Question

Two objects of masses $2\text{kg}$ and $3\text{kg}$ are dropped from a tower of height of $60\text{m}$. The ratio of their kinetic energy when they reach the ground is

• A. $1:2$
• B. $2:3$
• C. $2:1$
• D. $1:1$

Answer 1

The correct option is B $2:3$

Take the direction $+x$ towards right and $+y$ towards down.

As initial velocity $u=0\text{m/s}$, let the final velocity be $v$ is given by the equation of motion, $v^2=u^2+2as$ where $s$ is the displacement which is the height of the tower $h$

$v^2=u^2+2gh$
$v^2=2gh$
$\Rightarrow v=\sqrt{2gh}$

Final velocity will be same for both masses, hence the ratio of their kinetic energy having the masses, $m_1=2\text{kg}$ and $m_2=3\text{kg}$ is given by,

$\frac{K.E_1}{K.E_2}=\frac{\frac{1}{2}{m}_1{v}^2}{\frac{1}{2}{m}_2{v}^2}=\frac{m_1}{m_2}=\frac{2}{3}$

Hence option B is the correct answer