Impulse on Rubber and Metal Hammers

[Ballbearing101]

Homework Statement

Between a metal or wooden hammer of equal mass, which is more effective at hammering a nail into the ground.

Relevant Equations

Impulse = Ft
a=f/m

The rubber hammer is more elastic so the time of contact will increase. Ft=m(v-u)
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?

 

[berkeman]

Homework Statement:: Between a metal or wooden hammer of equal mass, which is more effective at hammering a nail into the ground.
Relevant Equations:: Impulse = Ft
a=f/m

The rubber hammer is more elastic so the time of contact will increase. Ft=m(v-u)
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?

Rubber or wooden?

 

[haruspex]

Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?

Which are you as user supplying, force or momentum?

 

[Ballbearing101]

Rubber or wooden?

Rubber

 

[berkeman]

Rubber

Thanks for the clarification.

So what is your intuition? And given that intuition, what do you think the quantitative difference is that makes it so much more ineffective to hammer nails with a rubber hammer?

 

[Ballbearing101]

My intuition went to bed at 11.

 

[haruspex]

My intuition went to bed at 11.

If it is awake now, please try to answer my question in post #3.

 

[Adesh]

The important points to be kept in mind:

1. Newton’s third Law will be followed. Nail will push on the hammer with same force.

2. The deformation caused in the rubber hammer would try to come at its origins shape.

 

[Adesh]

Try to think which ball bounces back higher: rubber/tennis ball or metal ball (when dropped from terrace).

 

[haruspex]

The important points to be kept in mind:
1. Newton’s third Law will be followed. Nail will push on the hammer with same force.
2. The deformation caused in the rubber hammer would try to come at its origins shape.

Try to think which ball bounces back higher: rubber/tennis ball or metal ball (when dropped from terrace).

No, as @Ballbearing101 surmised in post #1 it is time of contact that is key. But that has to be coupled with an appreciation of what is the same for both hammers.

 

[Delta2]

It turns out that the physics of hammering might not be so simple after all.

I am thinking that since the rubber hammer will bounce back more it will transfer more momentum to the nail, so seems to be more effective in that way, but of course the bouncing back will make harder the life and cause arm pain to the guy that hammers with it. Where am i wrong @haruspex cause we know i am wrong otherwise people would use rubber hammer from the middle ages...

 

[Adesh]

I am thinking that since the rubber hammer will bounce back more it will transfer more momentum to the nail, so seems to be more effective in that way,

I too think that, because the deformation that will be caused in the rubber hammer by the reaction force of the nail would try to come at its original shape and hence would apply force on the nail again.

 

[Delta2]

I too think that, because the deformation that will be caused in the rubber hammer by the reaction force of the nail would try to come at its original shape and hence would apply force on the nail again.

Not sure about that, i was thinking in terms of the momentum-impulse theorem, since the momentum of the rubber hammer is say MV before it hits the nail, after the hit it will bounce back with velocity -v, so the momentum impulse theorem says that $$-Mv-MV=J$$ where J is the impulse of the force from the nail to the hammer. By Newton's 3rd law the impulse of the force from the hammer to the nail will be $-J=M(v+V)$. The metal hammer will not bounce back rather it will continue with some smaller velocity v' towards the same direction so the impulse $-J'=M(V-v')$ (we assume before the hit metal hammer has the same momentum MV as the rubber hammer) which obviously is smaller than $-J=M(v+V)$.

 

[haruspex]

Not sure about that, i was thinking in terms of the momentum-impulse theorem, since the momentum of the rubber hammer is say MV before it hits the nail, after the hit it will bounce back with velocity -v, so the momentum impulse theorem says that $$-Mv-MV=J$$ where J is the impulse of the force from the nail to the hammer. By Newton's 3rd law the impulse of the force from the hammer to the nail will be $-J=M(v+V)$. The metal hammer will not bounce back rather it will continue with some smaller velocity v' towards the same direction so the impulse $-J'=M(V-v')$ (we assume before the hit metal hammer has the same momentum MV as the rubber hammer) which obviously is smaller than $-J=M(v+V)$.

As I wrote in post #10, this is not what is important. Please do not distract @Ballbearing101.

 

[Delta2]

As I wrote in post #10, this is not what is important. Please do not distract @Ballbearing101.

You are not saying that it is wrong, just that it isn't important. I wonder what is important then but ok i ll make a private conversation with you as to not reveal too much or distract the OP.

 

[Ballbearing101]

This is what evie and irene Curie must of felt like

 

[Ballbearing101]

@haruspex, your first post
What would be the difference? I assume your applying the force as your wielding the hammer and the nail is providing the change in momentum as its stopping the hammer.

 

[jbriggs444]

Post #10 by @haruspex is a good hint. What thing (or things) are the same for both hammers? What thing (or things) are different about the impacts that ensue?

@haruspex has mentioned time of impact. How does that differ for the two hammers? What is the effect of that difference?

 

[Ballbearing101]

The time of impact will increase for the rubber hammer, so for the same change in momentum the force decreases. So if the force needed to change the hammers momentum is less, +considering Newtons 3rd law, therefore, the force on the nail is less.
But if the change in momentum remains the same, (m(v-u)) doesn't that mean the nail will be displaced the same distance regardless of the hammer's property.

 

[haruspex]

if the change in momentum remains the same, (m(v-u)) doesn't that mean the nail will be displaced the same distance regardless of the hammer's property.

You are not considering what opposes the advance of the nail. We are driving it into a fixed block. To make progress we have to overcome the resistive force.

 

[Ballbearing101]

How does that affect it if we're considering a closed system.

 

[haruspex]

How does that affect it if we're considering a closed system.

The nail is being hammered into the ground, and the question relates to the relative movement of the two, so it is not a closed system.

 

[hutchphd]

I think this is a very dubious pedagogical example. For one thing soil is often not a Newtonian fluid (nor for that matter are the fascia in wood with a nail).
Maybe one should be trying to crack a walnut instead? Just a comment..

 

[haruspex]

soil is often not a Newtonian fluid

Why does that make it unsuitable for such a question? I would model it in terms of static and kinetic friction, increasing somewhat as the nail is driven in.

 

[jack action]

I don't understand why the discussion revolves around impulse, force and momentum. From my point of view, this is all about energy. Momentum is conserved, but kinetic energy isn't. There are two ways to look at it:

1. Inelastic collision and coefficient of restitution
2. Model the hammer and the ground as springs and determine the amount of elastic energy necessary to deform them (either elastic or plastic deformation). Whatever energy is used to deform one, will not be used to deform the other.

 

[haruspex]

From my point of view, this is all about energy.

Not so; see post #24. If the peak force does not overcome the resistance of the ground the nail makes no progress. The ground will deform and spring back.

 

[jbriggs444]

Not so; see post #24. If the peak force does not overcome the resistance of the ground the nail makes no progress. The ground will deform and spring back.

Speaking as someone who has swung a hammer into a nail in a soft target (unbacked board) and a hard one (board backed with plank end-on), the difference is dramatic. The effect to which @haruspex refers applies. Nobody who has ever tried it would drive a nail with a rubber hammer. You want a sharp, well-backed impact.

In my mind's eye, I see an ideal impact of hammer on spike being driven into the work as an inelastic collision of hammer head with spike followed by a period of deceleration as the spike plus hammer head move slightly into the work.

 

[jrmichler]

The simple model of a nail being driven into anything is zero movement until a certain force on the nail, then continuous movement as long as that (constant) force is maintained.

Try breaking down the problem like this:

Step 1: What is the force vs time (qualitative plot - hand sketch) of rubber and steel hammers hitting a rigid, immovable object. If the two hammers have equal mass and velocity, how does the area under the two curves compare? How does the time in contact compare? And the peak force? Show us a pair of sketches on the same axes.

Step 2: Apply the knowledge from Step 1 to a nail that needs a certain force to start and keep moving. Look at two possibilities:
1) The hammer bounces off the nail because the peak force is not enough to move the nail.
2) The hammer contacts the nail with enough force to move, then stays in contact while moving the nail. The force is F = ma, where m is the mass of the hammer, and a is the hammer acceleration while in contact. The force F is the force to move the nail.

Speaking as a person who has used heavy, light, hard, and soft hammers. Also sledges, splitting mauls, baseball bats, and axes.

 

[hutchphd]

hy does that make it unsuitable for such a question? I would model it in terms of static and kinetic friction, increasing somewhat as the nail is driven in.

I guess I am thinking mostly about splitting firewood with a maul or wedges. It is not at all clear to me that simple static and moving friction model makes any sense for that process. In particular the resistance force often seems to rapidly increase with the speed of the maul or wedge so there seems a "natural speed limit" because this nonlinearity. Perhaps soil is a simpler system...I've not spent much time pounding in the dirt.
Seems breaking rocks (or walnuts) would provide a more straightforward idea.

 

[haruspex]

I see an ideal impact of hammer on spike being driven into the work as an inelastic collision of hammer head with spike

I don't think inelasticity is in itself an advantage. What matters is the stiffness. Treating the hammer as a massive spring, a high k during compression means a large force. If completely inelastic then there is zero k under decompression, whereas elasticity might give a sufficiently large k to achieve a bit more movement of the nail.

splitting firewood with a maul or wedges. It is not at all clear to me that simple static and moving friction model makes any sense for that process. In particular the resistance force often seems to rapidly increase with the speed of the maul or wedge so there seems a "natural speed limit"

I would have thought the resistance would increase as the split progresses (up to a point) and the same would be true for driving a nail into the ground or into a block of wood. Thus, the threshold force to be exceeded increases with each blow. But to address the question in the thread it suffices to consider a single blow.

breaking rocks (or walnuts) would provide a more straightforward idea.

It might be a simpler problem to analyse, but it is not the one in the thread.

 

[jbriggs444]

I don't think inelasticity is in itself an advantage.

Agreed, not an advantage. But it is somewhat an inevitability for a heavy and hard hammer driving a light nail.

Even if the collision of hammer with nail is highly elastic (and brief), the result is to propel the lighter nail into the work. The hammer will follow, of course. The nail quickly stops in the work and another collision ensues. The process continues until the hammer is brought to a stop. So elastic or inelastic, the hammer winds up at a stop either way. The bulk of the hammer's initial kinetic energy winds up dissipated in the work either way.

A series of elastic collisions may save some of the heat energy that would otherwise result from the inelastic collision of hammer with nail, dissipating it in the work rather than in the nail/hammer interface. But for a heavy hammer and light nail, this is guaranteed to be a minor loss.

Agreed also that stiffness matters -- so that the force of static friction holding nail in the work is exceeded.

 

[Ballbearing101]

This is conceptual. Its requiring an explanation no calculation. The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.
I sort of get 1. In order for it to be less effective when time of contact increases. F must decrease, which means that delta p for both hammers must be considered to be equal. So it supplies less force to nail.
I don't get 2. Is the change in momentum increase just due to the time increase. I wouldn't use a rubber hammer so why is it more effective.

 

[haruspex]

This is conceptual. Its requiring an explanation no calculation. The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.
I sort of get 1. In order for it to be less effective when time of contact increases. F must decrease, which means that delta p for both hammers must be considered to be equal. So it supplies less force to nail.
I don't get 2. Is the change in momentum increase just due to the time increase. I wouldn't use a rubber hammer so why is it more effective.

The extra momentum change is because of the bounce. The hammer bounces back, so has a greater change in momentum.
But as I posted, this will generally be much less significant than the drawn out contact time.

 

[jack action]

The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.

It is all about energy. I really don't like the choices you are given, even though 1. is the correct answer. I also prefer explaining the phenomena with springs, as it is easier to visualize. But once you see that, you can explain it with inelastic collision and the coefficient of restitution (post #25). I'll show you the concept with springs and let you relate it to the inelastic collision on you own.

As @haruspex told us before, the force of impact is determined by the force needed to deform the ground. Say it takes 100 N to deform the ground. If the impact force is less than 100 N, then the nail doesn't move. If it reaches 100 N, then the nail begins to move. Once it moves, the force required to break the ground further is still 100 N (let's assume it's constant for simplicity). How we reach that force is irrelevant, if it is 100 N, it will move.

Once it moves, it means that we have kinetic energy involve. This our goal: we want the nail to move. The less energy is transformed into kinetic energy for the nail, the less efficient the hammer will be. The amount of energy coming from the hammer is $\frac{1}{2}mv^2$ where $m$ is the same whether it is made of rubber or metal and $v$ is the velocity at the moment of impact, which we can assume to be the same in both cases (same input from the person holding the hammer).

The difference between the two materials is their stiffnesses. Because the hammers can be modeled as springs, they will deform under a force, any force. So before the force of impact reaches the required 100 N (if it does), the hammer will compress. At 10 N it will compress a little bit, at 50 N it will compress a little bit more and it will stop compressing at 100 N because then the nail will begin to move away from the hammer. And the force will never go over 100 N.

But compressing a spring requires energy. The amount of energy required is $\frac{1}{2}Kx^2$, where $K$ is the spring stiffness and $x$ the spring displacement. But we also know that the force required to compress a spring for a displacement $x$ is $Kx$. So the energy required for a known force $F$ (100 N in our case) is $\frac{1}{2}Fx$. What this tells us is that for a given force acting on a spring, the greater the displacement, the greater the amount of energy is stored in the spring. The rubber hammer will have a greater displacement because it is less stiff than the metal hammer.

Now, the energy coming from the hammer ($\frac{1}{2}mv^2$) is split into 3 ways:

1. First, the hammer deforms until the force reaches 100 N;
2. Then, some energy is required to deform the ground ($= Fd$ where $d$ is the displacement of the nail)
3. Finally, whatever is left is transformed into kinetic energy to move the nail into the ground.

And when all the energy has been disposed of, the nail stops, and the elastic energy stored into the hammer comes back out. But at that point, the impact force will go from 100 N to 0 N as the spring (i.e. hammer) goes back to its original shape. So the nail will not get driven into the ground, and the stored energy will be converted to pure kinetic energy for the hammer. The amount of energy stored in the spring will determine how high the hammer will bounce back.

Of course, the result is that the momentum that the hammer goes back with, was not transferred to the nail (conservation of momentum). The greater momentum difference can be explain by the time taken for the impact (because it takes time to compress a spring). But momentum alone cannot explain why there is such a transfer and why the nail stops at one point. Only conservation of energy can.

Now try to explain this with the principle of inelastic collision (which is all about momentum and energy and doesn't need the concept of spring).

 

[haruspex]

It is all about energy.

Thought experiment: push nail a short way into ground; drop a pingpong ball onto the nail, and repeat that 1,000,000 times. How much energy have you expended in pingpong KE? How much further has the nail gone in?