Elastic collision in 2D?
922775
If everything were lined up exactly inline and marble 1 hits 2 which hits 3, then both marbles 2 and 3 would move together at the same rate as marble 1.    (1~~>2~~>3)   You have probably seen that demonstrated with the newton's pendulum toy that has 5 chrome balls hanging by strings.   When you pull 1 of the balls back and let it swing into the other 4, 1 ball swings off the other end.  2 balls cause 2 to swing off the other end.\n\nIn your experiment, the 1st marble will hit the other 2 and release all of it's energy.  They will then go at the same rate as the first marble.\n\nV1 = 0\nV2 = 15.5 in a direction 60 degrees off to left.\nV3 = 15.5 in a direction 60 degrees off to the right.
312163
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
578272
Try downloading a I-mesh.\n\nThey may have it.
318413
OOO, We did this project in IPC, except our was 20 meters -.-;;. You're ideal angle is 45 degrees. That usually gets you the farthest. It would be pretty smart to use elastic potential energy by using piece of bamboo or just a flexible stick (haha, flexible stick) or just a rubber band. Don't put anything on the egg. Our model was made with a giant rubber band that I found on the floor, and a little seat attached to the crossbow model. If your teacher is strict about a catapult model, your best bet would be to use the bamboo stick and research trebuchets.
905915
You're going to have to resolve this two-dimensional problem into two one-dimensional problems.  I'd suggest using vertical and horizontal.  On the vertical side, we have gravity pulling down, and the normal force and one component of the tension from the rope pulling up.  This, by the way, is why the normal force is not equal to the weight - by pulling on the rope, you're partially lifting the sled off the snow.\n\nYou know that there is no acceleration vertically, so the net force vertically must be zero.  Knowing this, you can get the normal force as a function of the tension as the weight of the sled minus the vertical component of the tension.  Knowing the normal force and the coefficient of friction will give you the force necessary to pull the sled at constant velocity.  (If the velocity of the sled is constant, what can you say about the net force horizontally?)  Since the friction is only opposed by the horizontal component of the tension, you can find that.  Once you have that, you'll find that the only variable you have left is the tension in the rope, and you can solve for that.\n\nHope that helps.
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
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
614061
Not real sure what you want to do, but, go here;\nhttp://quake.wr.usgs.gov/research/deformation/modeling/eqmodel.html
711513
Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!
1006529
How do I animate an inanimate object?
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
827118
We can't use conservation of energy because this is an inelastic collision. Conservation of momentum does hold, so\n\nm1*u1 + m2*u2 = (m1+m2)*v\n\nu2 is zero. So \n\nv = m1*u1/(m1+m2) = .24 kg*u1/2.85 kg = .084*u1\n\nThe projectile's incident kinetic energy \nKEp = (1/2)*m1*u1^2 = 0.12*u1^2 joules\nThe final kinetic energy \nKEf = (1/2)*(m1+m2)*v^2 = 1.42*(.084*u1)^2 = .0100*u1^2 joules.\n\nPercentage = 100%*KEf/KEp
170385
There seems to be a missing varible of time. Without a time contraint there is an infinite number of times these balls will touch. I am not sure the minimum distance from the wall though. I guess it would have to be at least smaller than the size of the small ball so the two would actually collide.
238652
Assuming you have a rectangular object, then L x H x W will work. For other shapes you need something else. See sources.
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)
961828
Try clicking all the keys shown below;\n\n[on] [y=] [(-)] [1] [6] [x,t,o,n] [^] [2] [+] [9] [6] [x,t,o,n] [+] [5] [enter] [window] [0] [enter] [1] [0] [enter] [1] [enter] [0] [enter] [2] [5] [0] [enter] [graph]\n\nNow on screen you should have a negative parabola with the ball reaching a maximum height  when t = 3 seconds.
701523
Good ? fernando,\n\nI thought about this problem and remembered blowing up balloons long, long ago in my childhood.  (It actually is easier to blow when the balloon gets larger).  \n\nReason: the material stretches in the larger balloon and becomes thinner, so the elastic modulus is inversely proportional to size.  Ergo, when the 2 are connected, air will flow from the smaller to the larger, since the smaller one has more pressure.\n\nI realize this is not a formal proof, but it will have to do for now!