Question

A small block slides with velocity $0.5\sqrt{gr}$ on the horizontal frictionless surface as shown in the figure. The block leaves the surface at point C. Calculate angle $\theta$ in the figure.

• A. $\theta ={\cos }^{-1}\left(\frac{1}{4}\right)$
• B. $\theta ={\cos }^{-1}\left(\frac{1}{3}\right)$
• C. $\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$
• D. $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$

Answer 1

The correct option is C $\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$

From the law of conservation of energy, the loss in gravitational potential energy is equal to the gain in kinetic energy of the block.

$⟹mgr(1-cos\theta )=\frac{1}{2}m(v^2-v_0^2)$

$⟹v^2=2.25gr-2grcos\theta$

Thus, the centripetal force acting on the block outwards $=\frac{m{v}^2}{r}$

This is balanced by the gravitational force acting on block normal to the surface $=mgcos\theta$ when the block is about to leave the surface.

$⟹\frac{m}{r}(\mathrm{2.2.5}gr-2grcos\theta )=mgcos\theta$

$⟹cos\theta =\frac{2.25}{3}$

$⟹\theta =co{s}^{-1}\left(\frac{3}{4}\right)$