Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Vicky Garg on 06/02/2011 23:30:03
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Vicky Garg asked the Naked Scientists:
A bullet moving with velocity of 10m/s is brought to rest after penetrating the wooden plank of 4cm thickness. How do I calculate the acceleration of the bullet?
What do you think?
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You might try Googling "Equations of Motion".
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this equation should help you . vf=vi+at
vf = final velocity
vi = initial velocity
a = acceleration
t = time
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is this a high school physics question?
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is this a high school physics question?
What! [:0] [:0] [:0] [:0]
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Vicky Garg asked the Naked Scientists:
A bullet moving with velocity of 10m/s is brought to rest after penetrating the wooden plank of 4cm thickness. How do I calculate the acceleration of the bullet?
What do you think?
You can't, unless you already know, or you make hypothesis, about how the force (and so the acceleration) depends on the space traveled inside the plank (or how it depends on time).
If you make, for example, the simplest hypothesis that the force is constant, so the acceleration is constant = a, then s = 1/2 v2/a where s = space traveled by the bullet, v = bullet initial speed. Then a = v2/2s = 100/0.08 = 1250 m*s-2 = 127.55g.
With different informations about the force, the problem is more complicated, assuming that, however, you are interested in the *average* acceleration.
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Vicky Garg asked the Naked Scientists:
A bullet moving with velocity of 10m/s is brought to rest after penetrating the wooden plank of 4cm thickness. How do I calculate the acceleration of the bullet?
What do you think?
You can't, unless you already know, or you make hypothesis, about how the force (and so the acceleration) depends on the space traveled inside the plank (or how it depends on time).
If you make, for example, the simplest hypothesis that the force is constant, so the acceleration is constant = a, then s = 1/2 v2/a where s = space traveled by the bullet, v = bullet initial speed. Then a = v2/2s = 100/0.08 = 1250 m*s-2 = 127.55g.
With different informations about the force, the problem is more complicated, assuming that, however, you are interested in the *average* acceleration.
Based on the data, and applying v² = u²+2as, we find that,
a = -u²/2s
or -1250 m/s²
That's the acceleration, regardless of the the force. We only need to know the initial velocity (u) the final velocity (v) and the distance travelled (s) to determine the acceleration (a). The possibility that the acceleration was not uniform only means that we can't quantify the maximum or minimum acceleration.
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Based on the data, and applying v² = u²+2as,
And how do you get this without making the assumption that the motion is uniformly accelerated?
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you make no assumption that acceleration is uniform - just that over a distance the acceleration will be... you choose your equation of motion and you take your choice
you got v, you got u, and you got s, and you want a
v^2=u^2+2as
seems a pretty good choice
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you make no assumption that acceleration is uniform - just that over a distance the acceleration will be... you choose your equation of motion and you take your choice
you got v, you got u, and you got s, and you want a
v^2=u^2+2as
seems a pretty good choice
No.
Let's make an example.
Someone says he has good reasons to assume that the force can be considered proportional to the distance traveled x, because at high speeds (as in this case) wood behaves as a spring (it's just for sake of discussion, I personally have no reasons to make such an assumption, because I don't know what really happens). So F = -kx.
Then: mx'' + kx = 0
--> x(t) = A*sin(ωt) where ω = sqrt(k/m)
since x(0) = 0; x'(0) = v, you find A = v/ω. Knowing that at t = t1 the bullet stops and x(t1) = 0.04m, you find: t1 = π/2ω and: ω = v/s = 10/0.04 = 250 rad/s.
Now let's compute the average acceleration:
<x''(t)> = (v - 0)/(t1 - 0) = v/(π/2ω) = 2ω*v/π = 5000/π ≈ 1591.55 m*s-2 which is different from 1250 m*s-2.
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I think we did calculate the average acceleration. Average acceleration is Δv/Δt. I believe that's what we get from
v² = u²+2as
Are you perhaps assuming constant acceleration?
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Responding to Lightarrow - cos geezer responded in the meantime
So basically you are making an assumption that is not in the question.
Perhaps for extra credit you might introduce your assumptions but please note
1. not high speed - i have met people who can run faster
2. any reason, other than we know the formula, to model on a spring
3. the equations governing forces on a spring are limited to being within the linear xpansion of the spring - outside those limits the standard hooke's law does not apply thus the use of a shm to model is doubly suspect
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Just a small question, when that bullet stops moving due to its initial force, 'slowing', gravitation taking over as the main 'force'. Is it then 'decelerating' or 'accelerating'? :)
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One aspect of the bullet and its motion specified in the question is that it is brought to rest. After that it has an acceleration (and velocity) of zero (measured WRT the plank).
I bet you wouldn't get full marks for that answer.
Would anyone like to solve the problem under the assumption that the plank is suspended from a string and is set swinging by the bullet?
Is the plank assumed to be at rest in the first place?
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I think we did calculate the average acceleration. Average acceleration is Δv/Δt. I believe that's what we get from
v² = u²+2as
Are you perhaps assuming constant acceleration?
And how do you compute Δt without solving the equation of motion, which you don't have at all? [;)]
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Responding to Lightarrow - cos geezer responded in the meantime
So basically you are making an assumption that is not in the question.
Perhaps for extra credit you might introduce your assumptions but please note
1. not high speed - i have met people who can run faster
2. any reason, other than we know the formula, to model on a spring
3. the equations governing forces on a spring are limited to being within the linear xpansion of the spring - outside those limits the standard hooke's law does not apply thus the use of a shm to model is doubly suspect
No, it's you Geezer that makes the arbitrary assumption that the motion is uniformly accelerated. If you don't make any assumption about the equation of motion/how the force depends on time or distance, you *cannot* solve the problem.
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One aspect of the bullet and its motion specified in the question is that it is brought to rest. After that it has an acceleration (and velocity) of zero (measured WRT the plank).
I bet you wouldn't get full marks for that answer.
Would anyone like to solve the problem under the assumption that the plank is suspended from a string and is set swinging by the bullet?
Is the plank assumed to be at rest in the first place?
If you allow me to make 4 further assumptions, that is:
1. the "ballistic hypothesis"(*),
2. neglect the air's resistance,
3. the plank is homogeneous,
4. the bullet impacts in the plank's centre of mass,
then I can solve it (faster if the plank is initially at rest [:)]). Without some of those assumptions, the problem can become quite or extremely, more difficult.
(*) means that we neglect the plank's displacement during the bullet's penetration into it; such hypothesis is valid in the limit m/M << 1
m = bullet's mass, M = plank's mass.
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I calculated the acceleration by a rather indirect but conceptually simple manner, I visualised the bullet acquiring its velocity of 10m/s by falling in the Earths gravity and calculated how far it would have to travel.
D=V^2/2g =5.0916497 m
it is then stopped in a distance of .04m hence the average acceleration is D*g/.04=1250 m/s^2
= 127.55g
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I think we did calculate the average acceleration. Average acceleration is Δv/Δt. I believe that's what we get from
v² = u²+2as
Are you perhaps assuming constant acceleration?
And how do you compute Δt without solving the equation of motion, which you don't have at all? [;)]
Oh! That's easy.
Δt = 2s/(u+v)
= 2s/u
= 8 ms
[;D]
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One aspect of the bullet and its motion specified in the question is that it is brought to rest. After that it has an acceleration (and velocity) of zero (measured WRT the plank).
I bet you wouldn't get full marks for that answer.
Would anyone like to solve the problem under the assumption that the plank is suspended from a string and is set swinging by the bullet?
Is the plank assumed to be at rest in the first place?
If you allow me to make 4 further assumptions, that is:
1. the "ballistic hypothesis"(*),
2. neglect the air's resistance,
3. the plank is homogeneous,
4. the bullet impacts in the plank's centre of mass,
then I can solve it (faster if the plank is initially at rest [:)]). Without some of those assumptions, the problem can become quite or extremely, more difficult.
(*) means that we neglect the plank's displacement during the bullet's penetration into it; such hypothesis is valid in the limit m/M << 1
m = bullet's mass, M = plank's mass.
Since I stated that the bullet sets the plank swinging you cannot assume that the plank's mass is >> that of the bullet.
Or, to put it another way, I was talking about the extremely difficult version of the problem
Seriously, I was just pointing out that there are lots of unstated assumptions.
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Responding to Lightarrow - cos geezer responded in the meantime
So basically you are making an assumption that is not in the question.
Perhaps for extra credit you might introduce your assumptions but please note
1. not high speed - i have met people who can run faster
2. any reason, other than we know the formula, to model on a spring
3. the equations governing forces on a spring are limited to being within the linear xpansion of the spring - outside those limits the standard hooke's law does not apply thus the use of a shm to model is doubly suspect
No, it's you Geezer that makes the arbitrary assumption that the motion is uniformly accelerated. If you don't make any assumption about the equation of motion/how the force depends on time or distance, you *cannot* solve the problem.
That is true. It does depend on whether or not the wood exerts a constant force on the bullet. It's probably not a completely unreasonable assumption that it does, but it would take a rather fancy setup to find out. I suppose you could mount the wood on a pressure transducer. Anybody know of a better way?
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It might be simple to assume that the force is constant, but it's unlikely. As the bullet slows down the force on it is likely to decrease. After all, the force is zero when the bullet's speed is zero.
You could use a pressure transducer, but a force transducer would seem better.
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It might be simple to assume that the force is constant, but it's unlikely.
It is quite likely if it's largely a function of friction between the bullet and the wood.
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Since I stated that the bullet sets the plank swinging you cannot assume that the plank's mass is >> that of the bullet.
You are half right [:)] : the rate m/M can be << 1 and at the same time high enough to set the plank in motion. This is what we do with the ballistic pendulum:
http://en.wikipedia.org/wiki/Ballistic_pendulum
What counts is the bullet momentum. The initial speed V of the pendulum, as a function of the bullet's speed v, is then simply V = m*v/M (momentum conservation).
Since the bullet's speed is usually high, its mass can be much smaller than that of the pendulum and V is however not negligible.
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Concerning the force on the bullet, I wouldn't take anything for granted. Being wood, I'd expect, for example, that for a very short path the force to be high while the fibers in the cylinder in front of the bullet are being stretched and broken, then that the force decreases while the wooden cylinder is moved ahead and compressed but with little force, then that the cylinder is being compressed at high pressure and the force increases with the path. In case, instead, the bullet goes through all the plank escaping out, I expect the force increases for a short path, then decreases when the cylinder just moves inside the hole.
Anyway, I wouldn't bet on anything particularly simple.
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Anyway, I wouldn't bet on anything particularly simple.
Agree entirely. We cannot treat the wood - even if not in a free swinging pendulum - as static -wouldn't we also be seeing compression waves within the structure of the wood. Some substances tend to deform until certain limits are reached - which often coincides with the return of the compression wave.
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Concerning the force on the bullet, I wouldn't take anything for granted. Being wood, I'd expect, for example, that for a very short path the force to be high while the fibers in the cylinder in front of the bullet are being stretched and broken, then that the force decreases while the wooden cylinder is moved ahead and compressed but with little force, then that the cylinder is being compressed at high pressure and the force increases with the path. In case, instead, the bullet goes through all the plank escaping out, I expect the force increases for a short path, then decreases when the cylinder just moves inside the hole.
Anyway, I wouldn't bet on anything particularly simple.
There is no "wooden cylinder". Assuming we are shooting the bullet across the grain of the wood, the pointy end of the bullet parts the lengthwise fibres of the wood quite easily. It's very similar to what happens to a nail when you drive it into wood (hint - nails stay in place because of friction between the nail and the wood.)
When you drive a nail into wood, most of the resistance is created by friction between the nail and the wood rather than the force required to part the wood fibres. You can prove this quite easily.
Try driving a nail into wood as normal. Then drill a pilot hole in the same piece of wood and try driving an identical nail into the pilot hole. Unless you make the pilot hole so large that there is almost no friction between the wood and the nail, you will not notice any difference in the effort required to drive the nail with, or without, a pilot hole.
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Concerning the force on the bullet, I wouldn't take anything for granted. Being wood, I'd expect, for example, that for a very short path the force to be high while the fibers in the cylinder in front of the bullet are being stretched and broken, then that the force decreases while the wooden cylinder is moved ahead and compressed but with little force, then that the cylinder is being compressed at high pressure and the force increases with the path. In case, instead, the bullet goes through all the plank escaping out, I expect the force increases for a short path, then decreases when the cylinder just moves inside the hole.
Anyway, I wouldn't bet on anything particularly simple.
There is no "wooden cylinder". Assuming we are shooting the bullet across the grain of the wood, the pointy end of the bullet parts the lengthwise fibres of the wood quite easily. It's very similar to what happens to a nail when you drive it into wood (hint - nails stay in place because of friction between the nail and the wood.)
When you drive a nail into wood, most of the resistance is created by friction between the nail and the wood rather than the force required to part the wood fibres. You can prove this quite easily.
Try driving a nail into wood as normal. Then drill a pilot hole in the same piece of wood and try driving an identical nail into the pilot hole. Unless you make the pilot hole so large that there is almost no friction between the wood and the nail, you will not notice any difference in the effort required to drive the nail with, or without, a pilot hole.
But when you put a nail in the wood, it hasn't speeds of hundreds of metres per second. Shoot a nail in the wood and the wood's behaviour will be very different. Of course I'm not saying that friction disappears.
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It might be simple to assume that the force is constant, but it's unlikely.
It is quite likely if it's largely a function of friction between the bullet and the wood.
Do you think it takes the same force to push a bullet through wood quickly as it takes to push it through slowly?
Also, do you think the force will be the same with the pointed tip of the bullet just entering the wood as when the whole slug is moving through it?
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It might be simple to assume that the force is constant, but it's unlikely.
It is quite likely if it's largely a function of friction between the bullet and the wood.
Do you think it takes the same force to push a bullet through wood quickly as it takes to push it through slowly?
Also, do you think the force will be the same with the pointed tip of the bullet just entering the wood as when the whole slug is moving through it?
Pretty much.
It depends on how much force it takes to "cut" the wood versus the force produced by friction between the surface of the bullet and the wood. If the latter is large in relation to the former (which I believe it is), the accelerating force will be fairly constant. Friction force is independent of velocity and surface area. It's only proportional to the total force the moving objects exert on each other normal to the direction of motion.
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Concerning the force on the bullet, I wouldn't take anything for granted. Being wood, I'd expect, for example, that for a very short path the force to be high while the fibers in the cylinder in front of the bullet are being stretched and broken, then that the force decreases while the wooden cylinder is moved ahead and compressed but with little force, then that the cylinder is being compressed at high pressure and the force increases with the path. In case, instead, the bullet goes through all the plank escaping out, I expect the force increases for a short path, then decreases when the cylinder just moves inside the hole.
Anyway, I wouldn't bet on anything particularly simple.
There is no "wooden cylinder". Assuming we are shooting the bullet across the grain of the wood, the pointy end of the bullet parts the lengthwise fibres of the wood quite easily. It's very similar to what happens to a nail when you drive it into wood (hint - nails stay in place because of friction between the nail and the wood.)
When you drive a nail into wood, most of the resistance is created by friction between the nail and the wood rather than the force required to part the wood fibres. You can prove this quite easily.
Try driving a nail into wood as normal. Then drill a pilot hole in the same piece of wood and try driving an identical nail into the pilot hole. Unless you make the pilot hole so large that there is almost no friction between the wood and the nail, you will not notice any difference in the effort required to drive the nail with, or without, a pilot hole.
But when you put a nail in the wood, it hasn't speeds of hundreds of metres per second. Shoot a nail in the wood and the wood's behaviour will be very different. Of course I'm not saying that friction disappears.
That's the funny thing about friction force. It's independent of velocity. When you apply the brakes on your car, the deceleration is constant from any speed. It only depends on the force exerted between the brake surfaces (unless you generate so much heat that things start to melt.)
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When the Americans were experimenting with 1600 KPH rocket propelled vehicles in the early fifties they ran them on steel rails and found low friction at high speeds.
This was a similar effect to that experienced by an ice skater due to a liquid layer being established.
The purpose of these experiments was to subject volunteers to high g forces when the vehicles were stopped by water braking.
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The "normal" force is caused by the wood trying to spring back. When the bullet is just entering the plank then there's little wood trying to spring back so the force is small.
Once the bullet is fully in the plank then the force will be larger and more nearly constant.
The pressure under an ice skate is not usually high enough to melt the ice. It takes something like a hundred atmospheres to raise the MP by 1 degree C.
The only skates I have used had blades roughly a tenth of an inch thick and ten inches long- a square inch or so of area. At the time I was probably rather less than 147 pounds so that's less than 10 atmospheres of pressure. Some of the kids there were probably half that weight.
The freezing point depression could only have been something like 0.1C
However, people go skating when it's well below freezing and the ice's surface would be much colder than -0.1C
I know that professional's skates are thinner but the point is that the ones I wore were quite slippery enough for me to fall over.
The reason skating is possible is simply that ice is slippery. Plenty of people who have gone base over apex wearing ordinary shoes can vouch for that.
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Do you think it takes the same force to push a bullet through wood quickly as it takes to push it through slowly?
Also, do you think the force will be the same with the pointed tip of the bullet just entering the wood as when the whole slug is moving through it?
Pretty much.
It depends on how much force it takes to "cut" the wood versus the force produced by friction between the surface of the bullet and the wood. If the latter is large in relation to the former (which I believe it is), the accelerating force will be fairly constant. Friction force is independent of velocity and surface area. It's only proportional to the total force the moving objects exert on each other normal to the direction of motion.
Instead I think, as Bored says, that friction is quite little compared to wood compression by the bullet; furthermore, you are neglecting the friction of the displaced piece of wood against wood itself.
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That's the funny thing about friction force. It's independent of velocity. When you apply the brakes on your car, the deceleration is constant from any speed. It only depends on the force exerted between the brake surfaces (unless you generate so much heat that things start to melt.)
When you put a nail with a hammer in a wooden plank, I believe the main force is friction, because the nail's tip makes its way between the wood fibers, separating them. But at high speeds as the case of a bullet, and (but not only for this reason) because it has an higher rate front surface/lateral surface with respect to a nail, there is not only friction, but also the piece of wood in front of the bullet which is displaced and compressed. Also remember that if the bullet is not jacketed and in lead, its surface melts with high friction (for example this happens inside the barrel) and the bullet's friction with the wood is negligible, at the beginning.
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The "normal" force is caused by the wood trying to spring back. When the bullet is just entering the plank then there's little wood trying to spring back so the force is small.
Once the bullet is fully in the plank then the force will be larger and more nearly constant.
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It's true that the friction force does not immediately reach its maximum value, but if the laws of friction hold true, the friction force will be at a maximum by the time the point of the bullet has created a hole in the wood that is the same diameter as the bullet.
Of course, we are assuming the bullet is hard enough that it does not deform in any way. That may not be a valid assumption. We are also assuming the structure of the wood is consistent. That's probably not a bad assumption.
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That's the funny thing about friction force. It's independent of velocity. When you apply the brakes on your car, the deceleration is constant from any speed. It only depends on the force exerted between the brake surfaces (unless you generate so much heat that things start to melt.)
When you put a nail with a hammer in a wooden plank, I believe the main force is friction, because the nail's tip makes its way between the wood fibers, separating them. But at high speeds as the case of a bullet, and (but not only for this reason) because it has an higher rate front surface/lateral surface with respect to a nail, there is not only friction, but also the piece of wood in front of the bullet which is displaced and compressed. Also remember that if the bullet is not jacketed and in lead, its surface melts with high friction (for example this happens inside the barrel) and the bullet's friction with the wood is negligible, at the beginning.
I agree that the profile, and the material of the bullet could make a big difference, although we could cheat and use a bullet that looks a lot more like a nail. I guess what you are saying is that at the initial speeds involved, even the relatively small mass of the wood that has to be "relocated" consumes quite a lot of energy because it experiences a huge acceleration. That must have some effect.
Wonder how large it is in relation to the friction force? It could be a bit like what happens when a plane tries to exceed the speed of sound! If I'm at a loss for anyting to do today, I'll take a shot at a calculation [;D]
We could also do an experiment by measuring the distance a bullet travels when it has to drill its own hole versus the distance it travels when the hole is pre-drilled. Unfortunatey, I only have a BB gun, which probably would not be much good! I do have an explosive powered nail gun though. That might work, although I have a feeling it operates more like a hammer than a gun. I'll need to take a look at it.
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We could also do an experiment by measuring the distance a bullet travels when it has to drill its own hole versus the distance it travels when the hole is pre-drilled.
Yes, I thought something similar: the bullet in the pre-drilled hole and the bullet ortogonally against the base-surface of a long cylinder of wood. In which case force on the bullet is stronger, on average? And how does the force depends on distance traveled?