Question

Example Problem

The potential energy of an object is given by U(x) = 8x2 - x4, where U is in joules and x is in
(a) Determine the force acting on this object.
(b) At what positions is this object in equilibrium?
(c) Which of these equilibrium positions are stable and which are unstable?
metres.
111 Unit 2 Concepts and Definitions Prof Mark Lester

Exam Part B Example
A neutron of mass m moving with velocity v collides head-on and elastically with a stationary nucleus of mass M.
(a) Show that the velocity of the nucleus after the collision, U, is given by
U= 2m v (m+M)
(b) Hence show that the neutron loses a fraction f of its energy where
f= 4mM (m+M)
10marks 5 marks
(c) A fast neutron enters a target of carbon nuclei which may be assumed to have masses 12 times that of the neutron. How many head-on collisions will it take
before the neutron loses 95% of its energy?
4 marks
(d) Suggest one reason why in a real reactor a neutron is likely to make more
collisions with the moderator nuclei before losing this much energy
2
1 mark

Answer 1

Part A

a) F = -16x + 4, b) x = 0.25 m, c) STABLE

Step-by-step explanation:

Part A

a) Potential energy and force are related

F =
`- (dU)/(dx)`- dU / dx

F = - (8 2x -4)

F = -16x + 4

b) The object is in equilibrium when the forces are zero

0 = -16x + 4

x = 4/16

x = 0.25 m

c) An equilibrium position is called stable if with a small change in position, the forces make it return to the initial position, in case the forces make it move away it is called unstable.

In this case there is only one equilibrium point

by changing the position a bit

x ’= x + Δx

we substitute

F ’= - 16 x’ + 4

F ’= - 16 (x + Δx) + 4

F ’= (-16x +4) - 16 Δx

at equilibrium position F = 0

F ’= 0 - 16 Δx

we can see that the body returns to the equilibrium position, therefore it is STABLE

PART B

This is an exercise in body collisions, let's define the system formed by the two bodies in such a way that the forces during the collisions are internal and the moment is conserved

initial instant. Before the shock

p₀ = m v

final instant. After the crash

p_f = (m + M) v_f

We have two possibilities: an elastic collision in which the bodies separate, each one maintaining its plus, and an INELASTIC collision where the neutron is absorbed by the nucleus and the final mass is M '= m + M, in this case they indicate that the collision is elastic

p₀ = pf

mv = mv ’+ M v_f

in the case of the elastic collision, the kinetic energy is conserved

K₀ = K_f

½ m v² = ½ m v’² + ½ M v_f²

we write the system of equations

mv = mv ’+ M v_f (1)

m (v² -v'²) = M v_f ²

m (v - v ’) = M v_f

m (v-v ’) (v + v’) = M v_f

v + v ’= v_f

we substitute in equation 1 and solve

v ’=
`(m -M )/(m+M ) \ vo`

v_f =
`(2m)/(m+M) \ v_o`

the mechanical energy of the neutron is

initial

Em₀ = K = ½ m v²

final moment

Em_f = K + U = ½ m v_f ² + U

U is the energy lost in the collision

total energy is conserved

Em₀ = Em_f

½ m v² = ½ m v_f ² + U

U = ½ m (v² -v_f ²)

U = ½ m [v² - (
`(m-M)/(m+M)` v)² ]

U = ½ m v² [1- (
`(m-M)/(m+M)` )² ]

U = ½ m v2 [
`(2M)/(m+M)`]

U =
`(2 mM)/(m +M ) \ v^2`

Let's do the same calculations for the nucleus

initial Em₀ = 0

final Em_f = K + U = ½ M v_f ² + U

Em₀ = Em_f

0 = K + U

U = -K

U = - ½ M v_f ²

U = - ½ M [
`(2m)/(m+M) \ v` ]²

U =
`(2 m M )/(m+M) \ v^2`

We can see that we obtain the same result, that is, the potential energy lost by the neutron is equal to the potential energy gained by the nucleus.

b) the fraction of energy lost

f = U / Em₀

f = 4 m M / m + M

c) let's calculate the fraction of energy lost in a collision

m = 1.67 10⁻²⁷ kg

M = 12 1.67 10⁻²⁷= 20 10⁻²⁷ kg

f = 4 1.6 20 / (1.6+ 20) 10⁻²⁷

f = 5.92 10⁻²⁷ J

the energy of a fast neutron is greater than 1 eV

Eo = 1 eV (1.67 10⁻¹⁹ J / 1eV) = 1.67 10⁻¹⁹ J

Let's use a direct portion rule if in a collision f loses in how many collisions it loses 0.95Eo

#_collisions = 0.95 Eo / f

#_collisions = 0.95 1.67 10⁻¹⁹ / 5.92 10⁻²⁷

#_collisions = 2.7 10⁷ collisions