Question

A stone hanging from a massless string of length $15 \, m$ is projected horizontally with speed $12 \, m \, s^{- 1}$ . The speed of the particle, at the point where the tension in the string is equal to the weight of the particle, is close to

• A. $10\,ms^{- 1}$
• B. $7 \, m \, s^{- 1}$
• C. $12\,ms^{- 1}$
• D. $5\,ms^{- 1}$

Answer 1

Answer: (B)

$u=12\,ms^{- 1}$
$T \, - \, mg \, cos \theta =\frac{m v^{2}}{l}$
As, $T=mg$ we get
$mg\left(1 - cos \theta \right)=\frac{m v^{2}}{l}$
Also we know that
$\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}= \, - \, mgl\left(l - cos \theta \right)$
$\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}=-mv^{2}$
$\Rightarrow v=\frac{u}{\sqrt{3}}=\frac{12}{\sqrt{3}}=6.93\,ms^{- 1}\approx 7\,ms^{- 1}$