What is conserved when 2 objects collide in a closed system?
100027
In a closed system, everything is conserved as nothing can escape...by virtue of it being "closed".
197223
It is in the form of potential energy. The two plates could be pushing against each other equally but not going anywhere. As soon as one of the plates give out, the other one will go sliding over or under causing all kinds of vibrations released as kinetic energy. "potential" is the energy of objects at rest and "kinetic" is the energy of objects in motion.
735311
the potential energy of the ball:\nPE = mgh = (.6 kg)(g)(.7m)\nwhen the ball collides with the block all the Potential Energy has been converted to Kinetic Energy so,\nPE = KE = (.6 kg)(g)(.7m) = (1/2)(m)(v^2)        (v^2 is v squared)\nall that kinetic energy is transfered to the block so\nKE(block) = 4.12 Newtons\n\nin one dimension the solution is that all momentum and kinetic energy is transfered to the block such that,\n(a) ball speed after collision = 0 m/s\nand \n(b) block speed after collision\n(1/2)(m)(v^2) = 4.12 Newtons\n(1/2)(2.5 kg)(v^2) = 4.12 Newtons\nv = 1.8 m/s
156917
It depends on how it came to rest. If it had Striked any other ball then its K.E will be transfered to the other. If u were to catch it , then imagine how u will bring your hand towards the ground so as to reduce its impact.Thus its k.E is transferred to you.\n               But if it were to strike the ground then the case is different.From conservation of momentum ,M1V1=M2V2.If M2 be the mass of earth, it is very large when compared to that of the ball. So to satisfy the above condition, V2 should be negligbly small. so we do not face amy impact. Now consider the case of an asteroid or an Hydrogen bomb.We r sure of expriencing the impact.
641698
The total amount of energy in a closed system remains constant.\nThis, however, was found to be untrue. It was later substituted for the conservation of Mass-Energy.\nThis was also disproven. In fact, if a tiny bit of energy is "borrowed" from nothing for a really short amount of time, it can create mass according to the equation:\nE*t = h/(2*pi) = 1.06*10^-34 Js\nOf course, for our pruposes, the conservation of energy works.
95497
This a momentum question.\n\nRemember momentum is:\n\nm * V\n\nm = mass\nV = velocity\n\nSo the conservation of momentum is:\n\n(m1 * V1) + (m2 * V2) = (m1 * V3) + (m1 * V4)\n\nMomentum (before collision) = Momentum (after collision)\n\n(Notice mass does not change after the collision, but velocity does.)\n\nSo let's plug in the variables:\n\nm1 = 2.0 kg\nV1 = 3.5 m/s\n\nm2 = 4.5 kg\nV2 = -5.0 m/s \n(negative indicates object is traveling to the left)\n\nV3 = ?\nV4 = -1.8 m/s\n\n(2.0 * 3.5) - (4.5 * 5.0) = (2.0 * V3) - (4.5 * 1.8)\n\n(7) - (22.5) = 2(V3) - (8.1)\n\n-15.5 = 2(V3) - (8.1)\n\n-7.4 = 2(V3)\n\nV3 = -3.7 m/s \n\n(the negative sign just indicates direction with relation to the axis; in this problem, negative indicates the object is traveling to the left)\n\nSo Object A will be traveling to the left at 3.7 m/s after the collision.
404136
Calculate the total energy of the system before and after and make an energy balance.\n\nFor example, lets say you have an inelastic collision between two cars of equal mass m:\n\nInitial State:\nCar 1 moves at v1 to the right\nCar 2 moves at v2 to the left\n\nAssume v1 > v2\nThey hit and stick together completely.  Use conservation of momentum to find the total resultant velocity:\n\np1 + p2 = ptotal (p is the momentum of car 1 and car 2 respectively)\n\nm*v1 - m*v2 = (m+m)*vf\n\nvf is the final velocity of the clump\nthe minus sign is because I am choosing the direction "right" as positive (momentum is a vector)\n\nvf = (v1-v2)/2\n\nNow do an energy balance:\n\nEbeforecollision = 1/2*m*(v1^2) + 1/2*m*(v2)^2\nEaftercollision = 1/2*(m+m)*(vf)^2 + Efrictionlosses\n\nEbefore = Eafter (this is conservation of energy, and it is true as long as you account for all forms of energy...it wont be true if you forget about the frictional losses in Eafter)\n\nNow solve for Efrictionlosses\n\nTo get the total fractional energy loss, take the ratio Efrictionlosses/Ebefore
816105
Generally, some of their collective kinetic energy is converted into heat, so the KE after the collision is less than the initial KE.  Momentum is conserved.
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
240568
Conservation of momentum
173849
Not necessarily. Because the collision is elastic, the net kinetic energy of the rubber ball and the sidewalk is conserved. But that doesn't mean that each object on its own has the same initial and final kinetic energy.
1134239
When iron rusts it oxidizes.  New atoms are being added to the iron, so of course the mass would increase.  In a closed system, lacking oxygen, iron wouldn't rust, so the law of conservation of mass and energy still holds.  When a match burns, it's losing mass as energy in the form of heat; again the law of conservation holds.
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
932111
No energy is lost in a perfectly elastic collision, so the total energy in the system is the same before and after the collision.
276072
A closed system is a system that does not gain or loose energy.  It only allows the state of the energy to change.\n\nTherefore a closed system follows the conservation of energy.\n\nAn open system on the other hand can take in energy and or release it.  If all you are concerned with is the one open system, for example an oven, then you wouldn't have to concern yourself with where the energy was coming from (gas, electric, combustion) or where it was going (leaking into the kitchen and raising the temperature of your house).
746965
You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
742829
Momentum is conserved. Figure out the total momentum of all three carrianges before the collision, taking into account the direction of the momentums.\n\nAfter the collision, the momentum of the three linked carrianges must be equal to the momentum of the system before the collision.\n\nIf the other two carriages are not moving before the collision, then the velocity of the linked system will be 40/3.