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Energy in Collision

A ball is thr...

Question

A ball is thrown up from the earth by a person and then caught by him on its return. When the ball falls towards the earth, the ratio of kinetic energies gained by the ball $(K_{B})$ and the earth $(K_{E})$ is

\frac{K_{E}}{K_{B}} = zero

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\frac{K_{E}}{K_{B}} = \frac{m_{E}}{m_{B}}

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\frac{K_{E}}{K_{B}} = \frac{m_{B}}{m_{E}}

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\frac{K_{E}}{K_{B}} = Infinite

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Solution

The correct option is C $\frac{K_{E}}{K_{B}} = \frac{m_{B}}{m_{E}}$
Here the man is throwing the ball upward is a part of earth since he does not lose contact from the earth. The gravitational forces between the parts of the system (man + Earth) are internal forces.
The momentum of the system is therefore conserved.
Since the momentum of the system is zero initially, therefore
$m_{B} {\to v}_{B} + m_{E} {\to v}_{E} = 0$
$\Rightarrow m_{B} {\to v}_{B} = - m_{E} {\to v}_{E}$
Now,
$\frac{K_{E}}{K_{B}} = \frac{\frac{1}{2} m_{E} v_{E}^{2}}{\frac{1}{2} m_{B} v_{B}^{2}} = \frac{(m_{E} v_{E})^{2}}{(m_{B} v_{B})^{2}} (\frac{m_{B}}{m_{E}}) = \frac{m_{B}}{m_{E}}$
Hence, the correct answer is option (c).

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A ball is thrown up from the earth by a person and then caught by him on its return. When the ball falls towards the earth, the ratio of kinetic energies gained by the ball (KB) and the earth (KE) is

A ball is thrown up from the earth by a person and then caught by him on its return. When the ball falls towards the earth, the ratio of kinetic energies gained by the ball $(K_{B})$ and the earth $(K_{E})$ is