# A How much energy dissipated in a nail gun?

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1. Apr 2, 2016

### Paul Duncan

Hi all,

I'm having trouble working out how much energy will have to be dissipated in a test rig for a nail gun.

From previous testing, I know that a test slug fired from the gun has a mean energy of 100J.

I now want to design a test rig which can fire the gun repeatedly for 100,000 shots.

I have decided that firing a test slug vertically upwards in to the bottom of a heavy piston in a steel tube will be the most suitable design.

Each time the slug is fired from the gun, it collides with a rubber-bottomed piston in a steel tube, thus transferring its energy to the piston by pushing it up the tube. The piston can then bounce up and down in the tube until its energy is entirely dissipated as heat and sound. The process is then repeated 100,000 times.

The problem I have with designing the rig, is working out how high the piston will travel within the tube. I know I'm making a flawed assumption somewhere so please let me know where my flawed reasoning is!

Here is my working and reasoning:

Assumptions:
- neglect energy dissipation due to air drag
- since the rubber bottom of the piston has a very high spring constant, assume that once the piston and slug collide, the slug and the piston both travel with the same velocity

m_s = mass of the test slug = 0.09 kg
m_p = mass of the piston = 10 kg
E_i = energy of the test slug exiting the gun = 100 J
E_f = combined kinetic energy of the test slug and piston immediately after the collision
v_s = velocity of the test slug before the collision
v_f = velocity of both the slug and the piston immediately after the collision

Working out the velocity of the test slug:

v_s = sqrt(2*E_i / m_s) (from the kinetic energy equation)
= sqrt(2*100 / 0.09)
= 47.1 m/s

Using the conservation of momentum principle, determine the velocity of the slug and piston immediately after the collision:

m_s * v_s = (m_s + m_p) * v_f
v_f = m_s/(m_s + m_p) * v_s
= 0.09/(0.09 + 10) * 47.1
= 0.42 m/s

Combined kinetic energy of the slug and piston immediately after the collision:

E_f = 0.5 * (m_s + m_p) * v_f^2
= 0.5 * 10.09 * 0.42^2
= 0.9 J

Thus the maximum height that the piston will reach:

h = E_f / ((m_s + m_p) * g)
= 0.9 / (10.09 * 9.81)
= 9 mm

I then double checked the maximum height of the piston using the conservation of energy equation and the initial energy of the test slug. I assumed that the kinetic energy of the test slug would be entirely converted to gravitational potential energy of the piston once the piston reached its highest point. Thus energy losses due to sound and drag etc were neglected.

h = E_i / ((m_s + m_p) * g)
= 100 / (10.09 * 9.81)
= 1 metre

Thus the maximum height differs significantly depending on the calculation method I use.

Is my conservation of momentum approach correct?
Is my conservation of energy approach correct?

My intuition tells me that the conservation of momentum approach is correct, however, I'm suspicious since I do not believe that the kinetic energy of the system would be reduced from 100J to 0.9J from the collision (i.e. huge energy losses to sound and heat).

Thanks,

Paul

2. Apr 3, 2016

### Andrew Mason

This is the coorect approach.

No. This is not an elastic collision. You cannot use conservation of kinetic energy. Most of the energy of the slug is dissipated as heat flow into the target. You can see this from your first calculation.

AM

3. Apr 3, 2016

### Drakkith

Staff Emeritus
That is indeed the case in a completely inelastic collision like yours here. Note that to "catch" the test slug, you need a material that will bend and crumple as it slows the slug down without storing energy as a spring would, otherwise it will just push the slug back the other way and it won't be caught. This requires that most of the kinetic energy be converted into heat or deformation of either the piston or the slug.

4. Apr 4, 2016

### Paul Duncan

Thanks for the responses Andrew and Drakkith! I can't believe I forgot about the difference between an elastic and inelastic collision.

Due to your input, I have decided to use a 90mm long pin which screws in to a flexible rubber isolator which then screws in to a 2.5kg 'piston'. Thus when the gun fires, its striking pin hits the 90mm long pin (which fits inside the gun's barrel where the nail would sit) which causes the 90mm pin/isolator/'piston' to travel 90mm vertically up inside a seamless steel pipe. The pin/isolator/'piston' assembly then falls back in to the gun's barrel (since it is guided by a lead-in) and bounces until its energy has been dissipated as heat.

I have allowed for the 'piston's weight to be adjustable to account for the assumptions that I've used in the conservation of momentum equation. I will also check the temperature of the assembly to ensure that it isn't getting too hot.