A 1055 kg van, stopped at a traffic light, is hit directly in the rear by a 655 kg car traveling with a veloci
963771
conservation of momentum (mv): mv_final=mv_initial:\n655kg*2.25m/s =1055kg*v_van_f +655kg*v_car_f (sign of velocity to be determined)\nconservation of energy: Ek_car_initial =Ek_van_final +Ek_car_final\n655kg*(2.25m/s)^2 =1055kg*v_van_f^2 +655kg*v_car_f^2. sub v_van or v_car from momentum eq-on into energy eq-on:\n\n655kg*(2.25m/s)^2 =655kg*v_car_f^2 + 1055kg * (655kg*2.25m/s/1055kg -655kg/1055kg*v_car_f)^2, expand and solve quadratic equation to get\nVcar=2.25m/s or -0.526m/s. the first one is nonrealistic, since it implies no collision ,ie velocity didn't change.\nso Vcar is -0.526m/s or opposite to where it was travelling before\nVvan is found from momentum eq-on: v_van = 655kg* 2.25m/s/1055kg - 655kg/1055kg*v_car_f=1.723m/s\nI checked this by plugging into the momentum formula.\nso \na) -0.526m/s (ie in opposite direction)\nb) 1.723 m/s\n\n[To Michael: the point of Yahoo Answers is that I answer the question and explain as good as I can/have time for. It's not up to me or you to judge educational value of my answer or to assume that this question is for homework.]
4088
because the air inside the car is moving with the car
4133
Yes. The car acts as a Faraday cage, so the current goes around the surface and does not harm the occupants.
394496
they use the point of impact of the crash and where the vehicle stopped and measure the distance. using math they can figure out just how fast the car was moving.
70853
You sure can.  Let's say that the speed of the offending car is "v."  If your car is at rest, then you just need to know the mass of your car (M), the mass of the car that hit you (m), the coefficient of friction of the road ("y"... in other words, how much the road "pulls" on your tires.  See the below link for some different coefficients of friction; it seems that the one we'll be using here is y~.8), and the distance you skidded "d."  If we call the gravitational constant "g," then it will go like this:\n\nThe kinetic energy of the offending car is KE=.5*m*v^2\n\nWhen the offending car hits yours, that energy is transferred to you, so your energy is now KE=.5*M*V^2\n\nThus, .5*m*v^2=.5*M*V^2.\n\nSolve: V^2=(m/M)*v^2.  V is your initial velocity after the collision.  Don't worry about the square root.\n\nYour car's decceleration is g*y.  The relationship between deceleration, distance, mass, and initial velocity is\n\nV^2=2*g*y*d\n\nso\n\n(m/M)v^2=2*g*y*d\n\nsolve\n\n---> v^2=2mgyd/M <--- the final equation\n\nnow just take the square root, and you have the offender's speed.\n\nCheck my work, though.  I probably made a mistake.\n\n***One final note - I'm not sure how familiar you are with physics, but you'll have to standardize your units when you do this problem.  For example, if you're going to use metric units, I'd advise using kilograms and meters, with 9.8 = g.  This will give you an answer in meters/sec, which you can convert.  If you're going to use english units, you'll want to use... uh... pounds and feet, with g = 32.
558814
that's called relative velocity.....for the bee the car was still.....
91362
It's tranferred to heat/friction from the impact
539548
A car that drives on a bridge.
771612
hello\n\na)take the momentum of the sports car and divide by the total mass of  both cars. 25×920÷3220=7.14m/s\n\nb)using the same method as a), we can find the answer. but use a little logic here, if the total mass doubles then the final speed must be halved for conservation of momentum to be followed. v=12.5m/s\n\nc)KE=0.5mv². speed is halved. v² means kinetic energy should be quartered (0.5²). but mass was doubled due to there being 2 cars now. so overall kinetic energy has only been halved (0.5²x2). the other half is lost.
768321
It's like a car accident. You just have to slow down and look....
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
1074577
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.
1183555
Probably pollution caused by motor vehicles.
884398
The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
816334
your body is accelerating at the same rate of the car so no matter what you do, you will impact the water at 9.8 meters per second squared. that should kill you. the same happens in falling elevators. if you jump before it hits the bottom, you will break your legs or die.
265074
Reckless driving resulting in an accident with property damage.
1007337
The KE of the moving vehicle is dissipated as heat in the brakes as the brakes use friction.\n\nThe answer to your second question is that no work has been done and therefore there has been no change in KE.\n\nIn terms of transfer of momentum, the really huge ball called the earth and the car did interact.  If the car started out going West and then ended up going East, the total momentum (angular and linear) between the car and the earth must be conserved.  This means that the earth was caused to change its rotation when the car turned around.  Obviously, the earth is so much bigger than the car that you don't even notice the change in momentum of the earth.