I am doing a school project on shotgun shells. I am attempting to determine the amount of force, possibly in pounds, exerted on an object given the individual pellet's weight and speed. Also, once I figure out what force one has, how does that apply to multiple pellets? For instance a single 1-ounce pellet going 1500 feet per second hits with so much impact, how would I figure the force of 8 pellets? Do you multiply the origional answer by the number of pellets? Any help finding the equation or web links related to this would be greatly appreciated. Thank you.
132087
"M x V to the power of 2 ÷ 450400 = foot pounds energy"\n"Example: The muzzle energy of a 300 Remington Ultra Mag 180gr Core-Lokt Ultra bullet propelled at 3250 feet per second"\n\n"Step 1: Multiply M (M = bullet weight in grains) times V2 (V2 = the square of bullet velocity in feet per second): 180 x 3250 x 3250 = 1,901,250,000"\n\n"Step 2: Divide the product of step 1 by 450400: 1,901,250,000 ÷ 450400 = 4221 foot pounds of energy."
461705
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
503183
Conservation of momentum so mv is constant.\n\nTotal final mass = 5333+14.6 = 5347.6g\n\nTotal final momentum = 5347 x 1.7 gm/s = 9090.92 gm/s\n\nBullet's initial momentum = 9090 gm/s\n\nMomentum/mass = velocity\n\n9090.92/14.6 = 622.666m/s
318043
1. the object is pushed up by the weight of its equivalent volume of water, 4-3.2=0.8l\nSo the object's density is equal to its mass divided by its volume:\n4kg/0.8l=4kg/0.0008m3=40,000kg/8m3=5000kg/m3\n\n2. the force decreases with the square of distance, so it will be 3^2=9 times smaller, which is 6N.
237582
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)
251323
OK, just to make this easier, let's label the bullets 1-12.\n\nFirst, weigh bullets 1-4 on one side and 5-8 on the other side.  If they're even, then the odd bullet is one of the bullets 9-12.  In that case, weigh bullets 9-11 against bullets 1-3.  If they're even, the odd bullet is 12 and you can weigh it against any other bullet to see if it's light or heavy.\n\nIf the second weighing is not even, then the odd bullet is in 9-11, but you know whether it's heavy or light because you just weighed those three against three good bullets.    Say for the sake of argument that it's heavier.  Then weigh 9 against 10.  If they're even, the 11 is odd and heavy.  If they're not even, then the heavier of the two is odd and heavy.  You can do the same thing with light--just say the lighter of the two is the odd one if they're uneven.\n\nNow, go back and suppose the first weighing was uneven.  For the sake of argument, say that 1-4 was heavier than 5-8. Then it's either in 1-4 and heavy, or in 5-8 and light.  Now weigh 1-3 and 5 on one side, 4 and 9-11 on the other.  If 1-3 and 5 are heavier, then the odd bullet is in 1-3 and heavy.  Weigh 1 and 2 against each other to see which bullet is odd.  If the sides are even, then the odd bullet must be in 6-8 (the ones you removed) and light, so weigh 6 and 7 against each other.  If 4 and 9-11 are heavier, then either 4 is odd and heavy or 5 is odd and light (you basically switched their places on the last weighing).  Weigh one of them against a standard bullet to tell whether it's the odd one or not.
118914
The longer barrel of the rifle means that the bullet is being accelerated over a longer distance.  If you assume a constant acceleration (which isn't quite the case) the velocity formula is \nV = sqrt(2*A*L)\n(V=velocity, A=acceleration, L=length)\nSo, with the same assumption, a doubling of the barrel length would result in multiplying the velocity by sqrt(2).
460072
the idea is that the fireman is accelerating such that ma=mg-800N, the 800N is exerted by the person holding the fireman and is facing up, the force on fireman that is pointign down is mg (a is his acceleration). where ma is mass of fireman times his acceleration. 1. find a\n2 find v, given that d=v^2/(2a) [from: d=at^2/2, v=at] where d is 10m\n3. knowing v, the ave force of impact is calculated based on impact energy and dissipation of this energy in frictional force acting for 20cm: mv^2/2=F*d, find F..., where d=20cm
393978
I'm not certain there is enough information provided to answer the question. However, try this. The average force per ms is the quotient obtained when one divides: the sum of the product of the mass of the ball multiplied by the incoming impact speed of the ball plus the mass of the ball multiplied by the departing speed of the ball from the bat by the duration of the impact, 1.1ms. But, since force is measured in units of mass per sq 'unit' of size, it seems more information is necessary to solve the problem.
464093
Force(avg) = mass*acceleration(avg)\n\nAlso, acceleration(avg) = (delta)velocity/(delta)time, where delta represents change.\n\nNow, solve for these variables:\n\nacceleration(avg) = (3.2-0)/(.01-0) = 320 m/s^2.  Now calculate avg force:\n\nForce(avg) = (.084)(320) = 26.88 N\n\n---------\n\nHope this helps
394261
Hi, \n\nIn general, the equation describing the position of an object under the influence of a constant force is as follows:\n\nh(t) = at^2 + vot + xo \n\nwhere : \nh(t) = height at any time t\na = acceleration\nvo = initial velocity\nxo = initial height.\n\nSo the answer would be \nF= m*a (mass times accelleration) = (the mass kg)*(1.96m/s^2) \n\nI'm guessing that you want to find out the equation for acceleration, so what you should do is use the values for t, and   h(t) to solve for the constants (xo, vo)\n\nexample: \nat time t=0, height is 0.50m\n\nso h(0) = a(0)^2 + vo (0) + xo = xo\n\nbut that also equals 0.50m;\n\nso:\nxo = 0.50m.\n\nand the equation becomes \nh(t) = at^2 + vot + 0.50m. \n\nThen use the different values of t, and h(t) to do the same thing with the other constant, and you'll have it. \n\nHope that helps, \nMatt
5070
im going to assume this ia an elastic collision (when the objects bounce off of each other).\nyou just need the formula for conservation of momentum. p=mv (momentum = mass times velocity)  conservation of momentum says that (assuming no friction) the total momentum before the collision would equal the total momentum after the collision.  in other words mv=mv.  now, there are two balls so your formula looks like this:\n mv+mv = mv+mv.  the reason it has 2 "mv" 's  on each side, is because there are two balls.  now the second mv on the first side can be taken out since the velocity is 0, so mass times velocity would be 0. that means mv + 0 = mv + mv  so the mass times velocity of the first ball before the collision equals the mass times velocity of the first ball after plus the mass times velocity of the second ball after.  this would look like this:\nmv= mv + mv\n(1kg)(10m/s)=(1kg)v1 + (2kg)v2 (where the 1 and 2 just represent the velocities of the first and second balls respectively.)\nso basically:\n10=v1 + 2v2\nyou would need one of the velocities in order to solve for the other. there has to be more to the problem to either give you one of these velocities, or enough information to solve for it somehow... the balls would move in opposite directions.\n\nhowever, we can solve this problem if it was an inelastic collision(meaning the balls stick together).  if this were the case, then (if this is a problem for class) it probably would have said it was inelastic and it probably isnt because balls generally bounce off of each other unless they're sticky or balls of clay or something, but lets say it was inelastic just in case so i can solve it this way for you.  if it was inelastic then mv + mv=mv so the mv of the first ball plus the mv of the second ball(which is zero like before because the ball is at rest so v is zero) equals the combined masses of the balls plus the new velocity. so mv + 0 = mv, because we have only one object(because they move as one object if they stick).  the combined masses of the balls is 1kg + 2kg = 3kg so now:\nmv=mv\n(1kg)(10m/s) = (3kg)v\nsolving for v we get 10/3 or 3.3333.  they would be moving in the same direction as the initial ball.
381081
The force applied to the beast (by the bullet) would have to be calculated by the angle/arc of it's descent, this new force differential may cause a new vector.  ie  If it was falling straight down and you fired straight up...no change.....if it were coming in at 45^ angle down, the bullets force will slow it slightly, and it's arc down will have increased due to the continuing force of gravity...with less momentum.  That is...also only if the head stays attached....if not then...the two bodies must be calculaated by the individual resistance/force of the bullet....    In the end...take one step to the right...after your shot...and watch out for falling bits and pieces.
363238
(A) The impulse on the ball due to the kick is the integral of the force of the time. Let a = 6x10^6 and b = 2x10^9, then\n\nImpulse = integral(F(t))dt = integral(a*t - b*t^2)dt\n= (0.5*a*t^2 - b*t^3/3) | (3x10-3,0)\n\nwhere the last part indicate evaluated from 0 to 3x10^-3 seconds.\n\n(B) The average force is the same integral divided by the time interval, or the impulse divided by the time over which the impulse is applied.\n\nImpulse/dT = Impulse/(3x10^-3 - 0)\n\n(C) The time of maximum force can be found by taking the derivative of the force and looking at which time the derivative equals zero.\n\ndF/dt = a - 2*b*tmax = 0\n\nThis leads to \n\ntmax = a/(2*b)\n\nUsing this time in the force equation will give you the maximum force on the ball.\n\n(D) The average force can be expressed as\n\n<F> = m*dV/dT\n\nwhere dV is the change in velocity and dT is the time over which the force is applied. Solving for dV\n\ndV = <F>*dT/m\n\nwill give the speed immediately after the impulse force.
40752
Energy is conserved so the gravitational potential energy (ge) of the projectile is equal to the spring potential energy (se) expended by the spring.\n\nge=mgh and se=1/2 k(x2-x1)^2, where x1 is the spring position when compressed and x2 is the equilibrium spring position. (x1-x2) is spring displacement.\n\nGiven: (x1-x1)=0.120m, m=0.035kg, h=20.0m.\n\nA) Find spring constant, k.\nee=se\nmgh=1/2 k(x2-x1)^2\nk=2mgh/(x2-x1)^2\n2*0.035kg*9.8m/s^2*20m/.12m)^2\nk=953N/m\n\nB) Find projective speed as it moves through spring equilibrium position.\n\nIf you set your 0 on the x-axis to the position where the spring is compressed, then x1 would be 0 and x2 would be 0.12m. So the spring equilibrium position is at x2=0.12m.\n\nNow use the kinematic equations of motion to find the projectile speed when spring displacement is 0.12m. You can assume constant acceleration to be 9.8m/s^2.\n\n(vf)^2=(vi)^2+2ad where vf and vi are final and initial velocity, respectively, vi=0, and d=(x2-x1)=x2.\n\nvf=sqrt(2*a*x2)\nvf=sqrt[2(9.8m/s^2)(.12m)]\nvf=2.325m/s
468149
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
424800
a=Fnet/m\n.5x1.5=0.75N net force\n\nFnet=Fa+Ff\nFf=.75-1.2\nFf=-.45\n\nI think this is the answeri, but in my class we only dealt with object on ramps and hills and cliffs.