Question

A ball of mass $12kg$ and another of mass $6kg$ are dropped from $60feet$ tall building. What will be the ratio of their kinetic energies after a fall of 30 feet each, towards earth?
A. $\sqrt 2 :1$
B. $1:4$
C. $2:1$
D. $1:\sqrt 2$

Answer 1

Hint As the mass of both the balls are given and their velocities for both the balls will be the same as they fall from the same 60 feet tall building. Then, evaluate kinetic energies of both the balls respectively. Now, find the ratio of their kinetic energies of both the balls.

Complete step-by-step solution :
Let the mass of the first ball be ${m_1}$ and the second ball be ${m_2}$.
Therefore, according to the question, it is given that –
${m_1} = 12kg$ and
${m_2} = 6kg$
The velocity of both the balls will be the same. The two balls will attain the same velocity.
Let the velocity of both the balls be $V$ $m/\sec$.
To find the kinetic energy we have to use the expression –
$KE = \dfrac{1}{2}m{V^2}$
where, $m$ is the mass of the object and $V$ is the velocity of the object.
Now, calculating the kinetic energy for first ball –
$K{E_1} = \dfrac{1}{2}{m_1}{V^2}$
Putting the value of mass of first ball in the above equation –
$K{E_1} = \dfrac{1}{2} \times 12{V^2} \\\ K{E_1} = 6{V^2} \\\$
Next, evaluating the kinetic energy for second ball –
$K{E_2} = \dfrac{1}{2}{m_2}{V^2}$
Putting the value of mass of second ball in the equation of kinetic energy of second ball –
$K{E_2} = \dfrac{1}{2} \times 6{V^2} \\\ K{E_2} = 3{V^2} \\\$
Finding the ratio of kinetic energy of both the balls –
$\dfrac{{K{E_1}}}{{K{E_2}}} = \dfrac{{6{V^2}}}{{3{V^2}}}$
By further solving, we get –
$\dfrac{{K{E_1}}}{{K{E_2}}} = \dfrac{2}{1}$
$K{E_1}:K{E_2} = 2:1$
Therefore, the option (C) is the correct option.

Note:- Kinetic energy is the energy which is attained by an object due to the motion of an object. Kinetic energy can be transferred between objects and transformed into the other kinds of energy. The S.I unit of kinetic energy is Joule which is equal to $1kg.{m^2}.{s^{ - 2}}$.