The density of the electron states in a metal can be written as g(E) = AE^(1/2), where is a constant and is measured from the bottom of the conduction band.
Derive the total number of states in this material and calculate the fraction of conduction electrons ƒ within KT of the Fermi energy for molybdenum at 300 K. Hints: Use the approximation (1- KT/Eꜰ)^(3/2) ≈ 1-(3/2)KT Eꜰ^(1/2).
Calculate the Fermi energy for this metal at 0 K using the expression for the density of electron states. Assume the electronic structure contains 1 electron per atom.

The speed was not exceeded so it was slow

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s