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#1
Old 07-10-2006, 12:52 PM
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Metal hammer vs rubber mallet

Simply put, which one works better?
Let's suppose I have a rubber mallet and a metal hammer, essentially identical in every aspect other than the stuff they're made of. As I see it, when the metallic one comes in contact with the nail, the collision can be modelled as a perfectly inelastic one, so all of the hammer's momentum and kinetic energy would be transferred to the nail. On the other hand, the rubber mallet hits the nail, it's a perfectly elastic collision we're talking about, since the mallet recoils at about the same speed as it had when it hit the nail; so twice its momentum, but no kinetic energy would be transferred to the nail?!?

Please help me here, guys.
#2
Old 07-10-2006, 01:14 PM
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I guess what you're getting at must really be a physics question, because they're not identical in any aspect, and you don't use a rubber mallet to drive a nail.

The model you describe for the rubber mallet only works if the nail is in an impenetrable substance. The mallet would bounce off and some energy would be transferred to the nail since the nail did exeprience a force, and that energy would take the form of heat. Other energy would take the form of heat in the mallet from the deformation and reformation. If the surface were ordinary wood then the nail would drive into the wood to some degree, even with an elastic collision with the mallet, which would also result in some heat.

BTW steel is rather elastic. I guess when you say "inelastic" you are talking about the permanent deformation of the nail and the wood as a system, i.e., the nail goes into the wood.

All of this notwithstanding I'm not sure what you're question is. Which one works better for driving a nail? The metal hammer. But your analysis shows you knew that already.
#3
Old 07-10-2006, 01:17 PM
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Still, the momentum transferred from the rubber mallet is greater than that from the metal hammer (let's suppose they have the same mass), for the recoil is greater. Why is the metal hammer more effective, then?
#4
Old 07-10-2006, 01:22 PM
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The steel hammer will be more elastic than the rubber one, and therefore impart more momentum to the nail. Steel is much more elastic than rubber, in the physics sense of the term, and elastic collisions transfer more momentum.

And steel hammers are more effective at driving nails, but rubber ones are more effective at (for instance) driving chisels. They're intended for completely different purposes.
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#5
Old 07-10-2006, 01:28 PM
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Also, it can make a difference what you're hitting. If you're driving in metal tent stakes, you need to use a metal hammer or you're just going to chew the heck out of a rubber mallet. On the other hand, if you try to pound in plastic stakes with a metal hammer, you stand a much better chance of breaking the stake.
#6
Old 07-10-2006, 01:35 PM
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Quote:
Originally Posted by Chronos
The steel hammer will be more elastic than the rubber one, and therefore impart more momentum to the nail. Steel is much more elastic than rubber, in the physics sense of the term, and elastic collisions transfer more momentum.
But less kinetic energy?
Also, I've seen the rubber mallet recoil more than the steel hammer. How is it that steel is more elastic?

Quote:
Originally Posted by Chronos
And steel hammers are more effective at driving nails, but rubber ones are more effective at (for instance) driving chisels. They're intended for completely different purposes.
And what's the physical rationale behind that?

Quote:
Originally Posted by Ethilrist
Also, it can make a difference what you're hitting. If you're driving in metal tent stakes, you need to use a metal hammer or you're just going to chew the heck out of a rubber mallet. On the other hand, if you try to pound in plastic stakes with a metal hammer, you stand a much better chance of breaking the stake.
Let's suppose, just as in the physics textbooks, both hammer and object being driven are unbreakable...
#7
Old 07-10-2006, 04:53 PM
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Here's my guess as to what's going on:

So suppose that you have a rubber mallet and a steel mallet about to impact on a nail, both having the same kinetic energy. They will both contact the mallet and begin to deform, the nail doing work on each mallet as it comes to a rest. However, the rubber mallet will deform more (i.e. a greater distance) than the steel mallet. Since the same amount of work was done on each mallet (they had the same initial and final kinetic energies), since work is distance times force, and since the work done on the steel mallet was done over a shorter distance, we conclude that the steel mallet exerted a greater force on the nail.

Aha, you say, then why did the nail hit by the steel mallet go in further, when the force wasn't exerted over very much distance? Well, the truth is that the above analysis only really works while we're trying to get the nail started, and it's not yet moving. When you're driving a nail into wood, you're essentially trying to make the wood grains "fail" in some way. Once you've managed to make part of a structure fail (in this case, part of the wood grain), it's often not too hard to make the rest of the structure fail as well. We all saw a particularly ghastly example of that about five years ago.

So the basic reason you want to exert a steel mallet rather than a rubber one is that it exerts much more force on the nail, and this force is enough to drive the wood into "failure mode" while that exerted by a rubber mallet isn't.

Hope this helps...
#8
Old 07-10-2006, 05:01 PM
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Quote:
Originally Posted by MikeS
Since the same amount of work was done on each mallet (they had the same initial and final kinetic energies)
I don't think that assumption is valid. Since the rubber mallet bounces more after hitting the nail than the steel mallet, wouldn't that mean that its final kinetic energy is greater than that of the metallic one?
#9
Old 07-10-2006, 05:14 PM
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Metal hammer vs rubber mallet
Quote:
Originally Posted by FibonacciPrower
Simply put, which one works better?
Which one works better for what?
For practical uses...
Metal hammers are for driving nails, heavy duty staples, etc.
A rubber mallet is for use in fitting and assembly of wooden components.
A heavy mallet with rawhide faces for assembly of metal components.
And there at times that a small metal mallet with plastic or brass faces is handy too!
PS There is also a 'dead' hammer or mallet that doesn't bounce at all.
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#10
Old 07-10-2006, 05:29 PM
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Quote:
Originally Posted by FibonacciPrower
Still, the momentum transferred from the rubber mallet is greater than that from the metal hammer (let's suppose they have the same mass), for the recoil is greater.
I think this is incorrect; the recoil is greater because rather than being transferred to the nail, the energy is temporarily stored as potential energy in the elastic deformation of the rubber; this potential energy is released, acting to cause the recoil.

If more momentum was transferred, it would recoil less.
#11
Old 07-10-2006, 06:41 PM
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Quote:
Originally Posted by Mangetout
I think this is incorrect; the recoil is greater because rather than being transferred to the nail, the energy is temporarily stored as potential energy in the elastic deformation of the rubber; this potential energy is released, acting to cause the recoil.

If more momentum was transferred, it would recoil less.
Nope.

Before hitting the nails, both hammers have momentum p.
After hitting the nail, the metal hammer has momentum 0 (let's suppose it doesn't recoil at all), so the transferred momentum is p-0=p.
Likewise, the rubber mallet has momentum -p after hitting the nail. Therefore, the transferred momentum is p-(-p)=2p.

Which one is obviously greater?
#12
Old 07-10-2006, 06:50 PM
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If the nail was a theoretically immovable object, would any momentum be transferred?
#13
Old 07-10-2006, 08:03 PM
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What if Jesus was swinging the hammer?
#14
Old 07-10-2006, 09:01 PM
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Quote:
Originally Posted by FibonacciPrower
Nope.

Before hitting the nails, both hammers have momentum p.
After hitting the nail, the metal hammer has momentum 0 (let's suppose it doesn't recoil at all), so the transferred momentum is p-0=p.
Likewise, the rubber mallet has momentum -p after hitting the nail. Therefore, the transferred momentum is p-(-p)=2p.

Which one is obviously greater?
Wait a minute; so the hammer is transferring more momentum than it actually has? There should be a law against it.
#15
Old 07-10-2006, 10:26 PM
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Quote:
Originally Posted by Mangetout
If the nail was a theoretically immovable object, would any momentum be transferred?
Who said the nail was a theoretically unmovable object?
#16
Old 07-10-2006, 10:27 PM
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Quote:
Originally Posted by Mangetout
Wait a minute; so the hammer is transferring more momentum than it actually has? There should be a law against it.
More momentum than it has with respect to you, yes, and much more than it has with respect to itself. But not more than it has with respect to, say, a mosquito which is flying away from the nail when it sees the hammer.
#17
Old 07-10-2006, 10:52 PM
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As a machinist I use a dead blow hammer with one good thwack to set a workpiece in a vise before milling it. Using a steel hammer makes the workpiece bounce back against the bottom of the vise and it is loose in the vise.

I don't want to hurt my head with the math and or physics but in a real world application there is a reason for a dead blow mallet.
#18
Old 07-10-2006, 11:29 PM
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Quote:
Originally Posted by FibonacciPrower
I don't think that assumption is valid. Since the rubber mallet bounces more after hitting the nail than the steel mallet, wouldn't that mean that its final kinetic energy is greater than that of the metallic one?
I mean from the point just before the mallet contacts the hammer to the point where it comes to rest. Sorry if that wasn't clear.
#19
Old 07-11-2006, 12:16 AM
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Quote:
Originally Posted by MikeS
I mean from the point just before the mallet contacts the hammer to the point where it comes to rest. Sorry if that wasn't clear.
Does it actually come to rest? If the hammer bounces back, it would continue moving indefinitely (in the absence of air resistance, opposite walls, or bystanders which might be hit by the hammer) and not come to rest.

Or am I missing something?
#20
Old 07-11-2006, 01:34 AM
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Quote:
Originally Posted by FibonacciPrower
Does it actually come to rest? If the hammer bounces back, it would continue moving indefinitely (in the absence of air resistance, opposite walls, or bystanders which might be hit by the hammer) and not come to rest.
<CHL>What we have here is failure to communicate.</CHL>

The mallet comes in and hits the nail. As the material of the mallet deforms, it slows down. Eventually, it stops. Then the head of the mallet begins to restore to its shape, and eventually rebounds. It's the intermediate stopping point, where the mallet has reached its maximum deformation, that I'm referring to.
#21
Old 07-11-2006, 03:31 AM
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Quote:
Originally Posted by FibonacciPrower
Who said the nail was a theoretically unmovable object?
Nobody; it was a question - hence the word 'if' and the question mark.
#22
Old 07-11-2006, 03:36 AM
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Quote:
Originally Posted by FibonacciPrower
More momentum than it has with respect to you, yes, and much more than it has with respect to itself. But not more than it has with respect to, say, a mosquito which is flying away from the nail when it sees the hammer.
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)
#23
Old 07-11-2006, 04:00 AM
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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.
#24
Old 07-11-2006, 02:55 PM
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Quote:
Originally Posted by j_sum1
#1. The principle of conservation of momentum.
This only applies in an elastic collision.
If I recall correctly, the principle of conservation of momentum always applies. Apparently Wikipedia agrees.

Quote:
Originally Posted by j_sum1
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.
So the transferred momentum is close to 2p, right?

Quote:
Originally Posted by j_sum1
I don't believe that it is helpful to consider momentum as a vector in this case.
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.

Quote:
Originally Posted by j_sum1
The metal hammer in this case has an inelastic collision. The momentum is therefore not conserved.
It must be conserved for the system (hammer+nail).

Quote:
Originally Posted by j_sum1
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.
Not the nail, but the sum of the two.

Quote:
Originally Posted by j_sum1
#2 The principle of conservation of energy.
This applies in all collisions -- elastic and inelastic.
Instead, it was conservation of mechanical energy which didn't apply to inelastic collisions.

Quote:
Originally Posted by j_sum1
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.
No energy would be dissipated if the collision is assumed to be perfectly elastic.

Quote:
Originally Posted by j_sum1
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.
And how is that possible, if less momentum is transferred to the nail?
#25
Old 07-11-2006, 02:57 PM
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Quote:
Originally Posted by Mangetout
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)
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.
#26
Old 07-11-2006, 03:02 PM
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Quote:
Originally Posted by MikeS
The mallet comes in and hits the nail. As the material of the mallet deforms, it slows down. Eventually, it stops. Then the head of the mallet begins to restore to its shape, and eventually rebounds. It's the intermediate stopping point, where the mallet has reached its maximum deformation, that I'm referring to.
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?
#27
Old 07-11-2006, 03:03 PM
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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.
#28
Old 07-11-2006, 03:10 PM
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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.
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#29
Old 07-11-2006, 03:29 PM
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Quote:
Originally Posted by FibonacciPrower
If I recall correctly, the principle of conservation of momentum always applies.
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.
Quote:
Impulse is the force applied in a unit of time. force x time which equates to change in momentum.
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.
#30
Old 07-11-2006, 03:55 PM
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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.
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#31
Old 07-11-2006, 04:29 PM
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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.
#32
Old 07-11-2006, 05:38 PM
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Quote:
Originally Posted by Cheesesteak
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.
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.
#33
Old 07-11-2006, 05:39 PM
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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.
#34
Old 07-11-2006, 05:47 PM
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Quote:
Originally Posted by Chronos
Momentum, however, can't hide in this way. Where you have momentum, you have something moving.
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.

Quote:
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.
Not in elastic collisions, which we are discussing here
#35
Old 07-12-2006, 10:45 PM
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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.
#36
Old 07-12-2006, 11:27 PM
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Quote:
Originally Posted by FibonacciPrower
Before hitting the nails, both hammers have momentum p. After hitting the nail, the metal hammer has momentum 0 (let's suppose it doesn't recoil at all), so the transferred momentum is p-0=p.
Likewise, the rubber mallet has momentum -p after hitting the nail. Therefore, the transferred momentum is p-(-p)=2p.

Which one is obviously greater?
There seems to be a problem here.

Let's say the mass of the hammer is m1, and the mass of the nail is m2. The initial velocity of the hammer is v1, the final velocity of the nail is v2.

Momentum is mv, so conservation of momentum gives us:

m1v1 = pfinal + m2v2

where pfinal is the final momentum of the hammer. But in your example, pfinal is negative, meaning m2v2 > m1v1.

So far so good. But since hammers are bigger than nails, we have m1 > m2

In order to get m2v2 > m1v1 with m1 > m2, you have to have v2 > v1. (Note that we have v1 and v2 both greater than zero.)

From m2v2 > m1v1, and v2 > v1 > 0, we get:
m2v22 > m1v12

Dividing both sides by two gives:
(1/2)m2v22 > (1/2)m1v12

But (1/2)m1v12 is the initial kinetic energy of the hammer, and (1/2)m2v22 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.
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