The physics of dead blow hammers

Factual Questions
1

So, when we want to knock something a little bit to get into place, we reach for a lead hammer, a plastic hammer, or a … dead blow hammer, filled with shot.

I have always had an intuitive feel that a dead blow hammer transfers more energy to the object, since it pretty much stops when strike something with it. But the more I try to think of this from a physics standpoint I can’t make sense out of what is going on.

So, consider two equal weight plastic hammer heads, one solid other shot-filled, identical in all respects except that one has shot in it and the other doesn’t. What is going on here that makes the dead blow hammer seem so useful?

2

I am no physicist but it seems self-evident that shot/sand/whatever-filled hammer will have the effect of spreading out the impact over a longer period than a solid one.

3

Same mass, same force.

4

Agreed. If we make the hollow hammer weigh the same as the shot-filled one, the “area under the curve” for force applied would be the same for both hammers, but the peak force would be less for the shot-filled one. This is good for working on materials that are brittle, I suspect.

5

A dead blow hammer doesn’t bounce, so more of the energy goes into the material you hit instead of half it returning to the hammer and your arm. I understand that the loose shot in the hammer redirects the energy from the bounce so some of that energy must be wasted by being redirected sideways. They are also soft-faced hammers so you can hit harder without damaging the material, and heavier than other soft materials. I don’t know if it’s any different than facing a hard hammer with a slightly compressible soft material.

6

I’m not sure how accurate this is, but this is how I’ve always looked at it. You start to swing the hammer. The head and the shot within it start moving toward the target. The shot is at this point more or less in free fall within the head. Then the head hits the target. The shot is still moving towards the front face of the head. The head compresses, and then the elastic expansion tries to make it bounce back, but for several milliseconds the shot is still raining down on the inside of the face, preventing that from happening. By the time the shot stops hitting the head, the expansion has completed, so the head doesn’t bounce.

7

Assuming the mass and length are the same, the force should be the same.

The main difference is that the regular hammer has a mostly elastic collision- whatever’s being hammered gets hammered, and the hammer itself rebounds upward.

A dead-blow hammer converts some of that energy that would be used in the rebound (and presumably in the hammering) into motion/friction within the hammer itself. This prevents the rebound, but doesn’t necessarily transmit more energy to the hammered object.

8

I think, the answer will be obvious if you consider a third hammer in this experiment : one where the head is built as a coiled spring with the same mass as the other hammers.

In simple terms, the more “springy” the hammer is, the more energy it can store and not deliver to the target. You want the opposite of a spring I.e. a weight that cannot store energy as deformation.

9

To put this in physics terms, striking an object with a normal hard-faced steel hammer is a highly elastic impact; that is, the head of the hammer is one unified solid mass which will rebound with much of the energy that was put into it unless the surface being struck is very compliant and has high dampening, and the hammer rebounds with most of the momentum, just in the opposite direction. Striking an object with a dead-blow hammer, on the other hand, is extremely inelastic; nearly all of the energy is delivered to the target surface and the hammer has almost no residual momentum because the shot or slug internal to the hammer flies away from the striking surface so there is almost no mass to which to back-transfer momentum.

You can visualize this another way by dropping a rubber ball versus a beanbag or a ball of clay. The ball and beanbag/clay will impact with the same kinetic energy, but while the ball will bounce back up nearly to the height as which it was dropped, the beanbag/clay will just kind of flop in place. This isn’t because the individual sand particles are less elastic but because of all of the hysteresis losses involved in the interactions between sand particles or deformation in the clay, whereas the ball has extremely low hysteresis because it is a solid mass of nearly incompressible material (ν~0.5).

This has nothing to do with the shape or duration of the impulse; it is just whether the hammer head is rigid enough that it can accept momentum back from the object it just struck.

Stranger

10

A deadblow hammer will certainly have less energy in its rebound, but that doesn’t necessarily mean more transferred to the target. I’d guess that most of that energy ends up heating the shot.

11

This is where I got stuck when trying to think it out. I figured that all that was happening was that the shot was heating up a little.

Anyway, I need to gently persuade my lathe chuck to move a half of a thou this way or that before snugging down the bolts holding it in place, so the mysterious dead blow hammer will come into play.

12

No, when using a shot-filled dead-blow hammer to drive a wedge or fireside friend into a knotty round to split firewood it is definitely more effective than hitting with a solid head hammer of the same weight, and safer as well since it has virtually no rebound; I can often split a round in one or two strikes with a 3# dead-blow behind a friend that will typically take three or four with a 3# sledge. I’m sure there is some heating due to friction between the shot itself (although since it is in a plastic head it is hard to tell) but there are also solid slug dead-blow hammers that will only experience the internal hysteresis in the steel slug.

Stranger

13

These kinds of things are tricky, in that it’s not abundantly clear what you’re really looking for. I dealt with this kind of thing a lot when I was doing my bachelor’s thesis on the physics of a karate strike.

The simplest thing is to treat this as an elastic collision, or an inelastic collision, or something between. The last is probably the closest, but the first two cases are easier to treat.

Assume you’ve got a hammer with head weight m hitting a much heavier target of weight M. The hammer is moving with initial velocity v[sub]i[/sub]. Moment is conserved in any case, so for an inelastic collision you have:

mv[sub]i[/sub] = (m + M)v[sub]f[/sub]

so

v[sub]f[/sub] = (m/(m + M))v[sub]i[/sub]

or, if M >> m

**v[sub]f[/sub] ** is about v[sub]i[/sub] (m/M)
In an elastic collision, if M >> m, the final velocity of the hammer is just -v[sub]i[/sub], to good approximation, so conservation of momentum gives you

mv[sub]i[/sub] = Mv - mv[sub]i[/sub] or, for the final velocity of the heavy object that was hit:

v = 2v[sub]i/sub

So the big mass moves of with very nearly twice the velocity it does in the inelastic case. Of course, since M >> m, the final velocity is pretty small in either case.

So the elastic case is better, right? Well, not obviously. Stranger’s experience is the same as many others – you seem to get better results with a deforming hammer. The truth is, treating the collision as elastic, or inelastic, or partially elastic* is an oversimplification. as i found from examining high-speed photographs of those beams and blocks struck by the karate expert, the break occurs while the hand is in contact, so there isn’t a clear-cut “before” and “after” impact. The interesting stuff happens during impact.
If you calculate the force, which is the momentum change with respect to time, your average force is going to be smaller in an inelastic impact, if only because the impact time will be longer (I guarantee that the elastic interaction between a hard steel hammer and a hard metal surface is much shorter than the time it takes a ball of putty, or a ball bearing-filled hammer to fully respond). So, again, the “elastic” collision is better, right?

Well, it’s not clear that greater instantaneous force is more effective in splitting wood or forging metal or whatever you’re doing.

Momentum id conserved, but energy isn’t. Certainly the difference in energy before and after impact is due to deformation of the hammer (which will heat the ball bearing to a small degree, but will heat the stuff containing them more, I expect. And other things in the environment). You typically use a dead blow hammer to spread out the force in time and area to avoid leaving marks. I’ll have to try one for splitting wood – I’ve been doing a lot of than lately, and I wouldn’t mind using a tool that required less work on my part.

14

I never tried splitting wood with a dead blow hammer but I’m sure it’s easier on your hands, arms and shoulders, and your back, hips, legs and feet. As a result it should make the manual wood splitting process overall more efficient for the human-hammer machine. I still think there’s a good bit of wasted energy imparted to the sides of the hammer, like a bean bag hitting an object the contents will try to spread out but is constrained by the sides of the hammer body. And with loose shot the impact must be producing numerous little collision and bounces inside the hammer, some of which will still be redirected to the head I suppose, but much of it going in all other directions.

Is a loose shot dead blow hammer going to behave differently than one made of a solid material with the proper elastic property? Overall it seems similar to mounting a hammer head on the end of a shock absorber.

15

Could be something about the way wood splits- maybe there’s a component of force over time there that we might not see in other situations.

16

I suspect it has more to do with the fact that with a dead-blow hammer the hammer doesn’t participate in any kind of resonance with the wedge and so the wedge just tends to seat more firmly instead of springing back out. Sometimes with a wedge that isn’t firmly placed, hitting it with a small sledge (really more of a linesman hammer) will actually cause it to pop back out and need to be reset. I’ve never had that happen with a dead-blow hammer.

Stranger

17

And just how are you going to do that?

18

Make it out of a denser material, of course.

19

Make the walls of the empty hollow hammer a bit thicker than the walls of the hollow shot filled hammer. Or make the empty hammer out of something a little more dense than the shot filled one is.

20

When someone swings a dead-blow hammer, do they swing harder, because they know that there will be no dangerous rebound? (P.S. This is a sincere question - I’m just wondering).