A ball of mass 12kg and another of mass 6kg are dropped together from a 60 feet tall building. When these balls are 30 feet. Find the ratio of kinetic energies.
A.) $\sqrt{2}:1$
B.) 1 : 4
C.) 2 : 1
D.) $1:\sqrt{2}$
Answer
624.3k+ views
Hint: Use kinematic equation ${{v}^{2}}-{{u}^{2}}=2as$. Further use kinetic energy formula. Velocity of both the balls are the same, since the velocity of the body is independent of time. You can take the value of acceleration due to gravity as 10 to make calculation easier.
Complete step by step solution:
To solve this question consider two cases before falling from a 60 feet building and after falling from a 60 feet building.
Before falling from 60 feet:
Initial velocity will be zero i.e. u=0 for both the balls.
Whenever there are such situations, always use a kinematic equation.
We know that, kinematic equation is given by,
${{v}^{2}}-{{u}^{2}}=2as$
Where,
u = initial velocity
v = is final velocity
a = acceleration due to gravity
s = displacement
$\begin{align}
& {{v}^{2}}-0=2\times 10\times 30 \\
& {{v}^{2}}=600 \\
& \\
\end{align}$
Final velocity is 600 m/s for the ball.
Let $KE_1$ is the kinetic energy, ${{m}_{1}}$is mass and ${{v}_{1}}$ is the velocity of 12 kg ball and $KE_2$ is kinetic energy,
\[{{m}_{2}}\] is mass and ${{v}_{2}}$ is the velocity of a 6 kg ball.
Therefore ratio of kinetic energy of 12 kg ball and kinetic energy of 6 kg ball is given by,
$\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}}{\dfrac{1}{2}{{m}_{2}}{{v}_{2}}^{2}}$
Since velocity of a 12 kg ball and velocity of a 6kg ball are the same.
Therefore,
$\begin{align}
& \dfrac{KE_1}{KE_2}=\dfrac{{{m}_{1}}}{{{m}_{2}}} \\
& \dfrac{KE_1}{KE_2}=\dfrac{12}{6} \\
& \dfrac{KE_1}{KE_2}=2:1 \\
\end{align}$
$KE_1:KE_2 = 2:1$
Option c is correct.
Note: Whenever there are two cases like before and after falling body, always use the kinematic equation. Initial velocity of both the balls are the same as they were in rest. Final velocity must be the same as velocity is independent of mass since velocity is defined as displacement per unit time. To make calculation easier, take the value of acceleration due to gravity as 10 instead of 9.8. Ratio does not possess any kind of unit.
Complete step by step solution:
To solve this question consider two cases before falling from a 60 feet building and after falling from a 60 feet building.
Before falling from 60 feet:
Initial velocity will be zero i.e. u=0 for both the balls.
Whenever there are such situations, always use a kinematic equation.
We know that, kinematic equation is given by,
${{v}^{2}}-{{u}^{2}}=2as$
Where,
u = initial velocity
v = is final velocity
a = acceleration due to gravity
s = displacement
$\begin{align}
& {{v}^{2}}-0=2\times 10\times 30 \\
& {{v}^{2}}=600 \\
& \\
\end{align}$
Final velocity is 600 m/s for the ball.
Let $KE_1$ is the kinetic energy, ${{m}_{1}}$is mass and ${{v}_{1}}$ is the velocity of 12 kg ball and $KE_2$ is kinetic energy,
\[{{m}_{2}}\] is mass and ${{v}_{2}}$ is the velocity of a 6 kg ball.
Therefore ratio of kinetic energy of 12 kg ball and kinetic energy of 6 kg ball is given by,
$\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}{{m}_{1}}{{v}_{1}}^{2}}{\dfrac{1}{2}{{m}_{2}}{{v}_{2}}^{2}}$
Since velocity of a 12 kg ball and velocity of a 6kg ball are the same.
Therefore,
$\begin{align}
& \dfrac{KE_1}{KE_2}=\dfrac{{{m}_{1}}}{{{m}_{2}}} \\
& \dfrac{KE_1}{KE_2}=\dfrac{12}{6} \\
& \dfrac{KE_1}{KE_2}=2:1 \\
\end{align}$
$KE_1:KE_2 = 2:1$
Option c is correct.
Note: Whenever there are two cases like before and after falling body, always use the kinematic equation. Initial velocity of both the balls are the same as they were in rest. Final velocity must be the same as velocity is independent of mass since velocity is defined as displacement per unit time. To make calculation easier, take the value of acceleration due to gravity as 10 instead of 9.8. Ratio does not possess any kind of unit.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

