How does the socket's mass work better? I have two impact sockets. Both of them drive fixtures which have 19mm heads. Both are driven by a 1/2" drive impact gun. Here is an image of the two sockets:

The one on the left is an ordinary impact socket. It weighs in about 10oz (maybe?). The one on the right is one specifically designed to take off the hub bolt on Honda engines (will work in many other applications as well). It weighs in at over 3 lbs. If I use the left impact socket on the Honda hub bolt, in most cases (far in excess of 99+% of the time) it will not take the bolt loose. Yet, with the one on the right, takes only moments to loosen the stubborn bolt. This is with the same impact gun, using the same air pressure (IOW: all other things are equal except the socket itself).
My question is: Can someone explain why this works the way it does? I know the mass of the socket is the real reason ... what I'm looking for is the physics reasons why it works.
EDIT: A newer video on YouTube which tries to answer the question at large, yet stumps the physics instructor who gave them feedback. NOTE: I have no affiliation with the video ... just thought it was pertinent.
 A: It's the greater rigidity of the larger socket rather than its greater mass which makes it more effective.   The physics of the situation is that the change in angular velocity ($\Delta \omega$) of the socket and bolt (which is what we want) is equal to the  impulse delivered by the tool, which is torque ($\tau$) times the amount of time it is appled ($\Delta t$), divided by the combined moment of inertia ($I$) of the socket plus the bolt.
$$\Delta\omega=\frac{\tau\Delta t}{I}$$
The greater mass of the big socket makes $I$ bigger, which makes $\Delta \omega$ smaller.    That doesn't help.    and we can't make $\tau \Delta t$ bigger, because that's a constant determined by the tool.   But $\Delta \omega$ is zero anyway until we get it to start turning, so what we need is the most torque we can get.    What we can do is to make the $\tau$ part, the force part, of that constant impulse bigger if we can make the $\Delta t$ part smaller.   Changing either the force or the time part of an impulse to make the other one bigger or smaller is an engineering principle that gets used a lot, for instance, in air bags.    If the bulky socket is more rigid, then in slow motion we might see that it is considerably less springy than the small one.  So as the flywheel in the tool comes to a stop in a shorter time trying to turn it, it delivers a bigger jolt of force to get the bolt moving.
A: I don't think the explanation is to do with the rigidity of the bigger socket. The difference is that to work efficiently, the rotational inertia of the socket has to match the inertia of what is driving it.
What happens at each "stroke" of the impact is basically the same as a collision between the parts of the gun which would rotate if there was nothing to stop them, and the socket. If you have a large gun and a small socket, most of the energy is wasted trying to turn the gun, not the socket. The best you can achieve is when the two are correctly matched, and (theoretically) all the energy from the gun motor gets into the socket while the "recoil" of the gun is just enough to stop it turning at all. If the socket is too heavy for the gun, again you waste energy when the gun recoils and tries to reverse its direction of rotation.
Of course there is no guarantee your large socket is "correctly" matched to the gun you are using, but apparently it is a better match than the small one.
As an analogy of this, think about hitting a ball with a hammer to start it rolling. If the hammer is much heavier than the ball, most of the energy you put into the hammer remains in the hammer, which continues to swing after you hit the ball. If the hammer is much too light, it will "bounce off" the ball after the impact. The most efficient situation is somewhere in between those two extremes.
A: Look at the width of the casing on the big socket, compared to the smaller one. The torque is, in effect higher for the right hand side socket, especially at relatively high speeds.  While you are not applying the force in the same way as you would by using a long lever on a wheelbrace, once you start up the drill, you are getting some of the lever effect.
It's a guess for sure, but between that and the extra mass overcoming the intertia/initial stickiness of the bolt, I think that's the reason.
If not, I will be delighted if someone comes up with a better (or the correct) one.  
