determine the final velocities of each ball after collision in the event that the collision is elastic.
131126
the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s
178779
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
510928
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
641519
do the math get the weight of each obgect and the velocity there traveling plus graviety constent and figure it out.
351353
Well if you're asking about the terminal or final speed this is how you do it:\n\nThe initial energy of the system is equal to the elastic energy kept on the spring because the blocks are not moving so they do not have kinetical energy.\n\nso let's say Eo=1/2k*X^2 where k is the spring constant and X is the magnitude of its compression.\n\nat the final moment the spring has no energy kept within but both blocks are moving so they do have energy\n\nso let's say Ef=1/2*(m1)*(V1)^2+1/2*(m2)*(V2)^2\nwher m1 and m2 are the mass of each block\nand V1 and V2 are the speed of each block\nalso because there's no friction the mechanical energy will be equal at the initial and final moment so: Eo = Ef\n\nalso we know that the linnear momentum of the system will be kept constant and because the blocks were not initially moving the linnear momentum will be equal to zero\n\nso the linnear momentum at the final moment will also be zero so:\n\n Pf = 0 = (m1)*(V1)+(m2)*(V2)\n\nso we got a system with two equations and two incognits\n\n1/2*(m1)*(V1)^2 + 1/2*(m2)*(V2)^2 = 1/2k*X^2         (1)\n(m1)*(V1) + (m2)*(V2) = 0                                        (2)\n\nthen V2 = -(m1)*(V1)/(m2)                 (3)\n\nso (3) -------> (1)\n\n1/2*(m1)*(V1)^2 + 1/2*(m2)*(-(m1)*(V1)/(m2))^2 = 1/2k*X^2\n\nsolving the equation for V1 we got two solutions\n\n V1a=1.21 m/s and V1b= -1.21 m/s\n\nyou can take eihter depending on your referation system\n\nlets take the positive answer\n\nso V2= -(m1)*(V1)/(m2)   => V2= -0.69m/s\n\nthe minus indicates only that the second block is moving in the opossite direction of the first one so we got:\n\nV1 = 1.21 m/s      and V2 = 0.69\nV1 is the speed of the 15.5kg block and\nV2 is the speed of the 27.1kg block\n\nthey move in oposite directions
258331
the faster the ball goes the more it goes down
439685
Equation:\n\nV^2 = U^2 + 2 A (delX)\n\nThen solve F = M A\n\nwhere V = Final Velocity\nU = Initial Velocity\nA = Acceleration (unknown)\ndelX = distance of applied force (delX meaning delta X, or change in distance)\nF = Force\nM = Mass\n\nOf course you have to consider two things:\nThe force applied over the two feet would have to be constant (this is an assumption you have to make to solve)\nAll your units are in ft and weight.  you have to use the meters measurements and convert your ball's weight of 0.25 lb to kilograms (that's the mass measurement)\n\nyou now have more than enough information to solve this problem on your own.\n\nNOTE: Jim, that is a horrible misuse of the wrong equations for solving this kind of problem.  Nice try though.
195427
FOLLOW THE BOUNCING BALL
488150
If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
471898
If m1 is the mass of the moving particle and m2 is the mass of the particle at rest, then \n\nm2 / (m1+m2) of the kinetic energy of the moving particle will be lost after inelastic collision.\n\n\nBy law of conservation of momentum the final velocity of the two together is \n\nV = m1 v1/ (m1 +m2) becasuse v2=0.\n\nAfter collision the kinetice energy is 0.5 (m1+m2) V^2.\n\nBefore collision the kinetic energy was only 0.5 m1 v^2.\n\nUsing these relations we get the above result.
448562
good start.\n\ntotal inital momentum = 480-397.5\n\nthis is an inelasitc collision, and aferwards you have \nboth masses travelling at velocity v:\n\n(75+120)v = 480-397.5\n\nsolve for v.  This is the velocity at which they both slide.\nThe inital energy now is 1/2(75+120)v^2\n\nThis energy is converted to frictional energy:\n\ninital kinetic energy = total work of friction = Fr(distance)\n\nFr = uN = u(120+75)g = .3(195)(9.8)\n\nso we have\n\n.3(195)(9.8)(distance slided) = 1/2(195)v^2 \n\nYou know v form above.  So find distance slided.
276011
Bricks sliding\n\nI won't give you detailed worked solution, rather some hints to get you started.\n\na) To calculate the speed of the first brick at the bottom of the incline, use the equation:\n \ns = ut + at^2\n\nwhere \ns = displacement or distance (1.00 m)\nu = initial speed (0 in here because it starts from rest)\na = acceleration (= gsin30 or 9.81 x sin30)\nt = time (s) which you need to calculate so that you can find the final speed from:\n\nv = u + at\nwhere \nv = final speed, u, a and t as b4.\n\nb) this is a momentum problem. Remember that momentum before = momentum after for this type of collision.\n\nc) Friction is a retarding force and F = ma, so now you can work out a (which is negative) and hence how much time will elapse before the bricks come to rest using v = u + at (Note: v = 0 here because the bricks come to rest)\n\nd) From c) and the equation s = ut + at^2 you can get an answer to this.\n\nCars colliding:\na) Again, momentum b4 = momentum after in this type of collision. Now momentum or p = mv. Note that it is a vector, so you will have to do a vector addition to find momentum after. and hence speed of the wreck.\n\nb) Use the equation relating friction to mass and the coefficient of kinetic friction. Then find the deceleration, hence find d.\n\nI hope that helps. If you get stuck, please ask a more specific question on the part you can't do.
539898
Apply v^2 = u^2 + 2as vertically;\nwhere v=end velocity, u=start velocity, a=acceleration, s=distance\nWhen the object reaches the height of its path, vertical velocity is zero.\nhence: 0=(30Sin30)^2 + 2(-10).s\ns = 225/20 = 11.25m. \nHence it reaches a total height of 11.25m+1.2(height from which it is thrown).\n.\n.\nTime of flight:\nApply: S=ut + 1/2 at^2 vertically, where S=distance, u = start velocity, a=acceleration, t=time\nSince the ball is caught at the same height at which it is thrown, the overall vertical distance travelled is zero.\n0 = 30Sin30.t + 1/2(-10).t^2\nt = 3s.\n.\n.\nHorizontal distance = horizontal velocity x time\ndistance = 30Cos30 x 3= 77.9 metres
399204
It depends on whether the objects are completely elastic, completely inelastic, or somewhere in between. And also, did they hit square and were the velocities both on the same line or did they get deflected at a new angle. I'll assume they hit square and their velocities were on a line - like they were on a track.\n\nElastic:\nMomentum before = momentum after\nMa*Va + Ma*Vb = Ma*Vb + Mb*Va\n\nInelastic:\nM*Va + M*Vb = 2M(Va+Vb)/2
333972
I don't usually answer homework questions, but you're being led astray by kingkongchess, and you'll pay the price at test time anyways if you don't figure these out for yourself, so I'll answer one and give you some hints on where to look for with the others.\n\n1) F=ma=dp/dt=m x dv/dt = constant, so m x delta v=const. x delta t.  More simply, m x v=const. x time, but be warned, this WONT work unless the force is constant!  Anyways, v=(32N x .16s)/.127kg = 40.3 m/s (just divide away the mass for the velocity).  A little more likely than 1575 m/s (~3500 mph!), no?  That'd be some hockey player!\n\n2) kingkongchess sets it up exactly right; all answers come from the conservation on momentum, m1v1=m2v2.  This is also true for problem 3.\n\n3) Trickier.  Before collision, the system consists of 2 seperate momentums, p1 and p2.  After the collision, the system consists of the single body (with momentum (m1 + m2) x v3 ) stuck together.  Conservation of momentum states that the momentum of the system before the collision must equal the momentum of the system after collision.  Just make them equal, and solve for what you're looking for.\n\n4) The word 'accelerates' should give you a clue.  Think 'force' when you hear the word accelerates, and when you think force, think 'rate of change in momentum'.  Put these ideas together, you'll have your answer.
549492
Projectile Motion-  Examine the initial velocity and angle necessary to hit targets at different points.
370967
Assume that the cue ball has mass m and initial velocity v, and that after the collision, the cue ball has velocity v1 at an angle of x to its original path, and the 8 ball has velocity v2.\n\nx-direction: p1 = p2: mv = mv2 cos 35 + mv1 cos x\ny-direction: p1 = p2: 0 = v2 sin 35 - v1 sin x\n\nv = v2 cos 35 + v1 cos x\nv1 = v2 sin 35 / sin x\n\nv = v2 cos 35 + v2 sin 35 cot x\n\nKinetic energy: 1/2 mv^2 = 1/2 m (v1^2 + v2^2)\nv^2 = v1^2 + v2^2\n\n(v2 cos 35 + v2 sin 35 cot x)^2 = (v2 sin 35 / sin x)^2 + v2^2\n\n(cos 35 + sin 35 cot x)^2 = (sin 35 csc x)^2 + 1\n\ncos^2 35 + (2 cos 35 sin 35) cot x + (sin ^2 35) cot^2 x = (sin^2 35) csc^2 x + 1\n\ncos^2 35 + (2 cos 35 sin 35) cot x = (sin^2 35)( csc^2 x - cot^2 x) + 1\n\n[Remember that 1 + cot^2 x = csc^2 x]\n\ncos^2 35 + (2 cos 35 sin 35) cot x = (sin^2 35)( 1) + 1\n\n(2 cos 35 sin 35) cot x = sin^2 35 + (1 - cos^2 35)\n\n(2 cos 35 sin 35) cot x =  2 sin^2 35\n\ncot x = 2 sin^2 35 / (2 cos 35 sin 35) = tan 35 = cot 55\n\nx = 55 degrees\n\nThanks for giving me something to do.