MHT CET · Physics · Laws of Motion
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
- 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 \(14 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(14 \mathrm{~m} / \mathrm{s}\) and \(15 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
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,
\(v=\sqrt{\mathrm{Rg}}\)
Putting the values, \(R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^{2}\)
So, \(\quad v=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^{2}\)
Thus, the speed of the car at the top of the hill is
\(\text { between } 14 \mathrm{~m} / \mathrm{s} \text { and } 15 \mathrm{~m} / \mathrm{s} \text { . }\)
Note : The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

\(M g-N=M \frac{v^{2}}{R}\)
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,
\(v=\sqrt{\mathrm{Rg}}\)
Putting the values, \(R=20 \mathrm{~m}, g=10.0 \mathrm{~m} / \mathrm{s}^{2}\)
So, \(\quad v=\sqrt{20 \times 10.0}=14.14 \mathrm{~m} / \mathrm{s}^{2}\)
Thus, the speed of the car at the top of the hill is
\(\text { between } 14 \mathrm{~m} / \mathrm{s} \text { and } 15 \mathrm{~m} / \mathrm{s} \text { . }\)
Note : The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.
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