A 900 kg car is traveling 30 degrees South of East at 18 m/s collides with a 750 kg car traveling north at 23 m/s.  THe cars stick together.  What is the speed of the wreckage just after the collision?\n\n(a) 12.2 m/s\n(b) 10.2 m/s\n(c) 25.1 m/s \n(d) 17.3 m/s \n\nI know that the answer is (b) 10.2 m/s.  But, how do I solve this problem?  I know that if one car is going east and the other car is going north and they meet at an intersection...\n\ntheta = tan^-1(m2v2/m1v1)\nand that I would plug that value of theta into...\n\nvf = m1v1/(m1+m2) cos theta\n\nWhen I worked it out the problem that way... the answer I obtained is not listed as a valid choice. \n\nthNks
136065
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
511144
You're on the right track\nOne small clarification:\n\n\nm1gh = 1/2m1v1^2 + 1/2m2v2^2\n\nm1gh=5*9.8*5\n=245\n\nYou can also calculate the velocity of m1 at the moment of collision:\nm1gh=1/2m1v1^2\n=sqrt(9.8*5*2)\n=9.9\nso\n5*9.9=10*v2f+5*v1f\n9.9=2*v2f+v1f\nv1f=9.9-2*v2f\nSubstitute into the quadratic:\n245*2=5*(9.9-2v2f)^2+10*v2f^2\n245*2/5=\n9.9^2-4*9.9V2f+4V2f^2+2v2f^2\n6v2f^2-39.6V2f+.01=0\nSolve\nfor v2f\n0, 6.6\nusing 6.6\nv1f=9.9-2*6.6\nv1f=-3.3\nagain, \nusing mgh=1/2mv^2\nh=.5*4.4*3.3/9.8\nh=.74m\n\n\nj
403122
I solved the problem by stating that P would travel distance d in time t\nd=20*t+.5*.3*t*t\n\nand Q would travel d+20 in time t\nt is the time to collision\n\nd+20=30*t\nd=30*t-20\nSubtract the equations and solve for t:\n\n0=.15*t*t-10t+20\nt=2.064 seconds\n(the second root is superfluous)\nd=41.9m\n\nWhich proves collision\n\nj
576902
Point A and Point B are 600 miles apart. A train leaves Point A travelling toward Point B at 5:30AM at a constant speed of 65 kph. Another train leaves Point B heading toward Point A at 6:20AM at a constant speed of 40 kph. What time do the two trains meet?1 kilometer = 0.6214 miles\n\nx=kilometers     5/6(50 minutes/60 minutes)*65= 54.166...km\nx=.6214 (600 miles)\nx=372.84 kilometers apart\n5/6(50 minutes/60 minutes)*65= 54.166...km\n372.84-54.17= 318.67km between the 2 trains at 6:20\nthats as far as I can get (I don't remember how to do this... hope this helps!)
462450
what's the 30 g for? here's my answer:\n\nfrom F=m*a,\n\nF= (70 kg)*(3m/s^2)\nF=210 newtons...\n\nthis is the force needed to help the 70 kg person survive the crash...does that help?
43169
Fnet= MA so you acceleration is 36km/h, so convert that to m/s, which is 10 m/s. To get acceleration, we know that the car started out at 18km/s, which is 5m/s. So the acceleartion = Change of velocity/Change of time. So that comes out 1 m/s/s. So there is 1000 Newtons on the car. The other guy's answer is wrong. I'm an engineering major. The standard Units in science is MKS, meters kilograms and seconds
471068
Let the speed of the car be v. The centripetal force for the car to go in a circle is provided by the friction in this case. That is \nFrictional force = m*v^2 / r\nIf this works out greater than the max static frictional force, the car will skid. Hence the max. speed is given by\nm*v^2 / r = Us*m*g\nv = sqrt ( Us*r*g) = sqrt ( 0.8*77*10) = 24.8 m/s\n\nFrom the above equation for v it is clear that the max speed is independent of the mass.
608538
We have the same amount of energy in each case, however\n\n\n(A)Kinetic Energy (KE) car1 >0 , KE car2 =0\n50% damage. Energy is being shared equally (since the cars are identical) \n\n(B) KE car1 >0 , KE Wall =0\n100% damage. All energy is observed by the car since the wall is inelastic and immovable.  \n(C) KE car1 = KE car2 >0\n100% damage since both cars had the same amount of kinetic energy.\n\nIn reality the structure of the vehicles and the geometry of collision would determine the extent of damage.
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
375719
The cars are moving away from each other because they they are moving in opposite directions.\nThe distance between them in 1 hour would be 45+40 km=85 km\nso to be 385 km apart,they would take 385/85=77/179after reducing)or4 9/17 hours.
395913
F=ma.\n\nSo now you know the available force.\n\nApply it to the mass and use the tangent function to solve.\n\n(Strictly speaking, this is a trick question, since any hill will cause the car to "slow" somewhat.  I get the sense you're missing a parameter, like the minimum speed.)
353636
problem number 1\nlet B = brian and M = maya. to find out when they catch each other we need to make the equation for brian's distance equal to maya's because that is when they will catch up:\n8 mph (x+1) = 10mph (x)   [remember x+1 is how long brian's been riding cuz he's been riding for one extra our than maya]\n8x+8 = 10x\n2x = 8\nx = 4\nso it will take Maya 4 hours to catch up\n\n2. x = hot dogs y = hamburger\n2x+1y = 3.99  (this shows 2 hot dogs multiplied by its price plus 1 hamburger multiplied by its price is 3.99)\n3x+3y = 6.57\n\nso since we need to find the cost of one hot dog (x) we need to make one of the two above equations into all x's.  so using the first one it would be y = 3.99-2x.  no using this equation, let's replace Y for the second equation.  \n3x+3y = 6.57\n3x+3(3.99-2x) = 6.57\n3x+11.97-6x=6.57\n11.97-3x=6.57\n11.97-6.57=3x\nx=1.80\nso one hot dog costs $1.80
493187
1st step: OK we need to know how many times  faster the car went on the way to the service station than on the return trip:\n49/16=3.0625 times faster on the way up than the way back. \n\n2nd step: find the how much time it took on the way up and the way back. \n\nNow comes the Algebra:\nif we let X stand for the way up, the equation for the round trip would be:\n X+ 3.0625X= 4\n4.0625X=4\nX=4/4.0625\nX=.985 \n\nso the time up = .985 hours and the time back = 4 -.985, or 3.015 hours.\n\nso how far is the station? well, it's 49km/h x .985 h, (using the data for the trip to the gas station) or:\n\n48.265 km\n\nok we're DONE!!\n\nPS: I have a sneaking suspicion that if you look at the problem again, you might find it said 48 km/h on the way up, which would work out to be that the trip was EXACTLY 3 times faster on the way up. ok just a hunch. But from what you gave me, that is the answer.
375923
let's say ball A is moving along x-axis before the collision.\nafter collision ball B is moving along y-axis.\n\nconservation of linear momentum.\nbefore collision: pxi = M*Vai; and pyi = 0\nafter collision: pxf = M*Vaf*cos(theta); pyf = M*Vbf - M*Vaf*sin(theta)\n\nwhere:\n - Vai: speed of ball A before collision (5 m/s)\n - Vaf: speed of ball A after collision (?)\n - Vbi: speed of ball B before collision (zero)\n - Vbf: speed of ball B after collision (?)\n - theta: direction of ball A after collision (30°)\n\npxi = pxf\nM*Vai = M*Vaf*cos(theta)\nVaf = Vai/cos(theta) = (5m/s)/cos 30 = 5.77 m/s
456770
The extent of damage sustained is directly proportional to the amount of energy consumed. If in case a) the stationary car is allowed to move after the collision, then case a) is where the original car sustains the LEAST damage. \n\nDue to conservation of momentum, in case c) the velocity of the two vehicles will be zero after the collision. So vehicle 1 will have lost all of its energy and the energy will be converted into deformation (damage).\n\nIn case b) the velocity of the car will also be zero after the collision, so all of its enery will be converted into damage also.\n\nSo, in case a) the car sustains the least damage, in cases b) and c) the car sustains the same amount of damage.\n\nI am only looking at the damage to ONE vehicle in each case, not the damage to all vehicles.
504455
Let's assume your range of vision was 0.2 miles infront of you, and 0.1 miles behind with the mirror.  (I'm pulling these numbers out of my behind.)\n\nAt t=0, the car was at position x = -0.1 relative to you.\nAt t=5 seconds, it was at x = 0.2.\n\nSo the speed (relative to you) is 0.3 miles / 5 seconds = 0.06 miles per second = 216 miles per hour.  Since you were doing 85 mph, it real speed was 301 mph.\n\nAre you sure this took only 5 seconds?  I picked some pretty small numbers for the range of vision and this still results in some wicked speed.
330981
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.