Question

A heavy stone hanging from a massless string of length $15m$ is projected horizontally with speed $147m/s$. 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

Answer: (B)

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