# Metal hammer vs rubber mallet

Nobody; it was a question - hence the word ‘if’ and the question mark.

None of which seems particularly relevant; you seemed, in your p-(-p)=2p example, to be describing a system wherein the hammer imparts energy/momentum to the nail, but also ends up keeping it - where is the extra energy coming from? (Angular momentum from the spinning corpse of Isaac Newton, I’m starting to suspect)

A few things are getting confsed here. Let’s see if my rusty high school physics can help.

#1. The principle of conservation of momentum.
This only applies in an elastic collision. The rubber hammer behaves close to elastic in this case. The momentum before hitting the nail is close to the momentum afterwards. Much the same as a billiard ball hitting the cushion. I don’t believe that it is helpful to consider momentum as a vector in this case.
The metal hammer in this case has an inelastic collision. The momentum is therefore not conserved. The metal hammer may stop dead and the nail may move a small distance, but does not gain anything like the same momentum that the hammer had.

#2 The principle of conservation of energy.
This applies in all collisions – elastic and inelastic. With the rubber mallet, the kinetic energy is temporarily converted to elastic potential energy as the rubber is compressed then returned to the system as kinetic energy with the mallet moving in the opposite direction. As usual some energy is dissipated as heat in the mallet, nail and air.
With the metal hammer, a large proportion of the energy is transferred to the nail and breaks apart the fibres of the wood that the nail is being driven into.

The reaons for the difference? The metal has a higher modulus of elasticity than the rubber. Which means that for the same applied force over the same area the rubber will deform or “give” a lot more than the metal. (I am making some assumptions here – the nail does not deform, the metal and rubber both deform linearly with applied force, neither material undergoes plastic deformation – none of these assumptions are necessarily true.) Which means that the distance over which the collision occurs and the time that the force is applied is greater for the rubber mallet. The net result is that the rubber absorbs the energy of the collision wheras the metal tends to transfer the energy to the nail. And since the nail is able to absorb this energy by moving into the wood, that is what happens.
If, on the other hand, you were to hit the hammer against a thick piece of high tensile steel instead of a nail, you would find that the hammer would bounce back at least as well as the rubber mallet.

If I recall correctly, the principle of conservation of momentum always applies. Apparently Wikipedia agrees.

So the transferred momentum is close to 2p, right?

But momentum IS a vector. And even if you didn’t, you must have a way to tell the mallet moving forward from the mallet moving backwards - that would be, of course, plus and minus signs.

It must be conserved for the system (hammer+nail).

Not the nail, but the sum of the two.

Instead, it was conservation of mechanical energy which didn’t apply to inelastic collisions.

No energy would be dissipated if the collision is assumed to be perfectly elastic.

And how is that possible, if less momentum is transferred to the nail?

I don’t know about energy, but I do know how much momentum the mallet has before hitting the nail, and how much momentum it has after that. And the difference, as any mathematician can tell you, cannot be anything other than 2p.

Understood. But still not too convinced. When the mallet begins to un-deform, doesn’t it exert more force on the nail than we have so far taken into account?

The system is larger than the hammer and nail. The system is the hammer, nail, and Ground. With respect to this larger system, the rubber mallet is more elastic than the metal hammer, because more of the kinetic energy remains after the collision.

Believe it or not, this was the subject of a paper in American Journal of Physics way back about 20 years ago. It is better to use a light, deformable hjammer or a heavy rigid one. The only difference is that they were comparing brass and steel hammers.

It depends upon the job being done. Just for the record, Moment is always conserved, but energy may be lost to deformation. I the Real World, collisions are rarely perfectly elastic or inelastic. You can define a coefficient of restitution for the collision that is (IIRC) the ratio of the velcity difference after collision to that before collision. It varies from 0 (perfectly inelastic) to 1 (perfectly elastic).

By the way, you don’t want to use a rubber mallet to drive nails. Not if you want to keep your rubber mallet. The small heads will pit and deteriorate the rubber mallet.

The momentum transferred from a rifle butt to the shooter’s shoulder is identical to the momentum transferred from the bullet to the person shot. (assume no air resistance, no through & through shot) The effect is vastly different depending on the manner in which the momentum is transferred.

The steel hammer will deliver significantly higher forces because the time over which the momentum gets transferred is shorter than with the rubber mallet. This is identified by the impulse.

Total momentum transferred via hammer will be a high force x short time, via mallet - lower force x longer time. Since the forces never get high enough to break/split the wood fibers, the rubber mallet sucks at driving nails, but is good at setting wooden pieces together (think mortise and tenon) without damaging the wood.

Believe it or not, this was the subject of a paper in American Journal of Physics way back about 20 years ago. It is better to use a light, deformable hjammer or a heavy rigid one. The only difference is that they were comparing brass and steel hammers.

It depends upon the job being done. Just for the record, Moment is always conserved, but energy may be lost to deformation. I the Real World, collisions are rarely perfectly elastic or inelastic. You can define a coefficient of restitution for the collision that is (IIRC) the ratio of the velcity difference after collision to that before collision. It varies from 0 (perfectly inelastic) to 1 (perfectly elastic).

By the way, you don’t want to use a rubber mallet to drive nails. Not if you want to keep your rubber mallet. The small heads will pit and deteriorate the rubber mallet.

Both momentum and energy are always conserved. The difference is that energy can take many forms, some more useful or interesting than others. When two objects collide, some of the energy may end up in the motions of the objects afterwards, but some may also end up in heat, or sound, or other unusable forms. Momentum, however, can’t hide in this way. Where you have momentum, you have something moving. So while energy and momentum are both conserved, it’s only generally conservation of momentum that’s relevant in a collision, because the energy can hide in forms like heat that aren’t too relevant to the problem.

Wading through the posts on this thread, I’ll admit I was growing frustrated with some of the muddled explanations until I read this succinct, exact response on the issue from Cheesesteak.

The parameters of FibonacciProwler’s original problem–equal weight hammers, similar striking, etc.–semm to be tailored such that the hammers each strike the nail with the same momentum/energy. As Cheesesteak rightly points out, there is a trade off between the two hammers: Each apply the same energy/momentum, but the steel hammer does it with higher force/shorter time compared to the mallet’s smaller force/longer time. Since you need the driving force to exceed a certain threshold to start the nail moving against friction, the steel hammer is better for driving the nail even though the same amount of energy/momentum is transferred as with the rubber mallet.

Isn’t it possible that not all of the rubber mallet’s momentum is transferred directly DOWN on the nail? That is, as it deforms around the nail’s head, some of the momentum is pushing sideways, at an angle, and therefore working against the momentum pushing from the opposite side of the nail pushing at an angle. A steel hammer is not going to deform around the edges as much, and therefore transfer more of it’s momentum straight down, and not work against itself.

It can in this case. If you consider the nail and mallet as one closed system, your calculations will be off, because at the end, you will have the “opposite” momentum that you had in the beginning, because momentum is a vector. Some of the “hidden” momentum is transferred to the nail, but the rest is sent through to the rest of the board/house/planet.

Not in elastic collisions, which we are discussing here

To elaborate on conservation of momentum:

Within a system, momentum is conserved if there is no net external force acting on the system. That’s Newton’s Second Law: the net force acting on a system is equal to the rate of change in its total momentum.

There seems to be a problem here.

Let’s say the mass of the hammer is m[sub]1[/sub], and the mass of the nail is m[sub]2[/sub]. The initial velocity of the hammer is v[sub]1[/sub], the final velocity of the nail is v[sub]2[/sub].

Momentum is mv, so conservation of momentum gives us:

m[sub]1[/sub]v[sub]1[/sub] = p[sub]final[/sub] + m[sub]2[/sub]v[sub]2[/sub]

where p[sub]final[/sub] is the final momentum of the hammer. But in your example, p[sub]final[/sub] is negative, meaning m[sub]2[/sub]v[sub]2[/sub] > m[sub]1[/sub]v[sub]1[/sub].

So far so good. But since hammers are bigger than nails, we have m[sub]1[/sub] > m[sub]2[/sub]

In order to get m[sub]2[/sub]v[sub]2[/sub] > m[sub]1[/sub]v[sub]1[/sub] with m[sub]1[/sub] > m[sub]2[/sub], you have to have v[sub]2[/sub] > v[sub]1[/sub]. (Note that we have v[sub]1[/sub] and v[sub]2[/sub] both greater than zero.)

From m[sub]2[/sub]v[sub]2[/sub] > m[sub]1[/sub]v[sub]1[/sub], and v[sub]2[/sub] > v[sub]1[/sub] > 0, we get:
m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup] > m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]

Dividing both sides by two gives:
(1/2)m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup] > (1/2)m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]

But (1/2)m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup] is the initial kinetic energy of the hammer, and (1/2)m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup] is the final kinetic energy of the nail.

So the nail has more kinetic energy after the collision than the hammer had before it. Where did the extra energy come from? (Note that unlike momentum, the final kinetic energy of the hammer can’t be negative.)

If you answered “from the resistance provided by the board that the nail is being driven into”, then that’s an external force on the hammer-nail system, meaning the system’s total momentum is not conserved. Your assumption of conservation of momentum only works if you assume the board exerts negligible force on the nail.