{m_1+m_2}$.
So a greater fraction of KE is lost if the bull is stationary. However, there is more KE if the bull is moving, so these 2 factors may cancel out.
Initial velocity is the same, therefore $K \propto m$. The absolute loss in KE is therefore in the ratio
$\frac{\Delta K_1}{\Delta K_2}=\frac{K_1m_2}{K_2m_1}=\frac{m_1m_2}{m_2m_1}=1$.
As you expected, the absolute loss of KE is the same in both cases.
