Final answer:
To show that the fractional loss of original kinetic energy in a head-on inelastic collision between two objects of equal masses is given by the expression (1/2) + (V1V2)/(V1² + V2²), one must consider the conservation of momentum and the kinetic energies before and after the collision.
Step-by-step explanation:
The question pertains to the concept of kinetic energy loss during an inelastic collision in physics. In an inelastic collision between two objects of equal mass moving with opposite velocities, when they collide and join together, their combined mass is equal to the sum of their individual masses, but their combined velocity is the average of their individual velocities due to the conservation of momentum.
To compute the fractional loss of original kinetic energy (KE), we use the given expression E = (1/2) + (V1V2)/(V1² + V2²). This represents the energy after the collision divided by the energy before the collision subtracted from one, giving us the fractional loss.
The kinetic energy before the collision is the sum of the kinetic energies of each object, which is 1/2 mV1² + 1/2 mV2² (assuming m is the mass and V1 and V2 are the velocities of the two objects). After the collision, since the objects stick together, they move with a common velocity, V, where V is given by the conservation of momentum as (mV1 + mV2) / (m + m).
The kinetic energy after the collision is 1/2 (2m)V², where V is obtained from the conservation of momentum. By substituting V and simplifying the expression, we find an equation for calculating the fractional kinetic energy loss in terms of the initial velocities of the colliding objects, which ultimately leads to the provided expression.