Question

1. many roller coasters have loops, where people are briefly upside down. for safety reasons, the cart is required to have a minimum speed of 10.0 m/s at the top of the loop. what is the minimum height of the first hill if the loop is 16 m tall? assume that this is a closed system. (4 marks-a)

Explanation:

Step1: Apply conservation of mechanical energy

In a closed - system, the total mechanical energy \(E = K+U\) (kinetic energy \(K=\frac{1}{2}mv^{2}\) and potential energy \(U = mgh\)) is conserved. Let the mass of the cart be \(m\), the height of the first hill be \(h\), and the height of the top of the loop be \(H = 16m\) with speed \(v=10.0m/s\) at the top of the loop. At the top of the first hill, the initial kinetic energy \(K_{i}=0\) (minimum speed at the top of the first hill is assumed to be \(0\) for minimum - height calculation) and the initial potential energy \(U_{i}=mgh\). At the top of the loop, \(K_{f}=\frac{1}{2}mv^{2}\) and \(U_{f}=mgH\). By conservation of energy \(E_{i}=E_{f}\), so \(mgh=\frac{1}{2}mv^{2}+mgH\).

Step2: Solve for \(h\)

Divide the energy - conservation equation \(mgh=\frac{1}{2}mv^{2}+mgH\) by \(m\) (since \(m
eq0\)). We get \(gh=\frac{1}{2}v^{2}+gH\). Then \(h=\frac{v^{2}}{2g}+H\). Substitute \(v = 10.0m/s\) and \(g = 9.8m/s^{2}\), \(H=16m\).

$$
LATEXBLOCK0
$$

Answer:

\(h\approx21.1m\)