Question

A stone hanging from a massless string of length $15m$ is projected horizontally with speed $12m{s}^{-1}$ . 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. $10m{s}^{-1}$
• B. $7m{s}^{-1}$
• C. $12m{s}^{-1}$
• D. $5m{s}^{-1}$

Answer 1

Answer: (B)

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