Explanation

(d) KE at the lowest position = $\frac{1}{2}\ mv^{2}$
$= \frac{1}{2}\ m(3)^{2} = \frac{9}{2}m$
When the length makes an angle $\theta( = 60{^\circ})$ to the vertical, the bob of the pendulum will have both KE and PE. If $v$ is the velocity of bob at this position and $h$ is the height of the bob w.r.t. $B$, then total energy of the bob
$= \frac{1}{2}mv^{2} + mgh$
[image:14_image81.emf]
But $h = l - l\cos\theta$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = l(1 - \cos{\theta)}$
$= 0.5(1 - \cos{60{^\circ}) = 0.5\left( 1 - \frac{1}{2} \right) = \frac{1}{4}}$
$E = \frac{1}{2}mv^{2} + m \times 10 \times \frac{1}{4}$
$= \frac{1}{2}mv^{2} + \frac{5}{2}m$
According to law of conservation of energy
$\frac{1}{2}mv^{2} + \frac{5m}{2} = \frac{9}{2}m$
$\Rightarrow$ $\frac{1}{2}mv^{2} = \frac{9}{2}m - \frac{5}{2}m = 2m$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = 2\ ms^{- 1}$