Momentum: Is collision between 2 equal massed billard balls elastic or inelastic?
929040
It is elastic. I can explain it to you, and I guarantee someone can explain it better but a picture speaks 1000 words. \n\nhttp://en.wikipedia.org/wiki/Elastic_collision\nScroll halfway down.
878067
As long as neither of the balls starts spinning (or changes the rate at which it's already spinning) after the collision (in which case some of their linear momentum would have been converted to angular momentum), the total linear momentum of the system before and after the collision must be the same.\n\nLet p11 and p21 be the momentum of balls 1 and 2 before the collision; let p12 and p22 be the momenum of balls 1 and 2 after ths collision; and let m be the mass of the balls (which we are told have the same mass).  The momenta are all vector quantities, which I'll write as order pairs of numbers (i.e.,  p = (x,y) = x*i + y*j, where i and j are the unit vectors in the +x and +y directions, respectively).\n\nLet the +x axis be pointed in the east direction, and the +y axis be in the north direction.\n\nThen, writing each \n\np11 = m*(2 m/s, 0 m/s)\np21 = m*(-1.2 m/s, 0 m/s)\np22 = m*(0 m/s, 1.43 m/s)\np12 = m*v12 (v12 is the vector velocity of ball 1 after the collision)\n\nBy conservation of momentum, we have that:\n\np11 + p21 = p12 + p22\n\nm*(2 m/s, 0 m/s) + m*(-1.2 m/s, 0 m/s) = m*v12 + m*(0 m/s, 1.43 m/s)\n\nv12 = ((2 - 1.2 - 0) m/s, (0 + 0 -1.43) m/s)\n\nv12 = (0.8 m/s, -1.43 m/s)\n\nThe speed of ball 1 after the collision is given by:\n\nsqrt((0.8 m/s)^2 + (-1.43 m/s)^2) = 1.639 m/s\n\nThe angle the velocity vector of ball 1 makes with respect to the +x axis after the collision is given by:\n\narctan(-1.43/0.8) = -1.061 radians = -60.776 degrees\n\n(It helps to draw a picture of this to visualize why this relationship holds.)\n\nThe first ball is traveling 60.776 degrees south of east after the collision.
803921
Use conservation of energy.\n\na.  1/2mA*vA^2 - mA*g*h = 0 \n\nwhere h =  L - Lcos60   assuming L is the length of whatever is suspending the two balls.\n\nb. Use:  1/2mA*vAi^2  + 1/2mA*vAf^2 = 1/2mB*vBi^2  + 1/2mB*vBf^2 \n\nc.  1/2mA*vAf^2 - mA*g*ha2 = 0 \n     1/2mB*vBf^2 -  mB*g*hb2 = 0 \n\nsolve for ha2  and hb2
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.
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.
711024
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
1081829
Elastic collision.  An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms, as in the collision of atoms.  In this case, two bodies (the ball and the earth) collide.  Since  the mass of the ball is very small compared to the mass of the earth and kinetic energy must be conserved, the ball bounces back with a velocity.
799487
Alright... physics is fun if you get the right problems :) .  \n\nThe collision is elastic... your book should have a conservation of momentum equation.  It states that p1+p2 = Ptotal\n\nYou have to use component vectors for the velocity after the collision.  A good thing to do would be to draw the problem.  It would make it a lot easier.  So use conservation of momentum.\n\nAnd there are two equations that would really help you:\n\nFinal velocity of block A= ((MB-MA)/(MA+MB))*VAinitial\n\nFinal velocity of block B= ((2MA)/(MA+MB))*VAinitial\n\nHelp with physics:\n\nALWAYS draw pictures epecially on your exams.  Get on the good side of the professor, go to a tutoring center if you have one, do many problems and memorize how to do the basic ones if you want an A on the test.
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.
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
787981
let al' mass be m\n\nconserving momentum\nm*0.91=(158-m).829\nsolving for m we get\n         m=75.320kg
884452
Let's assume that the cue ball is traveling in the +x-direction\n\nConservation of energy says\n\n0.5*m*v1^2 = 0.5*m*v2^2 + 0.5*m*v3^2\n\nwith the condition that all the masses are equal this reduces to\n\n(1): v1^1 = v2^2 + v3^2\n\nI'll assume that v2 is the speed of the 8-ball and v3 is the speed, after collision, of the cue ball.\n\nConservation of momentum says\n\nm*v1 = m*v2*cos(q) + m*v3*cos(a)\n0 = m*v2*sin(q) - m*v3*sin(a)\n\nfor the x and y directions, respectively. Here q is the angle that the 8-ball moves away at (35 deg) and a is the angle that the cue ball moves away at. These reduce to\n\n(2): v1 = v2*cos(q) + v3*cos(a)\n(3): v2*sin(q) = v3*sin(a)\n\nWe now have the three expression (1), (2), and (3) for which to solve we can solve for the angle a.\n\nFrom expression (3)\n\nv3 = v2*sin(q)/sin(a)\n\ninserting this into equation (2)\n\nv1 = v2*cos(q) + v2*sin(q)/tan(a)\n\nTaking these two expression and plugging them into equation 1.\n\nv2^2*(cos(q) + sin(q)/tan(a))^2 = v2^2 + v2^2*(sin(q)/sin(a))^2\n\nfrom which we can eliminate v2 to get and expression just in terms of the angles. Expanding this expression we have\n\ncos(q)^2 +sin(q)^2/tan(a)^2 +2*sin(q)*cos(q)/tan(a) = 1 + sin(q)^2/sin(a)^2\n\nrearranging terms and using sin(x)^2 + cos(x)^2 = 1 we have\n\n-sin(q)^2 + sin(q)^2/tan(a)^2 + 2*sin(q)*cos(q)/tan(a) = sin(q)^2/sin(a)^2\n\nmultiplying by sin(a)^2\n\n-sin(q)^2*sin(a)^2 + sin(q)^2*cos(a)^2 + 2*sin(q)*cos(q)*sin(a)*cos(a) = sin(q)^2\n\ncollecting terms\n\nsin(q)^2*(cos(a)^2 - sin(a)^2 - 1) + 2*sin(q)*cos(q)*sin(a)*cos(a) = 0\n\nwhich becomes\n\n-2*sin(q)^2*sin(a)^2 = -2*sin(q)*cos(q)*sin(a)*cos(a)\n\nseparating the a and q terms and simplifying\n\ntan(a) = 1/tan(q)\n\nwhich is readily solved for the a, the angle at which the cue ball moves after the collision.
776706
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
221114
I understand what you're saying, but you forgot one small thing:\n\nThe bowling ball has more mass, and therefore needs more pull to reach the same speed because of momentum. You know, it's harder to push a car than push a cup. The momentum pull and the pull of gravity completely cancel each other out, so all things fall at 9.81 m/s/s.
458509
Conserve both momentum and total energy (i.e., potential energy plus kinetic energy).\n\nMy coordinate system will have the pendulum lifted to the left and the block on the right moving left, so that the initial momentum is - M2 Vo2.\n\nIt will also have the block at height 0, so the initial potential is M1 g h where h is the initial height of the pendulum, which looks like L(1 - sin(10 degrees)).\n\nThe initial kinetic energy is of course (1/2) M2 (Vo2)^2.\n\nSo ...\n\ninitial momentum: - M2 Vo2\ninitial energy: M1 g h + (1/2) M2 (Vo2)^2\n\nAfter the collision the potential energy has decreased to zero, and if w is the combined velocity of the two masses then ...\n\nfinal momentum: (M1 + M2)w\nfinal energy: (1/2)(M1 + M2)w^2 + [the energy "loss" from the slight heating and deformation of the two masses sticking].
744356
I'd say (A) because momentum is directly proportional to mass.  The smaller the mass, the smaller the momentum.\n\nMomentum is the product of mass and velocity\n\nsince the force is equally applied to both masses, their velocities will be the same\n\nthe only difference are the inertial forces generated, which would be less for the smaller mass and more for the larger mass
714752
This problem is a conservation of momentum problem:\n\n(a) In an elastic collision, the formulas for calculating the final velocities are derived from both the conservation of momentum formula and the conservation of kinetic energy formula. Check out my source for more information, but for now just trust me that they are valid:\n\nv1 = (v1(m1 - m2) + 2m2v2)/(m1 + m2)\nv1 = (1*(4 - 7.2) + 0)/(4 + 7.2) = -0.3 m/s\n\nv2 = (v2(m2 - m1) + 2m1v1)/(m1 + m2)\nv2 = (0 + 2*4*1)/(4 + 7.2) = 0.7 m/s\n\n(b) In an inelastic collision, just use the conservation of momentum formula:\n\nm1v1' + m2v2' = (m1+ m2)v\n\n4*1 = (4+7.2)*v\nv = 4/(4+7.2) = 0.36 m/s\n\nI hope this helps!