Conservation of momentum problems?
961481
When dealing with the conservation of momentum, remember that the total momentum before a collision is the same as the total momentum after a collision. Momentum is mass*velocity.\n\n1) 2*45 + -7*90 = -1*90 + v*45\nSolve for v to get:\nv = (2*45 - 7*90 +1*90)/45 = -10 m/s\n\n2) 10000*10 = (10000+620)*v\nSolve for v to get:\nv = (10000*10)/(10000+620) = 9.42 m/s\n\n3) \n(a) -3*75 - 5*60 = (75+60)*v\nSolve for v to get:\nv = (-3*75 - 5*60)/(75+60) = -3.89 m/s\n(b) 3*75 - 5*60 = (75+60)*v\nSolve for v to get:\nv = (3*75 - 5*60)/(75+60) = -0.56 m/s\n\nI hope this helps!
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
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
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
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
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.
701538
yup you can do this with momentum conservation laws.  I wont do it for you though.\n\nBut I will help you out.\n\nResolve in the direction of the slide after the crash.  Work out the angle each car was travelling relative to this line (a diagram will help). Then use momentum of the collected impact slide.  From that you can work out the velocities of the two separate vehicles using cos*angle*mass*velocity=momentum
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
709316
assuming second block is originally at rest\n\nconservation of momentum says\n\nv11m1=v2m2+v12m1  (v11 initial block1 velocity, v12 final block 1 velocity)\n\nnow 1/2 mv11^2 = 1/2m2v2^2+1/2m1v12^2\nor m1v11^2 = m2v2^2+m1v12^2\n\n2 equations and 2 unknowns
896193
This is an impulse/momentum problem.  Impulse is ∫f*dt[0 - t1] it must equal the change in momentum.  Impulse is given by Favg*t, in your case 0.012*16000 N*sec  Therefore the momentum of the student on impact was 0.012*16000 = m*v.  m is the mass of the student, 63kg.  Therefore his velocity on impact was v =0.012*16000/63.  He fell under acceleration g = 9.8m/sec^2.  Velocity after falling time t is g*t.  Distance traveled under uniform acceleration is h=.5*g*t^.    Use the velocity to get t, then compute h.
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].
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!?
757057
Conservation of momentum\nP = mv before must equal mv after\nmass remains the same\nvelocity is a vecor quantity (magnitude and direction)
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?
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.
95497
This a momentum question.\n\nRemember momentum is:\n\nm * V\n\nm = mass\nV = velocity\n\nSo the conservation of momentum is:\n\n(m1 * V1) + (m2 * V2) = (m1 * V3) + (m1 * V4)\n\nMomentum (before collision) = Momentum (after collision)\n\n(Notice mass does not change after the collision, but velocity does.)\n\nSo let's plug in the variables:\n\nm1 = 2.0 kg\nV1 = 3.5 m/s\n\nm2 = 4.5 kg\nV2 = -5.0 m/s \n(negative indicates object is traveling to the left)\n\nV3 = ?\nV4 = -1.8 m/s\n\n(2.0 * 3.5) - (4.5 * 5.0) = (2.0 * V3) - (4.5 * 1.8)\n\n(7) - (22.5) = 2(V3) - (8.1)\n\n-15.5 = 2(V3) - (8.1)\n\n-7.4 = 2(V3)\n\nV3 = -3.7 m/s \n\n(the negative sign just indicates direction with relation to the axis; in this problem, negative indicates the object is traveling to the left)\n\nSo Object A will be traveling to the left at 3.7 m/s after the collision.
321671
Ah, this brings me back!\nThis is conservation of momentum problem. That means that the initial momentum must equal the final. You must break up the values to x and y directions.\n\nThe 900kg mass is traveling 18mps * cos(30) = 15.6mps in +x\n                            18mps * sin(30) = 8.98mps in -y\nThe 750kg mass is traveling 0mps in x\n                            23mps in +y\n\nWith these values take the initial momentum as\n\ninit mom in x = 900kg(15.6mps) + 0 = final mom = 1650kg*(v)\nwhere v is final speed in x = 8.5mps\n\ninit mom in y = 900kg(-8.98mps) + 750kg(23mps) \n\nthat equals the final mom in y = 1650kg*(v)\n\nso v=+5.6\n\nIf you know the speed in the x and y directions, convert to vector notation by using s^2=vx^6+vy^2 = 10.2mps\nand the angle is atn(vy/vx)\n\nHope this helps