Inelastic collision problem- bullets and pendulum?
The height the bob rises is related to θ as h=L*(1-cos(θ)) the energy required to raise the bob is (M+m)*g*h this is related to the speed of the bob after collision as .5*(M+m)*vc^2=(M+m)*g*h and it is related to v0 using conservatin of momentum m*v0=(M+m)*vc combine it all together now v0=(M+m)*vc/m vc=sqrt(2*g*h) v0=(M+m)*sqrt(2*g*L*(1-cos(θ)))/m ============================= 44 caliber: =(10+12/1000)* sqrt(2*9.81*L*(1-cos(10.1)))*1000/12 9mm: =(10+6/1000)* sqrt(2*9.81*L*(1-cos(4.3)))* 1000/6 take the ratio of 9mm/44 caliber =(10+6/1000)*sqrt(1-cos(4.3))* 12/(6*(10+12/1000)*sqrt(1-cos(10.1))) 0.851877 =========================
Help with physics? Bullet hitting a pendulum?
So you have in inelastic collision in which momentum is conserved, but energy is not. You work this backwards. Use the height the block rises to, h, to determine the potential energy of the block plus bullet. Then this energy is conserved and will equal the kinetic energy of the block plus bullet after impact, thus we can find the speed just after impact. Since momentum is conserved, we can use that speed to compute the momentum of the block plus bullet and set that equal to the momentum of the bullet just before it hits the block. Then we can solve for the bullet's initial speed. h = 60 mm = 0.006 m, g = 9.8 m/s^2, m = 9.6g = 0.0096 kg, M = 4.5 kg, v1 = speed of bullet plus block just after collision and v0 = initial speed of bullet. Both the bullet and block ate at height h so the potential energy is U = (m+M) gh Conservation of energy requires 1/2 (m+M) v1^2 = (M+m)gh --> v1 = sqrt(2gh) Momentum of bullet plus block is P1 = (m + M) v1 = (m +M) sqrt(2gh) Conservation of momentum requires P0 = mv0 = P1 = (m +M) sqrt(2gh) ---> v0 = sqrt(2gh)*(m+M)/m v0 = 509.4 m/s
Bullet, ballistic pendulum, collisions?
The method of solution is to write down the energy, kinetic and potential, of the two bodies before the collision and then to write down the energy, kinetic and potential, of the two bodies after the collision. Then invoke the Law of the Conservation of Energy: total energy in the entire system before collision = total energy in the entire system after the collision. Notes: K.E. = ½(mv²); P.E. = m.g.h. After the collision, the total mass = M + m. This bit is often overlooked. For questions like "What was the speed of the bullet-bob combination immediately after the collision takes place", it is far easier to invoke the Law of the Conservation of Momentum: momentum = product of mass and velocity. Again, note that after collision, if the bodies combine, we have (m + M) for the total moving mass.
A ballistic pendulum(for speed of bullet) consists of a block of wood suspended by a cord. When the bullet is fired into the block, the block is free to rise. how high does a 5.0 kg block rise when a 12g bullet travelling at 350 m/s is fired into it?
Start With conservation of momentum.mass of the bullet times velocity of the bullet = the total mass times the final velocitymb * Vb = mt * Vfmt = mb +mlmt = total masmb = mass of the bulletml = mass of the blocksolve for VfVf = mb * Vb/(mb + ml)then it is simple conservation energy kinetic is converted to potential1/2 * mt * Vf^2 = mt*g*hh = 1/2* Vf^2/gaccording to pythonVf = 0.838 m/sh = 0.0358
Physics, Please Help!!! (Elastic Ballistic Pendulum)?
This is an example of an inelastic collision, in which momentum, but not kinetic energy, is conserved. (Some of the initial kinetic energy of the bullet goes into performing work as the bullet blasts a hole in the pendulum's block, and eventually ends up as either heat or strain energy in the block). Let m, v1, and v2 be the mass, initial velocity, and final velocity of the bullet. Let M and V be the mass and velocity of the block just as the bullet exits the block (the initial velocity of the block is zero). Momentum conservation requires that: m*v1 = m*v2 + M*V At the moment the bullet exits the block, it has transferred some of its kinetic energy to the block. As the block rises, this kinetic energy is transformed into gravitational potential energy, and when the block reaches its highest point, the kinetic energy is zero, and the potential energy of the block has increased by an amount equal to the initial kinetic energy of the block: 0.5 *M*V^2 = M*g*h where g is the acceleration of gravity, and h is the maximum height to which the block rises. Rearranging this equation to solve for V: V = sqrt(2*g*h) Plugging this back into the momentum-conservation equation, we have: m*v1 = m*v2 + M*sqrt(2*g*h) v1 = v2 + (M/m) * sqrt(2*g*h) Plugging in the values for this problem, we have: v1 = 200 m/s + (1500 gm/ 7 gm)*sqrt(2 * 9.81 m/s^2 * 0.127 m) v1 = 338.2 m/s
Physics problem, ballistic pendulum. Please help!?
This starts as a momentum problem: Total Momentum before = Total Momentum after ( 0.0025*425) + (0.215*0) = V( 0.215 + 0.0025) V = 1.0625 / 0.2175 V = 4.885m/s This is the speed of the block/bullet combination. This now becomes an energy problem: Kinetic energy of block/bullet system: Ek = 0.5*m*v² Ek = 0.5*0.2175*4.885² Ek = 2.595J The potential energy at the top of the pendulum swing is equal to the kinetic energy at the bottom of the swing Ep = m*g*h m*g*h = 2.595 0.2175 * 9.81*h = 2.595 h = 2.595/ 2.1315 h = 1.217m The block/bullet combination will rise to a height of 1.217m
A bullet of mass m/5 is fired with a velocity (v) to hit a pendulum, what is the maximum height reached by the pendulum?
What is the mass of the pendulum?What is the nature of collision- elastic, inelastic or perfectly inelastic?If the collision is inelastic, what is the coefficient of restitution and does the bullet stick into the bob after collision?Provided data inconsistent with the question.An attempt to the solution:-Conserve linear momentum just before and after the impact.Conserve mechanical energy by equating kinetic energy of the pendulum with its final potential energy which would determine it's height.
In case of a pendulum held in vertical direction a bullet is shot at it, would the momentum be conserved and why?
I am assuming the bullet is shot in the horizontal direction perpendicular to gravity, and as the bullet is moving very fast in the horizontal direction, we can neglect the small vertical velocity it will gain in the time between its firing and the time it hits the pendulum bob.In such a case momentum in the horizontal direction will be conserved just after the bullet hits the bob. Shortly after that a force will start acting on the bob in the horizontal direction also - tension in the pendulum string, so horizontal momentum is not conserved in the journey of the pendulum, only just before and just after the collision with the bullet.Case A: Just before collision Case B: Some time after collision
MOMENTUM AND A PENDULUM ,,,,, pHYSICS !?
I don't see your diagram and don't have your initial values but I can give you the basic solution to the ballistic pendulum problem. This is an inelastic collision. You use the conservation of momentum and the conservation of energy to solve. let's call the mass of the bullet m1 and its initial velocity v1 the mass of the pendulum is m2 and it's initial velocity is v2 (v2 obviously = 0 m/s) the initial momentum = the final momentum m1v1 + m2v2 = (m1+m2)v3 the momentum of the bullet when it is fired plus the momentum of the pendulum before its hit (0) = the momentum of the bullet+pendulum combo after the collision, v3 is the velocity of the bulled+pendulum combo (1/2) (m1 +m2) v3^2 = (1/2)(m1+m2)gh The total kinetic energy before the collision (the bullet + pendulum ) is = to the total potential energy of the bullet+pendulum combo at the top of it's swing) the mass (m1+m2) cancels out v3 = sqrt(2gh) substitute back into the first equation m1v1 + m2v2 = (m1+m2)v3 v2 is the velocity of the pendulum before the collision, and that's zero, so... m1v1 + = (m1+m2)v3 m1v1 = (m1+m2)[sqrt(2gh)] v1 = {(m1+m2)[sqrt(2gh)]}/m1 this gives you the initial velocity in terms of things you can measure, the mass of the bullet, the mass of the pendulum, the acceleration due to gravity, 9.8 m/s/s, and the height of the swing. On some ballistic pendulum devices, you have to calculate out the height of the swing using the angle that the pendulum was displaced. If this is the case h = l(1-cos a) where l is the length of the pendulum and a is the angle
Ballistic Pendulum help (Finding the height)?
The length of the pendulum is unimportant. (kinetic energy) KE = (1/2)mv^2 (potential engery)PE = mgh (momentum) p = mv Total energy is conserved in a collision, so is momentum. KE of bullet before = (1/2)(.012)(380)^2 = 1444 N I assume the bullet and pendulum "stick" after collision, so that m = 6.012kg KE of bullet+pedulum = 1444N directly after collision. Now the pendulum swings and converts all the KE into PE mgh = 1444 => h = 1444/(6.012 * 9.8) = 24.51m