Dot Physics

I was recently re-watching a MythBusters episode and I found something I had wanted to explore previously (but accidentally deleted the episode). Here is a short clip from the “shooting fish in a barrel” episode:

Did you see what I found interesting? That big barrel of water left the floor from being hit by a bullet.
The question here is: Does a bullet have enough energy to increase the gravitational potential energy of the barrel to that height?

First, let me gather some data. From the episode, the Mythbusters were firing a 9mm pistol round into the barrel. I don’t know much about different guns, but [according to this site](, a 9 mm pistol round has a mass of about 8 grams and a velocity of 358 m/s.

Now, what about the barrel? How big is it, what is the mass and how much water does it hold?
[]( – this site lists some barrel dimensions. I am not sure exactly which one the Mythbusters used, but of the ones listed, 3 of them have a height of 0.88 meters, a mass of 47 kg and a capacity of 228 Liters – so I will use that.

One last thing – how high did the barrel go up? I will use my favorite [Tracker Video Analysis]( and use the height of the barrel to scale the video.

Now for the action plan. I will compare the kinetic energy of the bullet to the energy needed to lift that barrel.

![Page 0 Blog Entry 37 1](

This gives a height of about 7 cm.

Now for the calculations. First the kinetic energy of the bullet:

![Page 0 Blog Entry 37 2](

Ok – done. Now how much energy would it take to lift that barrel? First, what is the mass of the barrel plus water? 228 liters is 0.228 m3. The density of water is 1000 kg per m3 so the total mass is

![Page 0 Blog Entry 37 3](

The energy needed to move the barrel plus water up can be found from the Work-Energy theorem:

![Page 0 Blog Entry 37 4](

In this case, the barrel starts and stops at rest so the change in KE is zero. For the change in gravitational potential energy, I will use U = mgy so that

![Page 0 Blog Entry 37 5](

Which is LESS than the energy of the bullet. Less is good, less can be explained. If the energy were more than the bullet – that would be a problem. Not all the energy of the bullet goes in to increase the potential energy of the barrel and water – the bullet may have gone all the way through the barrel (but maybe not).

It is interesting to see how this happens. When the bullet strikes the water in the barrel, the whole barrel flexes outward (like a compressed spring). When the barrel returns to its original shape it causes it to jump up.

NOTE: This is an energy effect, not momentum. The momentum of the bullet is:

![Page 0 Blog Entry 37 6](

(note that momentum is a vector – in this case its pointing down).
The momentum of the barrel and water can be found from the KE. If the barrel has 188 Joules of potential energy at the end it must have had 188 Joules of KE right at the beginning of its “flight”. The relationship between KE and momentum is:

![Page 0 Blog Entry 37 7](

So the magnitude of the momentum would be:

![Page 0 Blog Entry 37 8](

in the up direction. BIG difference. Why? Isn’t momentum supposed to be conserved? Momentum is conserved if there are no external forces on the “system”. In this case, the calculations are looking at the bullet, barrel and water as the system. There are external forces on this system (mostly the floor) so that momentum will not be conserved. Clearly momentum is not conserved because the bullet was moving down and the barrel moved UP.


  1. #1 asdfas asdfasf
    October 5, 2008

    The amount of energy in a bullet can be measured accurately with a ballistic pendulum:

  2. #2 Martin
    February 21, 2009

    What about the rotational kinetic energy in the bullet? Would that now also add onto the vectorized kinetic energy?

  3. #3 Rhett
    February 21, 2009


    Good question. I am not really sure how fast the bullets rotate. However, if I assume it is a cylinder with a mass of 8 grams and a diameter of 9 mm then the rotational energy would be (1/2)mR^2(omega)^2 where omega is the rotational rate of the bullet. Suppose I completely guess an omega of 1000 rad/s, then the rotation energy would be: (1/2)(0.007 kg)(0.0045 m)^2(1000 rad/sec)^2 = 0.07 Joules. This is not a lot compared to the translational KE. Of course, I could be wrong about the rotational speed.

  4. #4 Robert Lucien
    July 14, 2009

    If the bullet is doing 358 m/s and the barrel is 10 cm long with one turn, the rotational rate will be 3580 RPS. Multiply be 2 pi = 22,494 rad/s. Using your calculation that is 35 joules – 7% of the total energy of the bullet.
    The centripetal acceleration a = v^2 * r , v = 3580 * 2 * pi * .0045 = 101 m/s
    so a = 101^2 * .0045 = 45.9 m/s^2.