Given are three different stages of an ancient and deadly two-body system: the bow and the arrow. If m represents the mass of the 'bow-arrow system', how can you compare the different cases?

m_{1} = m_{2} = m_{3}

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m_{1}

m_{2} = m_{3}

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m_{1}

m_{2} = m_{3}

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m_{1}

m_{2}

m_{3}

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Solution

The correct option is B $m_{1}$ < $m_{2} = m_{3}$
We know that in all three cases, the rest masses of the objects: the bow and the arrow won't change. Rest masses are constant quantities. Let these masses be $m_{B} a n d m_{A}$ , its clear which is which. So will all three systems weigh the same? Not quite.
System 2 has more energy than System 1, simply because in System 2 there is potential energy in the thread of the bow, since it's nocked. System 3 will have the same net energy as System 2 because the potential energy of the nocked bow just gets converted into the kinetic energy of the arrow - there is no loss of energy in the system. Writing the energy content of each system -
$E_{1} = m_{B} c^{2} + m_{A} c^{2} E_{2} = m_{B} c^{2} + m_{A} c^{2} + p o t e n t i a l e n e r g y = E_{1} + p o t e n t i a l e n e r g y E_{3} = m_{B} c^{2} + m_{A} c^{2} + k i n e t i c e n e r g y = E_{1} + k i n e t i c e n e r g y$
and since energy is conserved in Systems 2 and 3, potential energy of bow = kinetic energy of arrow, we can say -
$E_{1}$ < $E_{2} = E_{3}$ $\Rightarrow \frac{E_{1}}{c^{2}} < \frac{E_{2}}{c^{2}} = \frac{E_{3}}{c^{2}}$
$\Rightarrow m_{1}$ < $m_{2} = m_{3}$
P.S.: Please keep in mind that this difference is non-zero, but very, very tiny. For all practical purposes in the daily world, we ignore this difference. But when we go to atomic and nuclear length scales, it becomes significant.

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