can someone help me complete these momentum problems?
487088
#1) 32(.16)=.127v_2 ==> v_2=40.3\n#2)i'm too lazy\n#3).105(350=.345*v_2 ==> v_2=10.6\n#4)0+F(.10)=.40(19)  =>>F=76N
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
546536
Take two objects of the same mass moving towards each other (in opposite directions) with the same speed. Let's say one object has momentum of p and the other -p. The total momentum of the system (p-p) is 0. When they collide, they both stop moving and the total momentum of the system is still 0.
1085576
F = d/dt(mv) [the time deravitive of the momentum] = dp/dt  where p is the momentum\n\nWork = Energy = F x distance\n\nPower = Energy/time\n\nPower = (F x distance)/time\n\nPower = [(dp/dt) x distance]/t\n\nremember dp/dt = d/dt(mv) = m dv/dt + v dm/dt
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)
747766
Yes, use conservation of momentum.  The angle comes into the problem by determining how much of the total momentum is in the x and y directions.  Really, you'll be solving two different equations:\n\nm1*v1x+m2*v2x=m3*v3x\nand\nm1*v1y+m2*v2y=m3*v3y\n\nSo, find each player's initial momentum in the x and y directions, and then solve these equations to find the final x and y velocities of the group.  \n\nGood luck!
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
763881
Take the x-axis as the line of action of the moving ball. and suppose the second ball moved at an angle theta wrt to the line of action with certain velocity that u need to find out. Now resolve the velocity components of both balls along x-axis and y-axis before and after collision. Now conserve the momentum along both x-axis and y-axis before and after collision. solve the equations and u will know all the unknowns.\n\nFor the second question find 'e' i.e. co-effecient of restitution and depending on the value it will be elastic or inelastic.
694807
Use conservation of impulse and conservation of kinetic energy.\n\nmv + 0 = mu + mw\n(1/2)mv2 = (1/2)mu2 + (1/2)mw2\n\nu + w = 1 (consider it as a straight line in a graph)\nu2 + w2 = 1 (consider it as a circle in that graph)\nWhen you solve this set of two equations, you get\nu = 0 and w = 1 (m/s)\n(Find the useful intersection. The other one is is the initial situation.)\n\nTh
587714
First Question:\n\na) Solving for the height of the 1st brick\n\nh=1.00m (sin 30.0)=0.50 m\n\nSolving for the velocity of the 1st brick at the bottom\n\nKE at the bottom = PE at the top\n1/2 v^2 = (weight) h\n1/2 v^2 = 24.3 (0.50)\nv = 4.93 m/s\n\nb) momentum before = momentum after; momentum is just m*v\n\n(24.3/9.8)(4.93)+(35.7/9.8)(0)\n=[(24.3/9.8)+(35.7/9.8)] [v]\n\nv = 2.0 m/s\n\nc) F = ma\n-5.0 = [(24.3/9.8)+(35.7/9.8)] a; F is opposite to motion\na = -0.817 m/s^2; deceleration\n\na =(vf-vi)/t; vf=final velocity; vi=initial velocity\n-0.817 = (0-2.0)/t\nt = 2.45 seconds\n\nd) 2ad = vf^2 - vi^2\n2(-0.817)s = 0^2 - 2^2\nd = 2.45 meters\n\nSecond Question\n\na)Make West East axis as the x-axis; N S axis as y-axis.\n\nx momentum before = x momentum after\n\n880(0)+1585(12) = (880+1585)(vx)\n\nvx = 7.72 m/s; x component of final velocity\n\ny momentum before = y momentum after\n\n880(-16)+1585(0) = (880+1585)(vy); compact car is headed in the negative y-axis.\n\nvy = -5.71 m/s\n\nv^2 = vx^2 + vy^2\n\nv^2 = 7.72^2 + (-5.71)^2\n\nv = 9.6 m/s answer\n\nangle = Arctan 5.71/7.72\n\nangle = 36.5 degress south of east\n\nb) F = coef of friction (weight of the cars)\n\nF = 0.58 (880+1585)(9.8)\n\nF = 14011 N\n\nF = ma\n-14011 = (880+1585) a; - because opposite motion\n\na = -5.68 m/s^2; decelerating (-)\n\n2ad = vf^2 - vi^2\n\n2(-5.68)d = 0^2 - 9.6^2\n\nd = 8.11 meter from the poin of impact
52010
Nancy's momentum is 270 kg*m/s west.\nTonya's momentum is 210 kg*m/s 35 degrees south of west.\n\n1. Break Tonya's momentum into west and south components using trig.\n\n2. Add their westward momenta to get the new westward momentum for the two bodies.  Tonya's southward momentum becomes the southward momentum for the two bodies.\n\n3. Divide their southward momentum and westward momentum by the combined mass of 110 kg to get the southward velocity and westward velocity.\n\n4. Use Pythagoras to get the net velocity.  Use trig to get the angle.
985678
Incoming momentum = p1 = mv1\nOutgoing momentum = p2  = mv2\nChange in momentum = p2 - p1 = m(v2-v1)\nThis change occurs in time, t.\nRate of change of momentum = m(v2-v1)/t = impressed force, f.\nm=0.149 kg, v1 = 43.4 m/s, v2 = -48.7 m/s, t = 1.1x10^-3 sec\n\nf = m(v2-v1)/t\nYou do the arithmetic.
947129
a) Force on the box due to max acceleration witout moving the box = the maximum frictional force = 0.24* Normal force = 0.24*Weightof box = 0.24*mg\n\nSo max acceleration = 0.24*g = 2.352m/s^2\ns = .5at^2 = 10.584\n\nb)\nthis is a vector problem\nthe boat needs to be driven towards west inorder to cross exactly to a point at north.\nresultant speed of boat = 140/20 = 7m/s\nso using pythagorian theorem, U^2 = 7^2 +4^2, or U=sqrt(65) = 8.0623\n\nc)displacement = (11i - 5j) - (7i - 8j)  = (4i + 3j) \nforce = (3i + 4j)\nwork done = Fs = (3i + 4j)(4i + 3j)  = 12 + 12=24\n\nSo change in Kinectic energy  = work done  = 24J
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.
843756
USE PRINCIPLE OF CONSERVATION OF linear MOMENTUM.\n\nearth gas lower KE because it has huge mass and huge inertia
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
927051
The given final speed doesn't seem to compute. I get 14.8 m/s. \nMa*Va + Mb*Vb = (Ma+Mb)*Vf \nis the formula that applies to show conservation of momentum in a completely inelastic collision, which is what you have. The only explanation I could suspect is some cargo was spilled. \n\nTo continue the assignment - kinetic energy is given by\n(1/2)*m*v^2\nThere is no requirement that all the original kinetic energy be measurable as kinetic energy after a completely inelastic collision. But you should be able to point to where the missing energy went. Some of the energy probably went into cart damage. Perhaps the cargo slid around.