A bullet is a kinetic projectile, a component of firearm ammunition that is shot from a gun barrel. The term is from Middle French, originating as the diminutive of the word boulle (boullet), which means "small ball". Bullets are made of a variety of materials, such as copper, lead, steel, polymer, rubber and even wax. Bullets are made in various shapes and constructions (depending on the intended applications), including specialized functions such as hunting, target shooting, training and combat. bullets are often tapered, making them more aerodynamic. Bullet sizes are expressed by their weights and diameters (referred to as "calibers") in both imperial and metric measurement systems. For example: 55 grain .223 caliber bullets are of the same weight and caliber as 3.56 gram 5.56mm caliber bullets. Bullets do not normally contain explosives (see Incendiary ammunition and Exploding bullet), but strike or damage the intended target by transferring kinetic energy upon impact and penetration (see terminal ballistics).
Bullets are available singly (as in muzzle-loading and cap and ball firearms), but are more often packaged with propellant as cartridges ("rounds" of ammunition). Bullets are components of paper cartridges, or (much more commonly) in the form of metallic cartridges. Although the word bullet is often used in colloquial language to refer to a cartridge round, a bullet is not a cartridge but rather a component of one. A cartridge is a combination package of the bullet (i.e., the projectile), the case (which holds everything together), the propellant (which provides the majority of the energy to launch the projectile) and the primer (which ignites the propellant). This use of the term bullet (when intending to describe a cartridge) often leads to confusion when a cartridge, and all its components, are specifically referred to. The cartridges, in turn, may be held in a magazine or a belt (for rapid-fire weapons).
The bullets used in many cartridges are fired at muzzle velocities faster than the speed of sound—about 343 metres per second (1,130 ft/s) in dry air at 20 °C (68 °F)—and thus can travel a substantial distance to a target before a nearby observer hears the sound of the shot. The sound of gunfire (i.e. the "muzzle report") is often accompanied with a loud bullwhip-like crack as the supersonic bullet pierces through the air creating a sonic boom. Bullet speeds at various stages of flight depend on intrinsic factors such as sectional density, aerodynamic profile and ballistic coefficient, and extrinsic factors such as barometric pressure, humidity, air temperature and wind speed. Subsonic cartridges fire bullets slower than the speed of sound, so there are no sonic booms. This means that a subsonic cartridge, such as .45 ACP, can be substantially quieter than a supersonic cartridge, such as the .223 Remington, even without the use of a suppressor.Bullets shot by firearms can be used for target practice or to injure or kill animals, or people. Death can be by blood loss or damage to vital organs, or even asphyxiation if blood enters the lungs. Bullets are not the only projectiles shot from firearm-like equipment: BBs are shot from BB guns, airsoft pellets are shot by airsoft guns, paintballs are shot by paintball markers, and small rocks can be hurtled from slingshots. There are also flare guns, potato guns (and spud guns), rubber bullets, tasers, bean bag rounds, grenade launchers, flash bangs, tear gas, RPGs, and missile launchers.
I cannot seem to find them, and I am confused as to what they are. I am trying to simulate a bullet in flight and need to calculate air resistance. I'd like to be accurate to real life and use a lookup table but I can't seem to find a graph of the drag coefficients that has an accompanying...
First off, I do know how to solve this problem. We use the principle of conservation of angular momentum about the centre of mass of the system which comprises of the loop and the bullet to obtain option B. My doubt is, why do we just not use the principle about the centre of the loop? Where is...
$$ \frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2) $$
v = 56.64
I just don't get how this is the wrong answer....it's just simple conservation of energy, right?
For this problem,
The solution is,
However, dose anybody please know why the graphs for ##V_1## and ##V_2## are discontinuous where they cross the time axis?
Many thanks!
Gun with a tachyon bullet paradox
if you had a device akin to some kind of ‘gun’, which created tachyons and the fired them, perhaps in a beam or some kind of ‘packet’, when they go faster than the speed of light would go back in time. Lets say the ‘bullet’ or beam, went out into the universe...
My 15 y/o daughter is in driver's ed. Her teacher passed on some information as fact that seems more like an urban legend. He told the class that an inexperienced driver got into a bad car wreck, and the sudden loss of speed caused her belly button ring to penetrate her entire midsection like...
Is the bullet cluster evidence for or against dark matter?
I understand the explanation that it is evidence in favor of the existence of dark matter, and it convinces me. However, some argue that it is evidence against its existence? Why?
How to find the collision number if the moving bullet hits a few wooden blocks and every collision takes 10 percent of its speed. In which block will the bullet stay?
My apologies if the prefix is too high of complexity. I don't know where this would fall, difficulty or academically speaking.
While it may be surprising to some given Hollywood's portrayal of it in movies, if a person in wearing hard bulletproof armor is struck by a projectile, the person is...
I can understand that using conservation of momentum, we can find v. But we need V for that. The equation for V involves h and so we need h. But I am not able to comprehend the equation involving l,h and a. The question doesn't specify what a is.
Please be kind to help
A bullet with mass m, velocity v perfectly elastically, vertically collide with one end of a rod on a slippery plane and the bullet stops moving after the collision. Find the mass of the stick M
the bullet stops moving after an elastic collision, so all energy is transformed to the rod. There...
The Bullet Cluster counts as 'smoking gun' for the dark matter. But what lacks in these calculations is the super massive black holes at the center of almost every galaxy. As the most massive and most compact objects in the collision, it's expected that they are less slowed down by the collision...
(a) ##u_{min}=\big(1+\frac{m_2}{m_1}\big)\sqrt{2\mu_k g d}##
(b) ##x_f=\sqrt{\frac{2h}{g}\Big(\big(\frac{m_1}{m_1+m_2}u\big)^2-2\mu_k g d\Big)}##
Can someone check please?
Hi,
I was reading this article, https://science.howstuffworks.com/question456.htm, about firing a bullet from the back of train which is moving as fast as the bullet.
The following is taken from the article mentioned above.
So, ideally, the bullet will fall straight to the ground as it exits...
Summary:: What is the temperature change of a bullet upon impact.
I have this problem to solve but I'm kinda stuck, would apricate any feedback.
We fire a silver bullet with a muzzle speed of 200 ms−1 into a sack of sand. What is the temperature change of the bullet, if 40 % of its kinetic...
A 7.80 g bullet is initially moving at 500 m/s just before it penetrates a block of solid rubber to a depth of 4.50 cm.
(a)What is the magnitude of the average frictional force (in N) that is exerted on the bullet while it is moving through the block of solid rubber? Use work and energy...
Hi, I was given this exercice where I have to calculate the penetration distance, knowing that m = 20g, v = 300km/h and C = 750 N and I have to give the results with two significant numbers. I really thought that all I had to do was replace the values so I did :
d = (20.10^-3) x (300/3,6)^2 /...
1) Applying conservation of linear momentum:
$$m.u = M.V + m.v$$
where ##V## is final linear speed of the rod
$$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion:
$$V=\omega . r$$
$$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$
Is this correct?And can this be...
I think that the kinetic energy is equal to heat right. Mv^2 / 2 = ms delta T
Delta T is 0,64
But I don’t really know if the temperature decreases or increases at the end
A gun is fired vertically into a block of wood(unknown mass) at rest directly above it. If the bullet has a mass of 24.0g and a speed of 310 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
I first calculated the velocity v:
√2.8^2+6.3^2= 6.8942
then i used it as the final velocity, so final velocity=6.8942
and the initial velocity=0
acceleration=9.8
Then i substituted them into this equation:
final velocity=initial velocity + accelerationxtime
then time=0.703489843
hence i...
Change in KE = Change in thermal energy
0.5 * (6)* vblock^2 = 0.4 * 6 * 9.81* 0.1
vblock = 0.885
By Conservation of Momentum,
(0.05)(854) = (0.05)*vbu + (6)(0.885)I am not sure whether Change in KE = Change in thermal energy is true coz there should be a change in internal energy of the block...
As ##P=ccte## we can find final velocity considering a plastic collision
##m_0 . v_0 =(2m+m_0).V##
But what about the angular velocity? Because, as the bullet hits the centre of mass of the string, it won't have angular velocity
I first plugged my given values into m1v1+m2v2=(m1+m2)vf.
(0.002)(600)+(5)(0)=((0.0020)+(5))vf
vf=0.24 m/s
Next, I plugged my given values into F=ma.
((0.002)+(5))(9.8)
F=49.02 N
Next, I plugged my given values into Fdeltat=mdeltav.
deltat=mdeltav/F
((0.002)+(5))(0.24)/(49.02)...
My initial thought was to use the conservation of energy law since there're no external forces acting on the system bullet + rod. The rod is in rest, the bullet is moving. Then after the collision, the bullet and the rod are rotating around the pivot together, so the kinetic energy of the bullet...
If I had two bullets one moving 0.000001 c and the other moving at .9 c for example, and they both went thru a tin foil square target, assuming they are traveling in vacuum, and that the tin foil square target will make a perfect aperture around the radius of the passing bullet, assuming the...
used a) change in momentum / time, time from from kinematics equation d = vi+vf/2 *t ... 0.1 = 300*t
=3.33*10^-4
then 600/0.00333 = 1.8*10 ^5
how to proceed?
Here are my workings, and I was wondering if I'm correct so far.
Let ##m## be mass of bullet and ##M## be mass of plate.
COM:
##mu_{bullet} = (m+M)v##
##\frac{1}{2}(m+M)v^2=(m+M)gh +\frac{1}{2}I\omega ^2##
where I is the inertia, so using parallel axis theorem,
##I = \frac{1}{12}bh^3 + md^2 =...
I have calculated KE_i and KE_f, took the difference between the initial and final kinetic energy of the bullet to be equal to the work spent to overcome the friction, and divided it by the distance traveled, but arrive at around 20000N. The solution should be 9.5*10^7.
Not sure what else to...
Hey everyone
I'm struggling on the last part of this assignment. I need to find the total work done by the block and the bullet, when the collision happens. The informations is:
mblock=0.3 kg
mbullet=0.01 kg
vg=700 m/s
Height=0.72m
The final speed after the collision is vf=22.6 m/s and the...
So, Basic premise is I have 1 input and one output, Energy friom the gunpower burning is input, and the output is the force which pushes the bullet after it is reduced by the friction.
m = 12.96 g
s = 0.6985 m (barrel lenght)
E = 6734 J
k = 0.5 (friction coeffcient)
I am sorry for lack of...
This is task from my textbook and it does not provide us with an answer. So I cannot verify if I did mistake. Can someone double check, please? My solution:
##E_{k_{0}} = \frac{(m+M)v_{0}^2}{2} \quad \land \quad U = \frac{kx^2}{2}##
##E_{k_{0}} = U##
##\Longrightarrow (m+M)v_{0}^2 = kx^2##...
Problem:
A and B decide to duel but, being poor, they have just one gun (a six-shot revolver) and only one bullet. Being dumb, as well, this does not deter them and they agree to "duel" as follows: They will insert the lone bullet into the gun's cylinder, A will then spin the cylinder and shoot...
Okay so the first thing I did was was find the height of the apex. I found this by finding the x and y components of Vo:
Vox = VoCos(θ) = 22cos(59) = 11.33 m/s
Voy = VoSin(θ) = 22sin(59) = 18.86 m/s
Using the initial velocity of the y component I found the time it took to reach the apex:
t =...
Solving using Linear Momentum:
M vb2/2 = M g 2L
vb = 2√(g L)
m v = m v/2 + M (2√(g L) )
v = 4 M √(g L) / m
Note: I see from the answers - that this is correct.
--------------
Next, I tried to solve it via Energy conservation point of view.
M vb2/2 = M g 2L
vb = 2√(g L)
m v2/2 = m v2/8 + k...
I tried getting to the solution by the principle of conservation of energy. What goes in, must go out. If the final velocity, ie. the final kinetic energy is lower than what we started with, that energy difference must've been used to overcome the friction that lasted over some distance L...
Homework Statement
A 10 g bullet traveling at 370 m/s strikes a 8.0 kg , 0.80-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.
What is the angular velocity of the door just after impact?
Express your answer to two significant...
This took a lot of time and effort and I understand if you wish to skip past everything and just read my questions about it in the The too long didn't read summary (TL;DR) at the bottom.
Homework Statement
The 10-g bullet having a velocity of v = 750 m/s is fired into the edge of the 6-kg...
Homework Statement
A 28-g bullet is shot vertically into an 6-kg block. The block lifts upward 5mm (see the figure). The bullet penetrates the block and comes to rest in it in a time interval of 0.0010s. Assume the force on the bullet is constant during penetration and that air resistance is...
Homework Statement
A man in a car moving with a speed ##20~ms^{-1}## shoots a bullet forward with a speed of ##150~ms^{-1}##. The bullet hits a wall and stops. Find the rise in the temperature of the bullet if its specific heat is ##500~Jkg^{-1}K^{-1}##.
Homework Equations
##K.E=1/2~mv^2## and...
Homework Statement
A rifle bullet of mass 10g is fired and takes 2.0 ms to travel down the rifle barrel of length 0.50m with constant acceleration
Calculate the muzzle velocity (bullet velocity on leaving the barrel)
Calculate the acceleration in the barrel
After traveling a short distance...
Does anyone know how this works? I hypothesize that the reason that the bullet shatters is that boron carbide is much harder than the copper of the bullet. Does anyone think that this may be made into a conventional armor design as some people are saying or is boron carbide too expensive for use...