Wow ! This is not simple. At first, it looks like there's not enough information, because we don't know the mass of the cars. But I"m pretty sure it turns out that we don't need to know it.
At the top of the first hill, the car's potential energy is
PE = (mass) x (gravity) x (height) .
At the bottom, the car's kinetic energy is
KE = (1/2) (mass) (speed²) .
You said that the car's speed is 70 m/s at the bottom of the hill,
and you also said that 10% of the energy will be lost on the way
down. So now, here comes the big jump. Put a comment under
my answer if you don't see where I got this equation:
KE = 0.9 PE
(1/2) (mass) (70 m/s)² = (0.9) (mass) (gravity) (height)
Divide each side by (mass):
(0.5) (4900 m²/s²) = (0.9) (9.8 m/s²) (height)
(There goes the mass. As long as the whole thing is 90% efficient,
the solution will be the same for any number of cars, loaded with
any number of passengers.)
Divide each side by (0.9):
(0.5/0.9) (4900 m²/s²) = (9.8 m/s²) (height)
Divide each side by (9.8 m/s²):
Height = (5/9)(4900 m²/s²) / (9.8 m/s²)
= (5 x 4900 m²/s²) / (9 x 9.8 m/s²)
= (24,500 / 88.2) (m²/s²) / (m/s²)
= 277-7/9 meters
(about 911 feet)
"Balanced" means that if there's something pulling one way, then there's also
something else pulling the other way.
-- If there's a kid sitting on one end of a see-saw, and another one with the
same weight sitting on the other end, then the see-saw is balanced, and
neither end goes up or down. It's just as if there's nobody sitting on it.
-- If there's a tug-of-war going on, and there are 300 freshmen pulling on one
end of a rope, and another 300 freshmen pulling in the opposite direction on
the other end of the rope, then the hanky hanging from the middle of the rope
doesn't move. The pulls on the rope are balanced, and it's just as if nobody
is pulling on it at all.
-- If a lady in the supermarket is pushing her shopping cart up the aisle, and her
two little kids are in front of the cart pushing it in the other direction, backwards,
toward her. If the kids are strong enough, then the forces on the cart can be
balanced. Then the cart doesn't move at all, and it's just as if nobody is pushing
on it at all.
From these examples, you can see a few things:
-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.
-- The group of forces is balanced if their strengths and directions are
just right so that each force is canceled out by one or more of the others.
-- When the group of forces on an object is balanced, then the effect on the
object is just as if there were no force on it at all.
Answer:
Explanation:
According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.
Being 
and 
the masses of pucks a and b respectively, the initial momentum of the system is
Since b is initially at rest
After the collision and being 
and 
the respective velocities, the total momentum is
Both momentums are equal, thus
Solving for 
The initial kinetic energy can be found as (provided puck b is at rest)
The final kinetic energy is
The change of kinetic energy is
