D

Step 1: In a head-on elastic collision between two bodies of equal mass, we can analyze the situation using the principles of conservation of momentum and kinetic energy.

Step 2: Let the masses of the two bodies be m, and let their initial velocities be u₁ and u₂. After the collision, their final velocities will be v₁ and v₂.

Step 3: From the conservation of momentum, we have:
$$ m u_1 + m u_2 = m v_1 + m v_2 $$
Simplifying this gives:
$$ u_1 + u_2 = v_1 + v_2 $$

Step 4: For kinetic energy conservation in an elastic collision, we have:
$$ \frac{1}{2} m u_1^2 + \frac{1}{2} m u_2^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 $$
This simplifies to:
$$ u_1^2 + u_2^2 = v_1^2 + v_2^2 $$

Step 5: In a perfectly elastic collision between two bodies of equal mass, we find that the velocities can interchange if they collide head-on. This means:
- The final velocity of the first body can become the initial velocity of the second body and vice versa.
- Thus, the options regarding velocities and speeds interchanged are valid.

Step 6: However, in terms of momentum, both bodies have equal mass and the total momentum is conserved. Therefore, their momenta can also be considered interchanged in magnitude but not in behavior (the direction might change depending on their speeds).

Step 7: The last option 'the faster body speeds up and the slower body slows down' does not hold true because in elastic collisions, if one body is faster than the other, there is no mechanism for the velocities to change their magnitudes such that a 'speeding up' or 'slowing down' occurs—the speeds are exchanged.

Conclusion: Hence, the correct answer is that it is not possible for the faster body to speed up and the slower body to slow down in a head-on elastic collision.
Therefore, the correct answer is D.