Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u. What is its kinetic energy, expressed as a fraction of its original kinetic energy? b) what is its kinetic energy expressed as a fraction of its original kinetic energy? c) how many such successive collisions will reduce the speed of a neuron to 1/59000 of its original value?

Question

Conor

Physics

9 May, 20:11

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u. What is its kinetic energy, expressed as a fraction of its original kinetic energy? b) what is its kinetic energy expressed as a fraction of its original kinetic energy? c) how many such successive collisions will reduce the speed of a neuron to 1/59000 of its original value?

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Dork · Answer 1

M*U + 0 = m*v'1 + 2m*v'2

the zero means deuteron has no velocity

where v'1 and v'2 are the post-collision velocities.

The equatio becomes

U = v'1 + 2v'2

U = v'2 - v'1

v'2 = U + v'1

U = v'1 + 2 (U + v'1) = 2U + 3v'1

U = - 3V

V = - U / 3

The speed ratio is 1/3

B) Since KE is proportional to the square of the speed, if the speed is 1/3, then KE is 1/9

C) (1/3) ⁿ = 1/729

3ⁿ = 729

n = 6