Heavy and light speeding bullets

I really am not a projectile physics expert at all, although I think I know a little bit about the laws of motion. Anyhow, I've been pondering a question and I thought I had it figured, but then I had an argument, so now I'm not so sure . . .

If two identically sized bullets each leave the barrel of a gun and reach the same velocity (and travelling in a straight trajectory) but one bullet has more mass (we'll say, double the mass) which one will travel further and which will hit the ground first?

I had some thoughts about this, but:

I asked a physics major in my town and he disagreed with my thinking about the problem and "threw" some equations at me that I really didn't listen to, because I have no physics background. He figured the heavy one would obvioulsly hit the ground first, and travel less distance - but is this as obvious as he says?

My thinking was that each bullet has just 2 forces acting on it: air resistance and gravity. I figure that gravity would "lower" each bullet at the same rate, which would result in each bullet hitting the ground at the same instant. I also figured that the bullet with greater mass would therefore have greater inertia and be able to overcome air resistance more easily, enabling it to travel further. I thought that due to this, the bullet with the greater mass (all other things being equal) should travel further, although they each hit the ground at the same time.

Is this right, or is my basic reasoning flawed? (which would not really be a shocker, since I don't really have a physics background)
Thanks for your help!

 

Log in or Sign up to hide all adverts.

The bullet with the greater mass should travel further when there is air resistance and all other things are equal. It should travel the same distance when there is no air resistance.

 

Log in or Sign up to hide all adverts.

Log in or Sign up to hide all adverts.

Hi quirk27

Welcome to SciForums. I personally think you and your physics friend are both part right and part wrong. I think the more massive bullet will hit the ground first and travel further.

Since gravity affects them both the same the only important different is the effect of air resistance. The more massive bullet experiences less air resistance, so it falls slightly faster and it flies slightly faster. Since it falls slightly faster obviously it will hit the ground first.

What is not clear is if the fact that it flies faster means that it travels more distance in this shorter time or if the fact that it hits the ground first means that it travels less distance despite the higher velocity. I think that the air resistance effects are probably much more important for the extremely fast horizontal motion than for the relatively slow vertical motion, so I think the "flys faster" effect wins out over the "falls faster" effect: longer distance even though it is a shorter time.

-Dale

 

If the bullets are in the same shape and size, why would the air resistance be different?

Since the acceleration toward Earth is the same they should fall within same time. But, since the Earth accelerates faster towards the heavier bullet, this one will contact Eart sooner(fall down sooner).

 

But if one is heavier and has the same diameter then it will have a greater sectional density, which means that it will lose velocity less rapidly than the lighter bullet. Both will hit the ground at the same time (See James R's answer), but the heavier one will maintain a higher speed throughout the flight and therefore travel further in that time.

 

You can look at both the gravity and air resistance effects in terms of their force (weight and drag) or their acceleration (gravitation and horizontal deceleration). In each case the forces are proportional to the accelerations by F=ma where m is different for each bullet. The weight is different for the two but the gravitation is the same. The drag is the same for the two so the horizontal deceleration is different.

Forget the earth's acceleration, it is negligible compared to air resistance and even in a vacuum it could be removed simply by firing the bullets simultaneously.

-Dale

 

Two so called heavyweights have given weasel answers which are not definitive answers, so I will not give a definite answer either .

Maybe the heavy one drops first and maybe not.

 

Did you qualify, by giving a weasel answer?

Or, did you plainly state which would drop to ground first? According to the thread question.

 

It would seem to me that the heavier bullet would travel further. Since they are the same size & shape, the atmospheric drag force would be the same on each. The same force applied to a lighter object would cause more deceleration in the horizontal direction. Hence the lighter bullet would not travel as far.

I am guessing that the atmospheric effects in the vertical direction would not be much different, so they would hit the Earth at close to the same time.

Note that if you drop an object vertially and propell one horizontally, they hit the ground at about the same time.

 

Try it and see... not with a gun and bullets, but with reasonable facsimiles. It will help a lot if the difference in mass is very large. Perhaps a ping-pong ball and a golf ball?

Actually, you don't even need to do the experiment. If you make the light bullet very light, the result is intuitively obvious.

Imagine this experiment:
Take two ping-pong balls. Replace the air inside one of them with helium until the ball is not much heavier than air.
Now throw each horizontally, hard.

The result is clear:
The heavier bullet travels further and hits the ground first.


Assumptions:
The curvature of the Earth isn't significant over the bullet's travel distance.
The bullets are fired horizontally from the same height over flat ground.
Neither bullet is lighter than air.
The air is still.
Aerodynamic effects aren't significant.

I don't know if the last assumption is reasonable for real bullets or not. It wouldn't be reasonable for golf balls struck by a driver, because a well struck golf ball get significant aerodynamic lift, meaning a heavier ball might not only travel further, but also stay up for longer.

 

I'm not convinced that the bullets would hit the ground at different times. I still think they should land at the same time with the heavier bullet travelling further.

I think that buoyancy in air may be responsible for the results explained in Pete's post, but I'm not sure exactly how buoyancy is determined.

When does buoyancy begin to have an effect? It is when the majority of an objects volume is of a density less (or perhaps equal to?) than the surrounding fluid (ie: air)?

If so, the lighter ball remains airborne longer due to buoyancy in air (the one ball has a density less than air does throughout the majority of its volume).

Are objects that are more dense than the surrounding fluid (air), throughout affected by buoyancy at all? (for example, the two bullets).

Would a balloon just barely more dense than air (say, filled with pure oxygen, which would be more dense than the dominant nitrogen/oxygen mix) throughout fall at the same rate as an object way more dense (such as balloon shaped large steel ball)? If not, when do objects begin to fall at the same speed in a vertical freefall?

I apologize for all the new questions at this point, but they seem relevant and interesting . . .

 

Ping Pong balls, balloons, et cetera seem to me to be a separate issue from speeding bullets. Atmospheric effects become very significant as an object approaches the density of air, while atmospheric effects on object like metal bullets are minimal.

In a vacuum, a small object propelled horizontally will hit the ground in the same amount of time as a small object that is dropped. This ignores the curvature of the Earth, which would be significant if the horizontal velocity was very high. This also ignores the effect due to the objects attracting the Earth. The Earth gets displaced due to this effect, but not enough to be measurable in the case of the two bullets.

In a vacuum, the above indicates that horizontal distance is a function of initial velocity, while the time of impact with the Earth is a function of initial height of the object.

For bullets, it seems to me that atmospheric effects would have a noticeable effect horizontally, but the effects on vertical movement would be almost negligible for bullets fired horizontally at about shoulder height. As mentioned in previous posts (by me and others), the drag force is a function of velocity for dense objects of the same size, shape, and frictional coefficient (close to true for the two bullets). Note that the horizontal velocity of the two bullets is high, while the vertical velocity starts at zero and does not increase much, assuming that the bullets are fired at shoulder height.

Based on the above, I would expect the two bullets to hit the ground at almost the same time, with the heavier bullet traveling farther. The drag force would have less effect on the velocity of the heavier object. For bullets fired at shoulder height, the vertical drag forces start at zero and increase very slowly since the vertical velocity never gets large. The horizontal velocity starts high and is probably a measurable effect.

Hence the two bullets hit the ground at almost the same time, with the heavier bullet having traveled farther.

 

Hi quirk27,

Interesting question you've posed! Let me try to provide some quantitative information for you. I propose to model the system as follows: each bullet experiences a gravitational force [mg] down and a resistive force [cv] that opposes the motion. If the bullets are identical except for their mass then c is the same constant for each bullet. If I call x the horizontal direction and z the vertical direction, then I can write the equations of motion (just Newton's 2nd Law):

m ax = - c vx
m az = - c vz - mg

Here ax is the acceleration in the x direction, vx is the velocity in the x direction, etc. I won't bother you with the details, but one can easily solve these equations. A special role is played by the time t0 = m/c (check the units to see that this is indeed a time). This is roughly the time it takes for the bullet to reach terminal velocity.

x(t) = x(0) + vx(0) t0 (1 - exp(-t/t0))
z(t) = z(0) + (vz(0) + g t0) t0 (1 - exp(-t/t0)) - g t t0

x(0) is the initial horizontal position, vz(0) is the initial vertical velocity, etc. Now there are several important things to notice at this point. First, the mass seems to have disappeared, but it is hiding in the time t0 that we defined. You can easily see that t0 gets larger with mass in accord with your physical intuition. Second, the time of flight is determined only by the vertical motion (z motion). Let me call the time of flight T, and following the details of your question, let me suppose that the bullet starts with no vertical velocity at a height h above the ground. The equation which determines T is then:

z(T) = 0 = h + g t0^2 (1 - exp(-T/t0)) - g t0 T, or equivalently
t0^2 (1 - exp(-T/t0)) - t0 T = - h/g

Since t0 is such an important time, it makes sense to count time in units of t0. Define the ratio r = T/t0; r is just the time of flight in units of t0. Our equation for r is now very simple:

t0^2 (1 - exp(-r) - r) = - h/g

This equation is unfortunately transcendental, and we cannot solve it exactly. Fortunately, if we are somewhat clever we can still get some useful information out of it. After all, what we really want to know is how the time of flight varies with mass, and this is the same as asking how it varies with respect to t0. Now we make use of some calculus, and differentiate both sides of the equation for r with respect to t0. h/g is independent of t0, so the right hand side gives zero. The left hand side contains two terms:

2 t0 (1 - exp(-r) - r) + t0^2 (exp(-r) - 1) dr/dt0 = 0

The first term is the explicit derivative with respect to t0, and the second term is the implicit derivative since r depends on t0. We can now solve for dr/dt0:

dr/dt0 = (2/t0)[(1 - exp(-r) - r)/(1 - exp(-r))] = (2/t0)[1 - r/(1-exp(-r))]

and from dr/dt0 we can calculate dT/dt0 = t0 dr/dt0 + r:

dT/dt0 = (2/(1-exp(-r))) (2 (1-exp(-r)) - r (1+exp(-r)))

Now the punchline. By graphing this function, or by any other method you can dream up, you can verify that dT/dt0 is always less than zero. But remember that t0 is proportional to m, so this tells us that the time of flight decreases as you increase the mass. This agrees with your physical intuition.

A similar calculation will reveal that the range (total horizontal distance covered) is a monotone increasing function of mass. The more massive object will travel further. This is interesting because it isn't necessarily obvious. As mass increases, the time of flight decreases, but the deceleration also decreases. This means the bullet flies faster but for less time and there is a competition of effects. It turns out that (unless I've made a mistake!) one effect always wins, and the more massive bullet does go further. This again agrees with your physical intuition in the large m limit.

The differential equation I derived for the dependence of time of flight on mass has a lot of physics in it. It can tell you that the time of flight only depends weakly on the mass for r much less than 1 (short time of flight). This is probably the case with your bullets. On the other hand, it says that the time of flight decreases as 1/m for r much greater than 1 (long time of flight). This can be understood by observing that for a long time of flight, the object would be traveling at constant velocity (terminal velocity) for most of the trip. You can check that the terminal velocity is proportional to m, so that the time of flight is roughly h/(terminal velocity) ~ 1/m.

Others have already given clear discussions of the relevant physics, so I will stop here. Hope you find this helpful and welcome to Sciforums.

 

Let's work the numbers!

Buoyancy is the uplifting force of the surrounding medium. It is equal to the weight of the displaced medium. The net force is Gravity - Buoyancy, and force = mass by acceleration, so the acceleration of an object falling through a perfect fliud is:
Mass / (Gravity - Buoyancy) = m / (mg - Pvg)
Where m = mass, g = acceleration due to gravity, P = density of fluid, v = volume.


For our bullets, the buoyant force is the weight of a volume of air the same size as a bullet.

Using round numbers:
A 0.005kg lead bullet has a volume of 0.0000005m³.
0.0000005m³ of air has a mass of 0.0000005kg, or one ten-thousandth the mass of the bullet.
This means that the buoyant force will be one ten-thousandth the force of gravity on the bullet, the net downward force will be 0.9999 the net downward force in vaccuum, and the downward acceleration will be reduced by the same proportion.

For a bullet of the same volume and half the mass, the buoyant force will be one five-thousandth the force of gravity on the bullet, and the net downward force will be 0.9995 the net downward force in vaccuum, and the downward acceleration will be reduced by the same proportion.

So due to buoyancy, there is about a 0.04% difference in the downward acceleration of the bullets. This corresponds to a time difference of about 0.02%. From shoulder height on Earth, bullets fired horizontally over flat ground will take about half a second to fall, which means the time difference is perhaps 0.1 milliseconds... Not much!

Air resistance will slow the bullets a trifle, increasing the time difference by a fraction, but I'd be surprised if the effect was any greater than the buoyancy effect.

 

Aerodynamics is most definitely NOT a 1+1=2 tea party. Size is crucially important. As always.

Please Register or Log in to view the hidden image!

Bullet sized projectiles, with the bullet range of velocities, may ramble back and forth between different regions of laminar and non laminar air flow drag characteristics, and also may be in various regions of subsonic, transonic, and supersonic aerodynamic theory.

If the thread question object is not uniquely specified as to mass, size, surface aerodynamic characteristics, imparted spin or lack thereof, etc., etc., etc., then the answerer must postulate all pertinent qualities and THEN try to produce an answer.

Any conscientious aerodynamicist will THEN (again) say that the answer MUST be tested in a wind tunnel. Because aerodynamics is an exact mathematical science. Partly. And the other half is black magic voodoo.

So, sitting in our kocy (cosy?) (cozy?) soft chairs, we cannot possibly accurately predict the solution to this thread question without empirical help from an actual science experiment.

 

Thanks for the explanation of buoyancy Pete - it helps!
You've convinced me that the heavier bullet hits the ground slightly (barely) before the lighter bullet in an envrionment with any type of an atmosphere, due to the effect of buoyancy on the bullet.

Do we all agree that in a vaccum they hit the ground at the same instant?

 

An envrionment of vaccum does what?

My minitor went ifflone.

P.S. Anyone convinced by Pete without a wind tunnel test is "easily entertained".

 

I agree.

In theory, yes. In practice, close enough. Probably close enough even with an atmosphere.