
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
604.5k+ 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.
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