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 these

Answer 1

Answer: (B)

7 m/s

Answer 2

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