text: need physics help!? a few physics questions? i'm working on completeing a physics assignments and have found myself lost. here is the information i have been given:  car 1 approached the intersection from the top of a 25-meter hill.  ∙ car 2 was on a flat stretch of road directly in front of car 1.  ∙ at the bottom of the hill, before braking for the stop sign, car 1 was going 20 m/s  and car 2 was going 30 m/s before it stopped at the stop sign.  ∙ from the skid marks on the road you can see that car 1 applied force on its brakes  for 3 seconds, 100 meters before the stop sign.  ∙ there were no skid marks left by car 2. the collision occurred at the stop sign,  where car 2 had stopped.  ∙ after the collision, both cars were moving together in the same direction at 10 m/s,  before slowly rolling to a stop.  you must now push car 2, using 800 n of force, 10 meters off to the side of the road so  no one else gets hurt.   i have run out of characters to use, so if you can help me, please email me! a fascinating history, but you haven't yet asked the question. you really can't run out of characters. just use the edit question/add details option on this question to continue, and then i can edit my answer.  it looks like, in addition to the question(s), some information is missing like what the braking force was, or what the cars' masses are. put all the facts in.  so far we know: v1a (car 1's velocity before braking) = 20 m/s. v1 (car 1's velocity before collision) is unknown. v2 (car 2's velocity before collision) = 0 m/s. vfinal (velocity of both cars after collision) = 10 m/s. edit: now we're on track. call the mass of each car m. working backward and using conservation of momentum, 2*m*vfinal=m*v1+m*v2, so v1 = (2*m*vfinal-m*v2)/m = 2*vfinal - v2 = 2*vfinal = 20 m/s. so car1 actually didn't do any braking! are you sure 20 m/s was the velocity before braking? i'm going to go ahead assuming the 20 m/s was after braking. going (actually coasting) down the hill, car 1 would have picked up 25*m*g n-m of energy, but we don't know its velocity v0 at the top of the hill so we don't know its total energy at the bottom of the hill. if v0=0, its velocity would be found from the potential-to-kinetic energy of the hill: 0.5*m*v^2 = 25*m*g, so v = sqrt(50*g) = 22.14 m/s. then the energy removed from the car by the braking =  0.5*m*22.14^2-0.5*m*20^2 = 0.5*m*(22.14^2-20^2). in the collision the car1 energy loss = 0.5*m*20^2-0.5*m*10^2, and the car2 energy gain = 0.5*m*10^2. the above assumes that the actual scenario was that car1 had zero velocity at the top of the hill and braked down to 20 m/s rather than going 20 m/s before braking. respond with more info if these assumptions are wrong. 2nd edit: car1 had 200,000 n-m energy before the collision  and then lost 150,000 n-m, and car2 had no energy and gained 50,000 n-m. energy wasn't conserved in the collision. only elastic collisions conserve energy, and if 2 cars collide and don't bounce apart it's not elastic. in fact real collisions bend metal and break things, and this converts mechanical (kinetic) energy to heat energy.