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Defining Kinetic Energy

A particle of...

Question

A particle of mass $m_{1}$ is moving with a velocity $v_{1}$ and another particle of mass $m_{2}$ is moving with a velocity $v_{2}$ . Both of them have same momentum but their different kinetic energies are $E_{1}$ and $E_{2}$ respectively. If $m_{1} > m_{2}$ then

A

E_{1} < E_{2}

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B

\frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}}

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C

E_{1} > E_{2}

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D

E_{1} = E_{2}

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Solution

The correct option is A $E_{1} < E_{2}$
First body :
mass= $m_{1}$ , velocity= $v_{1}$

Second body :
mass= $m_{2}$ , velocity= $v_{2}$

Using the conservation of momentum, $m_{1} v_{1} = m_{2} v_{2}$ ...(1)
multiplying $m_{1} v_{1}$ and $m_{2} v_{2}$ at LHS and RHS on eq. (1) respectively.

so, $m_{1}^{2} v_{1}^{2} = m_{2}^{2} v_{2}^{2}$ ...(2)
multiplying $\frac{1}{2}$ on both sides in eq.(2)
$\frac{1}{2} m_{1}^{2} v_{1}^{2} = \frac{1}{2} m_{2}^{2} v_{2}^{2}$
$(\frac{1}{2} m_{1} v_{1}^{2}) m_{1} = (\frac{1}{2} m_{2} v_{2}^{2}) m_{2}$

where, $\frac{1}{2} m v^{2} =$ kinetic energy ( $E$ )
then, $E_{1} m_{1} = E_{2} m_{2}$
$m_{1} > m_{2}$ (Given)
Hence, $E_{1} < E_{2}$

option A is correct.

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