Ballbearing101 said:
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:
- First, the hammer deforms until the force reaches 100 N;
- Then, some energy is required to deform the ground (##= Fd## where ##d## is the displacement of the nail)
- 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).