a few physics questions?

Question

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!

Thank you for the help!

So that means that Car 1 lost 4,290,179.1 of energy from the top to the bottom of the hill? Just want to make sure I understood that correctly.

Answer 1

This answer is kind of a repeat of another question you asked (see ref.), but I think I've earned the right to try to bag both answers. I've added the data as corrected and data missing from the above question. The unknowns asked for are:
a. Energy lost in braking after downhill coast from 0 initial velocity at top
b. Energy lost by car1 in collision.
c. Energy gained by car2 in collision.
Apparently the question doesn't ask anything about pushing car2 off the road.

Givens, work and answers:
V1 (car 1's velocity after braking and before collision) is 20 m/s.
V2 (car 2's velocity before collision) = 0 m/s.
VFinal (velocity of both cars after collision) = 10 m/s.
The mass m of each car is 1000 kg. Since both masses are equal, the 3 velocities above are consistent with conservation of momentum.
Coasting down the hill, car 1 acquires 25*m*g N-m of energy (potential energy converted to kinetic energy). So 0.5*m*V1^2 = 25*m*g, so V1 = sqrt(50*g) = 22.14 m/s. Then (answer to a) the energy removed from car1 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.
This was an inelastic collision. Car1 had 200,000 N-m energy before the collision and then lost 150,000 N-m (answer to b), and car2 had no energy and gained 50,000 N-m (answer to c). 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.

Answer 2

Maybe they can help you in the Physics section.