when a moving ball stops, where does its kinetic energy go.?
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.
643339
It's all about conservation of energy.   The contestant did some work on the ball.   Remember the definition of work?\n\nWork = Force * Distance\n\nSo, the contestant did a work:  \n              W = 810N * 2.0m = 1620Joule\n\nOk, since energy is conserved, the ball will have a kinetic energy of 1620Joule at the begining of the throw, but it will eventually lose it as it goes up, transformed into Potential energy.\n\nWhen the ball reaches the highest point its speed is equal to zero, and hence the kinetic energy is zero too.  Conservation of energy requires that all 1620Joules transformed into potential energy.  So, do you remember the formula for potential energy?\n\nP. energy =  mass * gravity * height ; \n\nwith gravity = 9.82m/sec^2\n\nSince the potential energy is equal to 1620Joules:\n\n1620Joules = (7.0Kg)*(9.82m/sec^2)*height\n\nSo:\n\nheight = (1620Joules)/(7.0Kg * 9.8m/sec^2) = 23.567m\n\nHope I helped
608223
For moving objects, calculate kinetic energy using the formula:\nKE = (mass x velocity2)/2 or 1/2 mv2\n\nhttp://www.gmhsscience.com/problems/kineticpotential.htm
846849
1) Yes if you neglect the effect of wind resistance\n2) Yup, because some of its momentum is transferred to the ball (Total remains constant)\n3) Yup, since it was in zero momentum before it was hit and now it has a speed\n4) Ya, all happens
892916
I think the key here is that the objects will be rolling down the plane; thus the gravitational potential energy they have is not only converted into translational kinetic energy, but also into rotational kinetic energy.\n\nAnyway, I would approach this from energy considerations.  Start by calculating the amount of potential energy the objects will lose by falling a distance of 3 m.  (For both, this will simply be PE = m g h).  Equate this to the total amount of kinetic energy the objects have at the bottom of the ramp (translational plus rotational, or 0.5 m v^2 + 0.5 I w^2 where I is the moment of inertia, and w is the angular velocity.)\n\nSo, you need to know the moment of inertia for a disk and a hoop.  The hoop I know off the top of my head, I(hoop) = m R^2.  The disk, I looked up to find I(disk)=0.25 m R^2.\n\nFinally, recognize that as they do not slip, you can relate the angular velocity w, to the translational velocity v:  v=2 Pi R w.\nOnce you do this, you find that the kinetic energy depends only on the velocity v, so we have one equation with one unknown.  Just solve for v with the two different values of I, and I think you ought to find that the disk will be moving faster than the hoop.\n\nSorry for being long winded.... but I hope it helps.
237074
Yes, but you notice the rubber ball is going slower now. Its because part of the momentum has been transfered from the ball to the sidewalk (eventually tranfered to the Earth itself).  This is called transfer of momentum.
503790
Gravity always accelerates the ball in the "Y" direction such that the force pulls the ball downward.\n\nEven if the ball stops in midair, it is still being accelerated at 9.8m/sec^2\n\nThis means no matter how hard you hit it in the x-direction, it will always be accelerating at 9.8m/sec^2 in the y-direction.\n\nThat is a major concept in kinematics.  The motion in the x and y directions are independent of one another!
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
241285
For a back row spike I found 26.7-27.6m/s (mean).  Sorry, but that was all I could find!  I was going to do a study on it - but then I switched out of biomechanics into exercise physiology.  You could do it by calculating time elapsed between contacting the ball and the ball reaching the floor.  You would have to know the height at which the ball was contacted and how far it traveled horizontally from the point of contact.  They you find the hypotenuse using Pythagorean theorem and divide this number by the time to get an estimated velocity.
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
958811
You physics types ALWAYS have to delete something, eh? Like mass, or air resistance or friction! lol\n\nAnyway, I think the ball would be a) either moving forward or stationary, depending on the wagon's rate of acceleration, relative to the ground, and b) rolling backward (due to inertia), relative to the wagon.
865948
KE = 1/2 mv^2\nfind the KE before the collision using the formula.\nfind the KE after the collision using the formula.\nsubtract the two values to find the change in kinetic energy
969296
Their acceleration becomes zero\nThey will maintain constant velocity from then on\nThey will no longer accelerate
698592
The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.
862752
This question deals with either conservation of enregy, or conservation of momentum. In this case it will be easier to go with momentum, so that's what I'll do.\n\nThe momentum before an event will equal the momentum after the event. In this case, the momentum before the event equaled zero, so the momentum after must equal zero also.\nIf the ball has a momentum of (.350kg)(28m/s)=9.8kgm/s, then the man must have an equal and opposite momentum to cancel it: (65kg)(v)=-9.8kgm/s. Since we now know the mans mass and momentum, figuring out his velocity is simple:\n(65kg)(v)=-9.8kgm/s\nv=-0.15m/s
201479
When you throw a ball in the air it gradually gets slower until it starts falling back towards the Earth.  The faster you throw it the higher it goes.\n\nImagine now if you throw the ball fast enough that the Earth's gravity will not be able to slow it down enough before it escapes into deep space (outside our solar system).\n\nThis speed is the escape speed or escape velocity.\n\n---------------------------------------------------------\n\nMore accurately when you through a ball in the air you give it a certain amount of kinetic energy, as it gets higher this kinetic energy is being converted into gravitational potential energy (gravitiational potential energy is greater the higher you are).  The escape velocity is the velocity at which it has enough kinetic energy that you would have to take it an infinite distance away from Earth before it was all converted into gravitational potential energy.
137603
Consider  one small part of the dancing string:\nelastic energy <==> kinetic energy.\nThe sum of both is mechanical energy.\nIt is supposed to be constant.\n\nOnly the friction of the air damps the oscillations.\nMechanical energy ==> (friction) heat\nThat's all foulks!