Physics - W.E.P. Question with Solution | TestHub

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, $R=20\mathrm{m},g=10.0\mathrm{m}/{\mathrm{s}}^2$ 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.