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. The angle q in the figure is-

• A. cos^–1(4/9)
• B. cos^–1(3/5)
• C. cos^–1(1/2)
• D. None of the above

Answer 1

Answer: (B)

cos^–1(3/5)

Answer 2

mg cos q – N= $\frac{mv^{2}}{R}$; N = mg cosq – $\frac{mv^{2}}{R}$

N = 0 ̃ mg cos q = $\frac{mv^{2}}{R}$

v = $\sqrt{gR\cos\theta}$ or cos q = $\frac{v^{2}}{gR}$ .….(i)

From energy conservation

$\frac{1}{2}$m$\left\lbrack v^{2} - \frac{gR}{4} \right\rbrack$ = mgR(1 – cos q)

̃ $\frac{v^{2}}{2}$– $\frac{gR}{8}$ = gR – gR cos q …(ii)

From (i) and (ii) q = cos^–1$\left( \frac{3}{5} \right)$