  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?  >   < <  < Solution

## The correct option is D < 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 mB and mA, 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 - E1=mBc2+mAc2 E2=mBc2+mAc2+potential energy=E1+potential energy E3=mBc2+mAc2+kinetic energy=E1+kinetic energy, and since energy is conserved in Systems 2 and 3, potential energy of bow=kinetic energy of arrow, we can say - E1 < E2=E3 ⇒E1c2 < E2c2=E3c2 m1 < m2=m3. 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.  Suggest corrections   