A roller coaster car of mass 500 kg starts at rest at the top of a hill 30 m high. It then goes down the hill and up a second hill that is 20 m high. What is the speed of the car at the top of the second hill, assuming negligible friction?

Core Concept

Using conservation of mechanical energy: initial energy = final energy. At the first hill: E₁ = mgh₁. At the second hill: E₂ = mgh₂ + ½mv². Setting them equal: mgh₁ = mgh₂ + ½mv². Solving for v: v = √[2g(h₁ - h₂)] = √[2 × 9.8 × (30 - 20)] = √196 ≈ 14 m/s. Option A would be correct if the height difference was only 5 m. Option C would be correct if the height difference was 20 m. Option D would be correct if the height difference was 40 m.