Physics experiment

[rubberchiken]

i have to do a physics investigation, but it's free choice. so, being the sorta guy i am, i chose to do it weapons-oriented. i will be testing different projectiles (eg. fruit, wooden 'bullets', golf-balls, ect.) to find which has the best penetration.

now thats the easy part; i can find ammuntion or manufacture it without any problems - my school has a full machine shop which i can use. My issue here is a lack of a target; i need something cheap, abundant, easily obtained, and obviously which i could fire objects into. styrofoam is out of the question - i don't particularly enjoy picking up many little peices of the stuff.

By the way, the gun im using for the investigation is of the pnuematic kind; an over-and-under modified solenoid-valve gun, made of schedule 40 PVC. i can use the school's HUGE compressor, set at a certain output, to reliably get an accurate pressure-output each time.

thanks for any help you guys can give, im sure there will be some good ideas.

[T_Pyro]

I'd measure the penetration of the projectile into some viscous liquid. The drag experienced by a body would thus be (approximately) directly proportional to the instantaneous velocity of the projectile (if the shape & size of the projectile is constant). Hence, the momentum of the incident projectile could be calculated.

Deformation of clay would be another option, but there's no easy way of "measuring" the deformations produced.

If you're doing a physics project, most probably your teacher will expect you to explain the results of your tests scientifically, too. Hence, instead of measuring just penetration (which would not be a linear dependent of momentum / kinetic energy), I would suggest that you measure the momentum and kinetic energy imparted by the projectiles. Momentum could be measured indirectly by measuring the angular displacement of a pendulum bob of known mass. Kinetic energy could be measured by noting the maximum compression of a spring kept in the path of the projectile.

[Anthony]

Surely the kinetic energy can be measure by the ballistic pendulum? As long as you know the vertical rise of the mass, the weight of the mass, and that there was (practically) total energy transfer (total capture of the projectile by the mass)?

[T_Pyro]

If energy is conserved during the collision with the pendulum, then the kinetic energy could also be calculated from it, but since such a collision invariably involves dissipation of energy, it would not be too accurate. Like you said, for an accurate reading, we'll need (practically) total energy transfer.

With the spring, however, provided the mass of the spring is negligible compared to that of the projectile, energy dissipated as energy of deformation can be minimised by using a rigid body as a projectile.

[Anthony]

I suppose that for a demonstration of good physics principles, you need to be totally accurate. My reasoning was that a sturdy projectile fired into a block of polystyrene (attached to the mass :) ) should give almost total energy transfer. Deformation of the projectile shouldn't occur, there'd be a small amount of noise, the deformation of the polystyrene and any little bits that fly off shouldn't add up to much.

I.e. the uncertainty probably wouldn't be much compared to other factors, such as shot-to-shot muzzle energy variation.

How practical would it be to have a spring of negligable mass compared to the projectile? Even a projectile of 20gm fired at spudgun type velocities carries a fair emount of energy. Presumably, the spring cannot become coil-bound during the test, so it'd have to be fairly strong - probably more than the projectile weight!

[Tuatara]

As an alternative to styrofoam you might try stacking and gluing corrugated cardboard into a block, then fire your projectile into the edges of the cardboard.

[T_Pyro]

The choice of the spring to be used would indeed be crucial. The two options would be to use a short spring with a high spring constant (such a spring would be quite heavy!), or to use a spring with a comparatively lower spring constant, but of a longer natural length. A spring with a lower spring constant could be made of a finer wire, hence could be lighter.

[nbk2000]

If I was your physics teacher, I'd give you a big juicy F, meaning FAILURE, for FAILING to capitalize the letter I when referring to yourself.

What hope have you for passing physics when you're failing 3rd grade grammer?

I've had it with morons who can't even follow the simplist grammer conventions! :mad:

Death to the chicken and all of his ilk! :D

New rule:

Failure to capitalize the letter I, when used to refer to yourself, is a banning offense. One warning (not likely) and then you're gone. Consistant failure to capitalize the first letter of a new sentence, run-on sentences, dozen sentence paragraphs, and the like, are also banning offenses after (maybe) one warning.

Newbies have to demonstrate proper grammatical skills before they can be forgiving the occasional mistake.

Though this is already covered under the "raping the english language" rule. ;)

[Marvin]

On ballistic pendulums.

Well, if grades are being given out then Anthony and T_Pyro get a D in basic mechanics.

The pendulum bob stops the bullet. This means that conservation of energy cannot be operating however well the catcher is made. Since the bob is more massive than the bullet, conservation of momentum (which always applies) and conservation of kinetic energy (which is optional) mandate the bullet would bounce off the bob in the other direction.

To get the right answer you need 2 steps. You know the mass of the bob, and the mass of the bullet. Conservation of momentum for the collision gets you the momentum, and thus the velocity and thus the kinetic energy of the combined mass. Conservation of energy then gets you the height the bob will swing to. Depending on how you measure the bob height you might need some trig.

So long as nothing falls off the bob, and it does catch the bullet in line with its center of gravity, it doesnt much matter how its made. Noise, heat, deformation of the material wont affect the result. Only time loss of energy matters is the air resistance as the bob rises, and this will be much less than the air resistance affecting the bullet in the first place.

Alternativly, if your school has one of those air float tracks that usually use a backwards vacuum cleaner, you can use that instead, measuring the momentum from the velocity of the train (they usually have electronics to do this).

[T_Pyro]

Ouch. That hurt. :( I said "If energy is conserved during the collision with the pendulum", not "Energy is conserved...". Meaning, that I accept that energy cannot be conserved in this method, as in any real experiment. Note that that's the same reason that I advocate the massless spring approach. If the problem is with exactly what is "massless" spring, refer to the motion (and hence the KE imparted) of the spring in such a situation.

Oh well, I suppose if we get a "D", at least you'd be there to share the glory with us. :D

Since the bob is more massive than the bullet, conservation of momentum (which always applies) and conservation of kinetic energy (which is optional) mandate the bullet would bounce off the bob in the other direction.

Law of conservation of momentum always applies, as everyone has stated. However, your presumption about the motion of the bullet after the collision is grossly wrong. The actual motion depends on the coefficient of restitution during the collision. As a simple proof, imagine a 40g Iron bob that is bombarded by a 1g lead bullet. Does the bullet bounce back? No. If the coefficient of restitution=0, the bullet sticks to the bob, and the velocity of the entire mass is found out be conserving linear momentum for time dt after the impact. In general, there are 2 unknowns: velocity of projectile after collision, and velocity of target after collision. There are also (surprise, surprise!) 2 equations, 1 from the conservation of linear momentum, and the other from the relation between the velocities, and the coefficient of restitution. Solve the 2 equations simultaneously, and you're done.