A) 14 m/s and 15 m/s
B) 15 m/s and 16 m/s
C) 16 m/s and 17 m/s
D) 13 m/s and 14 m/s
Correct Answer: A
Solution :
Balancing the forces, we get \[Mg-N=M\frac{{{v}^{2}}}{R}\] For weightlessness, N = 0 \[\therefore \] \[\frac{M{{v}^{2}}}{R}=Mg\] Where R is the radius of curvature and v is the speed of car. Therefore, \[v=\sqrt{Rg}\] Putting the values, \[R=20\text{ }m,\,g=10.0\text{ }m/{{s}^{2}}\] So, \[v=\sqrt{20\times 10.0}=14.14\,m/{{s}^{2}}\] Thus, the speed of the car at the top of the hill is between 14 m/s and 15 m/s.You need to login to perform this action.
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