The engine of a 1950 kg car driving at 25m/s suddenly dies out as it approaches a 25 m high hill. Will the car have enough energy to clear this hill?

Work out the kinetic and the potential energy needed, compare and there's your answer.

With no air or mechanical drag the mass and energy of the car is irrelevant. If you were to drop the car from 25m, would it be going 25m/s when it hit the ground?

I can't recall the math at the moment, other than acceleration due to gravity is 9.8m/s/s. A napkin calculation says it comes up a little short. I think you could clear a 30m hill at 30m/s, but at 25m/s you only make about 22m up, 20m/s you make 15m, etc.

You're making it too complex catgita.

At the bottom of the hill, the car has KE = \frac{1}{2}mv^2 = 0.5*1950*25*25 J
At the top of the hill, the car must have PE = mgh = 1950*25*9.8 J

As the KE is > the PE, the car makes it over the hill. As it has enough KE to convert into PE.

I don't understand your method of doing the sum. I don't see how you are able to do it that way without knowing more information.

For one, the mass cancels out when you have:
1/2mv^2>mgh
giving you
1/2v^2>gh

Second, maybe you have heard of someone called Newton? Drop a hammer, drop a feather from 25m, which hits the ground first? Make the car 1kg or 1,000,000kg, no difference.

Regardless, you are correct, the car makes it to the top going about 11.6m/s. Thanks for the equasions, much better than my estimates.;)

Of course you're right, but you're just confusing things for our high school friend here. :)