Question

A heavy stone hanging from a massless string of length $15\,m$ is projected horizontally with speed $147 \,m/s$. The speed of the particle at the point where the tension in the string equals the weight of the particle is?

• A. 10 m/s
• B. 7 m/s
• C. 12 m/s
• D. None of the above

Answer 1

The Correct Option is B

$m g-m g \cos \theta=\frac{m v^{2}}{l}$ or $\frac{v^{2}}{l}=g(1-\cos \theta)$ $v^{2}=g l(1-\cos \theta) \ldots$(1) Applying conservation of energy $\frac{1}{2} m g l =\frac{1}{2} m v^{2}+m g l(1-\cos \theta) $ $v^{2} =g l-2 g l(1-\cos \theta) \ldots$(2) Solving Eqs. (1) and (2), we get $\theta=\cos ^{-1} \frac{2}{3}$ From E (1) $v^{2}=10 \times 15\left(1-\frac{2}{3}\right)=150\left(\frac{1}{3}\right)=50 $ $\therefore v=\sqrt{50}=7\, m / s$