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5.Work, Energy, Power and Collision
hard
A small block slides down from rest at point $A$ on the surface of a smooth cylinder, as shown. At point $B$, the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by
A$sin\ \theta_2 = \frac{2}{3}sin\ \theta_1$
B$sin\ \theta_2 = \frac{3}{2}sin\ \theta_1$
C$cos\ \theta_2 = \frac{2}{3}cos\ \theta_1$
D$cos\ \theta_2 = \frac{3}{2}cos\ \theta_1$
Solution
$\mathrm{h}=\mathrm{R}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$\frac{1}{2} m v^{2}=m g h$
$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg} \cos \theta_{2} \Rightarrow \mathrm{v}=\sqrt{\mathrm{Rg} \cos \theta_{2}}$
$\frac{1}{2} \mathrm{m} \mathrm{Rg} \cos \theta_{2}=\mathrm{mgR}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$3 \cos \theta_{2}=2 \cos \theta_{1}$
$\frac{1}{2} m v^{2}=m g h$
$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg} \cos \theta_{2} \Rightarrow \mathrm{v}=\sqrt{\mathrm{Rg} \cos \theta_{2}}$
$\frac{1}{2} \mathrm{m} \mathrm{Rg} \cos \theta_{2}=\mathrm{mgR}\left(\cos \theta_{1}-\cos \theta_{2}\right)$
$3 \cos \theta_{2}=2 \cos \theta_{1}$
Standard 11
Physics
