A roller coaster is designed such that riders experience "weightlessness" as they go round the top of a hill whose radius of curvature is $20 \mathrm{~m}$. The speed of the car at the top of the hill is between

• Correct answer
  A

  $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$
• B

  $15 \mathrm{~m} / \mathrm{s}$ and $16 \mathrm{~m} / \mathrm{s}$
• C

  $16 \mathrm{~m} / \mathrm{s}$ and $17 \mathrm{~m} / \mathrm{s}$
• D

  $13 \mathrm{~m} / \mathrm{s}$ and $14 \mathrm{~m} / \mathrm{s}$

Solution

Balancing the forces, we get $M g-N=M \frac{v^2}{R}$ For weightlessness, $N=0$ $\therefore \frac{M v^2}{R}=M g$ where $R$ is the radius of curvature and $v$ is the speed of car. Therefore, $v=\sqrt{R g}$ Putting the values, $R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^2$ So, $y=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^2$ Thus, the speed of the car at the top of the hill is between $14 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$. Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

Hence, the correct answer is option: 'A'.