Physics. Elastic collisions.?
960298
m1vi1+m2vi2 = m1vf1 + m2vf2\n\nWhere:\n\nm1 = mass of first object\nvi1 = initial velocity of object 1\nm2 = mass of object 2\nvi2 = initial velocity of object 2\nvf1 = final velocity of object 1\nvf2 = final velocity of object 2\n\nKeep in mind though that velocity is a vector quantity. That means that there is directional intent with each velocity. Overall, that means when you add the velocities, you have to take in account the sign (+/-) and the direction. The easiest way to add velocity vectors is to place the end of one velocity at the head of another one. It's difficult to explain the concept of vector addition (or subtraction) in text, so look it up on the web, or the link I'm providing.\n\nOh, and one more thing. Since it's an elastic collision, kinetic energy is conserved. that means K1i + K2i = K1f + K2f, where K = (1/2)(mv^2)\n\nIf kinetic energy is not conserved, then it's an inelastic collision.\n\nInelastic collisions only have momentum conserved.
716426
If the objects are perfectly elastic (that is they don't deform), then you can use the laws of conservation of momentum and conservation of energy to solve the problem. If they (or one of them is) are plastic...that is they deform during the collision, then conservation of momentum still applies, but it is more difficult to use conservation of energy unless you know how energy is lost in the deformation process.\n\nI don't mean that to sound complicated, but that's it in a nutshell.
1119730
ahh, good theory problem...\nlet's see, there's energy, friction, force (normal, tension, as well as the other ones mentioned), gravitation, mass, pendulum motion, gotta be more that I've missed, but no doubt someone'll be here to correct me.
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.
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.
635128
Rubber, like many other materials has a property called ELASTICITY, that is the ability of an object to recover its shape after a deformation.\n\nSo, when the ball bounces, the force of the hit deforms the ball, and it recovers its shape because of elasticity, "pushing" ground in the opposite way, as Newton´s Second Law states.
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
322347
it is the collision of two preasure sysems and the difference of the tempeture
735972
Conservation of energy and momentum basically means that energy is never created or destroyed, and same with momentum. It has to come with somewhere. If energy or momentum leaves one system, it must enter another system. If energy appears in a system, it had to have come from one. Momentum is never really gained or lost, and same with energy -- it is simply transfered from one body to another. An elastic collision is one in which no kinetic energy is lost, all of it is transfered directly from one system to another.. In an inelastic collision, some of the energy is transfered to a third system (via sound waves, heat energy, or other such energy loss). In this case of an inelastic collision, the total kinetic energy of the two bodies before, is not the same as it is afterwards. The difference in these values would be the energy "lost" from these systems that was tranfered elsewhere.
763881
Take the x-axis as the line of action of the moving ball. and suppose the second ball moved at an angle theta wrt to the line of action with certain velocity that u need to find out. Now resolve the velocity components of both balls along x-axis and y-axis before and after collision. Now conserve the momentum along both x-axis and y-axis before and after collision. solve the equations and u will know all the unknowns.\n\nFor the second question find 'e' i.e. co-effecient of restitution and depending on the value it will be elastic or inelastic.
936755
Momentum is defined as mass x velocity.  When two objects, m1 and m2 collide with velocities v1 and v2 the equations is:\n\nm1 x v1 = m2 x v2 showing that like energy momentum is conserved.  Use this equation to solve your porblem and go to the head of the class.
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)
834476
No energy is lost as a result of heat or material deformation in an elastic collision.  It simplifies the mathematics by not having to keep track of non-motion energy, but it does not reflect practical results.\n\nIf automobile collisions were totally elastic, body shops would be out of business but funeral homes and hospitals would have a boom.
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
800469
Well, the way the balls collide, the ball that is being hit always goes in the direction that is perpendicular to the tangent between the two balls at contact. And you know that it is an elastic collision, and that momentum is conserved, so once you know the direction of one ball, you can figure out the direction of the other ball, and prove that their paths create a 90 degree angle (other then head on collisions). Hope this helps.
501193
In physics a collision is said to take place \n * When two bodies physically collide against each other or\n * When the path of one body is changed by the influence of the other body.\n \nEXAMPLES :\n  We have several examples for collision\n * Marbles or carom coins colliding with each other\n * Molecules of a gas colliding with one another etc.\n\nThere are two types of collisions.\n1. ELASTIC COLLISION\n2. IN-ELASTIC COLLISION\n\n1. ELASTIC COLLISION:\n   If there is no loss of kinetic energy during a collision, it is said to be elastic collision.\nExample: collision between atomic and nuclear particles are elastic.\n\n2. INELASTIC COLLISION:\n   If there is loss of kinetic energy during a collision it is said to be in-elastic collision.\nExample: If a ball is dropped from a certain height and the ball is unable to rise completely to its original height, then it would mean that the ball has lost some kinetic energy due to collision.\n\nHEAD-ON ELASTIC COLLISION:\n  It is that elastic collision in which the colliding bodies move along the same straight line path before and after collision.