A 6.00 gram bullet moving at 320 m/s embeds itself in a wood block of mass 2994 g suspended as a ballistic?
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
851042
Weight of the bullet..IE. grain  how far it'll travel @ what velocity
245381
5.9736×1024 kg\n\ncheck the link for more info
821414
The bullet's vertical motion is governed by the vertical forces only and not governed by horizontal and vice versa.\nIn each case the bullet's vertical motion is governed by gravity acting on itself. The gravity force gives an acceleration of 9.81 met/sec squared. Both have the same distance to go through and both have a initial vertical velocity of 0. Therefore they both reach at the same time, neglecting air friction.
478202
If is really no wind, it should fall right back into your gun.\n\nBut an experimental proof I think can hardly be done because of friction effects between air and bullet: so you would need an evacuated tube of some 100 m height, higher than any building. And the accuracy of this tube also woud be hard to be realized.\n\nNevertheless, such an accuracy problem I think was encountered when was built the SLAC (STANFORD Linear Accelerator) in the 1950ies; but this installation was not upright.
1102604
This is an oddly specific question just to be asked out of curiosity.\n\nThe mass of the cannon ball is 13.6 kg, thus it has a weight of about 133.4 Newtons (assume g = 9.81 m/s^2).\nThe force of weight is pointed down...however, since the cannon ball in not in a vacuum and/or not a point mass, it has a non-zero buoyant force acting upward on it.\n\nThe buoyancy force acting on the ball is dependent on the weight of the fluid (water in this case) it displaces.\nIron has a density of about 7874 kg/m^3, thus a 13.6 kg sphere would have a volume of about 1.73 E-3 m^3.\nIf we assume the salt water of the ocean has a uniform density of about 1028 kg/m^3, then the weight of the displaced water exerts a force of about 17.42 Newton’s upward.\n\nThus, the net force acting on (stationary) cannon ball is about 116 Newtons downward.\nAccording to Newton’s 2nd law,\nF = ma\nThis force will cause the 13.6 kg cannon ball to accelerate at a rate of 8.53 m/s^2.\n\nAssuming there is no "water" resistance / drag, then using the formula,\nd = 1/2 at^2,\nwhere the ball stars from rest and experiences a uniform acceleration for a time t as the ball covers d distance,\nThe cannon ball should take about 50.57 seconds to reach the bottom.\n\nHOWEVER,\nJust as there would be air friction if this ball was launched into the air, there is most defiantly "water" friction as this ball accelerated downward....eventually reaching a terminal velocity.\nThe water will exert a drag force upward on the ball, causing its acceleration to be non-uniform.\n\nThe drag force can be calculated as,\nF_d = C_d * ((p * v^2) / 2) * A\nwhere F_d is the drag force, C_d is the drag coefficient, p is the density of the medium which the ball is falling in, v is the instantaneous velocity of the ball, and A is the cross sectional surface area of the ball in the direction of travel.\nIn order to calculate the drag force, one must know the drag coefficient of the cannon ball in water, something which usually must be calculated experimentally.\nThe ball will accelerate non-uniformly until it reaches it terminal velocity in the water, at which time, acceleration will cease and the ball will continue falling at its terminal velocity.\n\nUnless more information is know (the drag coefficient), the best (guess)/estimate of the how long the ball will take is sometime longer than 50.57 seconds...probably on the order of well over a minute since the ball will not be accelerating nearly as long as it did in our mathematical model calculate earlier.
387792
Literally a bullet with a hollow tip. They're able to pierce armor.
1155452
Only impulse (force x duration, or the quantity of momentum transferred to the pin) is calculable from the data, not force by itself, which is variable in time and not directly meaningful.\n\nIf the moving object has mass M and velocity V, and the object being struck has mass m, then assuming the collision is elastic and all motion is in the forward direction (all these assumptions are at least approximately true for the bowling ball and pin) then the impulse is:\n\nImpulse = 2*M*m*V/(M+m)\n\nEach pound value must be divided by g to get proper units of mass, so this is:\n\n=(g/g^2)*[2*16*3.5/19.5]*V = 5.74*(V/g)\n\nSince:\n\nV/g = 23mph/(32ft/sec^2) = 1.05 seconds \n\nthe result is:\n\n6 pound-seconds\n\nIt's the equivalent of 6 pounds acting for 1 second, or 1 pound acting for 6 seconds, or etc..
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
514815
OK, the barrel´s mass is 40Kg (notice: its mass is 40 Kg, not its weight). \n\na) So, your formula is Weight = mass(m)*gravity´s acceleration(g)= 40Kg*10m/s^2 = 400 KG*m/s^2 (=400 Newton)\n\nb) Work = force * distance\nSince work is calculated only for the vertical, and the force equals the weight, you have:\nWork = 400N*5m = 2000 Joule\n\nc) Eff = 250N*5m/2000J*100 = 0.625*100 = 62.5%\n(You got 12.5% because you forgot to multiply 250N for 5 meters. Remember that efficiency is a relation of works, and work is force * DISTANCE).\n;o)
757769
Here's how to solve this.\nThe velocity is a vector, and right-angle components of velocity operate independently.\nThe bullet is fired with 2000 ft/s horizontal velocity but zero vertical velocity. So you know the horizontal and vertical components of velocity at the instant the bullet leaves the barrel.\n\nHowever, once it leaves the gun's barrel, the bullet experiences no horizontal forces, but vertically it is acted upon by gravity.\n\nSo: 1) how long does it take for the bullet to fall 5 ft, starting with zero vertical velocity? (hint: s = 1/2 a t^2\n\n2) Once you figure out how long until the bullet hits the ground, then figure out how far the bullet has travelled horizontally during that time, given 2000 m/s initial horizontal velocity and no horizontal forces.\n\nDoes this help?
451765
A shell.\n\nFrom a tracer bullet.\n\nWhich fell from a Royal Air Force Spitfire in the Second World War in a dogfight with a German aircraft.\n\nI dug it up in my garden sixty years after it had been fired by some brave airman in defence of his country.
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
649550
It is the reaction of shooting force on gun's body. let m=bullet mass and M=gun mass. Then if bullet is shot from gun by velocity of v, the recoil velocity would be V=m*v/(m+M) and its shock is M*V=m*M*v/(m+M)\n\nIt does not depend of gun/bullet type.\n\nIf you want to measure it, you must first measure the mass of bullet and empty gun, then measure the bullet shooting speed. This measurement could be done using a common method of shooting to a tough wooden block, which is hung by four equal size lines from a support. You must measure the maximum height (vertically) of displacement of block after shooting (name it H). Then if the mass of block be MB, then the speed of bullet when going out of gun can be calculated as follows:\n\n(MB+m)*g*H=1/2*m*v^2 then v=sqrt[2*(MB+m)*g*H/m]\nwhere m=bullet mass and g=9.81N/m^2 (gravity factor)\n\nAll units are metric (mass=grams; H=meters; v=meters/second)\n\nGood luck!
866141
You have to calculate p=massXspeed before the "collision".  The quantity remains constant, so if the mass increases, the speed must decreases by the same ratio for the p to stay the same.  (First, you have to convert the 8N weight to a mass using the 9,8N/kg gravity constant)
374463
Assuming no air resistance, the first equation is:\n\n400=1/2 a t^2  (a, as you said, is 32.2 ft/sec^2)\n\nt, the time calculates to be abot 4.98 seconds\n\nthen, v=at, gives v=(32.2)*(4.98)=160 ft/sec\n\nIn the real world, the speed after 400 feet of fall will be less than that due to air resistance, which would slow a brick considerably at 160 ft/sec (over 100 miles/hr)
574466
A heavy pendulum with a ballpoint