Determine the velocity at the bottom of the first hill and the velocity at the top of the second hill

Question

A person is traveling in a rollercoaster. The car is lifted to a point 20 m above the ground before

the cart is released. The car travels to the bottom of the first hill which is 10 m above theground, then goes over the top of the next hill which is 15 m above the ground. Determine the

velocity at the bottom of the first hill and the velocity at the top of the second hill

Answer 1

We can solve this problem by looking at it in terms of total energy_. E_T = PE (Potential) + KE (Kinetic). At the top of the rollercoaster (20m above the ground) there is no KE since KE = (1/2)mv² and the cart is not moving (v = 0). Thus all the energy is PE. When the cart travels to the bottom of the first hill (10m above the ground), the change in PE is mg(Δh), which is equal to mg(10) in this case. Since energy is conserved, and PE decreases by mg(10), KE must increase by mg(10). Since KE was originally 0, it is now mg(10). However, we know that KE = (1/2)mv², so mg(10) = (1/2)mv². Dividing by m and simplifying yields v²=20g, so v=2√(5g) at the bottom of the first hill. We can do the same process for the top of the 2nd hill. The change in PE (from the top of 20m) is mg(5), so KE must increase by mg(5) from 0 to mg(5). Thus, mg(5) = (1/2)mv², so v²=10g, so v=√(10g) at the top of the next hill.