Question

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?

• A.
• B. <
• C. >
• D. < <

Answer 1

The correct option is B <

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 $m_B$ and $m_A,$. 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 -
$E_1=m_B{C}^2+m_A{C}^2$
$E_2=m_B{C}^2+m_A{C}^2+\text{potential energy}=E_1+\text{potential energy}$
$E_3=m_B{C}^2+m_A{C}^2+\text{kinetic energy}=E_1+\text{kinetic energy},$
and since energy is conserved in Systems 2 and 3, the potential energy of bow = kinetic energy of arrow, we can say -
$E_1$ < $E_2=E_3$
$\Rightarrow \frac{E_1}{C^2}$ < $\frac{E_2}{C^2}=\frac{E_3}{C^2}$
$m_1$ < $m_2=m_3.$
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.