### Work Power Energy Question 6

#### Question 6 - 2024 (29 Jan Shift 2)

A bob of mass ’ $m$ ’ is suspended by a light string of length ’ $L$ ‘. It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position $B$. The ratio of kinetic energies $\frac{(\text { K.E. }) _A}{(\text { K.E. }) _B}$ is :

(1) $3: 2$

(2) $5: 1$

(3) $2: 5$

(4) $1: 5$

## Show Answer

#### Answer: (2)

#### Solution:

Apply energy conservation between A \b & B

$\frac{1}{2} mV _L^{2}=\frac{1}{2} mV _H^{2}+mg(2 L)$

$\because V _L=\sqrt{5 gL}$

So, $V _H=\sqrt{gL}$

$\frac{(K . E) _A}{(K . E) _B}=\frac{\frac{1}{2} m(\sqrt{5 gL})^{2}}{\frac{1}{2} m(\sqrt{gL})^{2}}=\frac{5}{1}$